\documentclass[12pt]{amsart} %\documentclass[]{amsart} \usepackage{amsmath} \usepackage{amssymb} \theoremstyle{plain} \newtheorem{theor}{Theorem}[section] \newtheorem{Theor}{Theorem}[section]{} \newtheorem{lemma}{Lemma}[section] \newtheorem{propo}{Proposition}[section] \newtheorem{corol}{Corollary}[section] \newtheorem*{claim}{Claim}{} \theoremstyle{definition} \newtheorem{defin}{Definition}[section] \newtheorem{conje}{Conjecture}[section] \newtheorem{notat}{Notation}[section] \newtheorem{quest}{Question}[section] \newtheorem{examp}{Example}[section] \newtheorem{condi}{Condition}[section] \newtheorem*{case1}{Case One} \newtheorem*{case2}{Case Two} \newtheorem*{case3}{Case Three} \newtheorem*{case4}{Case Four} \newtheorem*{case5}{Case Five} \theoremstyle{remark} \newtheorem{rem}{Remark} \newtheorem{note}{Note} \numberwithin{equation}{section} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%$$%%%%% THIS PREAMBLE CONSIST OF MACROS I SOMETIMES USE %%%%%%%%%%% %%%%%%%%% NOT ALL OF THEM WILL BE USED IN THIS PAPER %%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \DeclareMathOperator{\cov}{cov} \DeclareMathOperator{\non}{non} \DeclareMathOperator{\add}{add} \DeclareMathOperator{\unif}{unif} \DeclareMathOperator{\dom}{domain} \DeclareMathOperator{\cof}{cof} \DeclareMathOperator{\ran}{range} \DeclareMathOperator{\hght}{HT} \DeclareMathOperator{\sakne}{RT} \DeclareMathOperator{\suc}{SC} \newcommand{\IF}{\text{ if }} \newcommand{\OR}{\text{ or }} \newcommand{\AND}{\text{ and }} \newcommand{\CH}{$2^{\aleph_0} = \aleph_1$} \newcommand{\forces}[2]{\Vdash_{#1} \mbox{``} #2 \mbox{''}} \newcommand{\notforces}[2]{\not\Vdash_{#1} \mbox{``} #2 \mbox{''}} \newcommand{\Model}{{\mathfrak M}} \newcommand{\Sacks}{{\mathbb S}} \newcommand{\Reals}{{\mathbb R}} \newcommand{\Rationals}{{\mathbb Q}} \newcommand{\Poset}{{\mathbb P}} \newcommand{\Ameoba}{{\mathbb A}} \newcommand{\Integers}{{\mathbb Z}} \newcommand{\Naturals}{{\mathbb N}} \newcommand{\BA}{{\mathbb N}} \newcommand{\presup}[2]{\, ^{#1} \! #2} \newcommand{\betan}{\beta{\mathbb N} \setminus {\mathbb N}} \newcommand{\fomom}{\presup{\omega}{\omega}} \newcommand{\wfomom}{\presup{\stackrel{\omega}{\smile}}{\omega}} \newcommand{\pomega}{{\cal P}(\omega)} \newcommand{\pomegaf}{{\cal P}(\omega)/[\omega]^{<\aleph_0}} \newcommand{\card}[1]{\lvert #1 \rvert} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%% END OF PREAMBLE %%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \DeclareMathOperator{\ess}{Essential} \DeclareMathOperator{\diam}{Diam} \DeclareMathOperator{\smll}{SM} \DeclareMathOperator{\nonmax}{Bot} \DeclareMathOperator{\tsuc}{TS} \DeclareMathOperator{\Top}{Top} \newcommand{\norm}[1]{{\lVert #1 \rVert}_{\text{unif}}} \newcommand{\SingleSeq}[1]{\vec{(#1)}} \newcommand{\Angbr}[1]{\langle #1 \rangle} \newcommand{\angbr}[1]{\langle\langle #1 \rangle\rangle} \title[Unions of Rectifiable Curves]{Unions of Rectifiable Curves in Euclidean Space and the Covering Number of the Meagre Ideal} \author[J. Stepr\={a}ns]{Juris Stepr\={a}ns} \address{Department of Mathematics, York University\\ 4700 Keele Street, North York, Ontario\\ Canada \ \ \ M3J 1P3} %\curraddr{Department of Mathematics, University of Latvia\\ %Riga, Latvia} \email{steprans@mathstat.yorku.ca} \dedicatory{} \date{} \thanks{Research for this paper was partially supported by NSERC of Canada} \keywords{meagre set, rectifiable curve, proper forcing} \subjclass{Primary 03E35; Secondary 26A45} \begin{document} \begin{abstract} To any metric space it is possible to associate the cardinal invariant corresponding to the least number of rectifiable curves in the space whose union is not meagre. It is shown that this invariant can vary with the metric space considered, even when restricted to the class of convex subspaces of separable Banach spaces. As a corollary it is obtained that it is consistent with set theory that that any set of reals of size $\aleph_1$ is meagre yet there are $\aleph_1$ rectifiable curves in $\Reals^3$ whose union is not meagre. The consistency of this statement when the phrase ``rectifiable curves'' is replaced by ``straight lines'' remains open. \end{abstract} \maketitle \tableofcontents \bibliographystyle{plain} \section{Introduction} Cardinal invariants of the continuum have been studied implicitly for almost a century and explicitly, with the help of forcing techniques, for the past thirty years. One of the occasionally overlooked features of the well known cardinal invariants --- such as the covering number, additivity, cofinality and uniformity of measure and category --- is their invariance with respect to the Polish space used to define them. This is not true for other cardinal invariants associated with classical real analysis. A striking result in this direction is due to Cichon and Morayne \cite{MR87m:26002,MR86i:26008,MR88h:26002} who showed that the statement that $\aleph_n$ is a bound on the cardinality of the continuum is equivalent to the assertion that there is a function $f:\Reals^n \to \Reals^{n+m}$ which is onto and such that at each point at least one of the coordinate functions is differentiable. For example, the cardinal invariants associated with covering Euclidean space with smooth surfaces depend on the dimension of the Euclidean space being considered \cite{step.38}. Cardinalities of maximal almost disjoint families of paths lattices of integers also depend on the dimension of the lattice --- although in this case results are known \cite{step.19} only in dimensions 1 and 2. The same can be said of the problem of finding homeomorphisms of Euclidean space which take a given $\aleph_1$-dense set to some other $\aleph_1$-dense set \cite{step.09}. Less is known about the question of the covering number of the $n$-dimensional Mycielski ideal \cite{step.25} but there is reason to think that this invariant may also be related to the dimension. The present paper will consider another class of cardinal invariants which occur naturally in analysis and whose values are not independent of the metric space on which they are defined, even if the metric spaces are taken to be separable Banach spaces. Let ${\mathfrak Z}$ be a metric space and define ${\mathcal R}({\mathfrak Z})$ to be the $\sigma$-ideal generated by all rectifiable curves in ${\mathfrak Z}$. It is an easy exercise to show that there is a $\sigma$-centred forcing which, for any $n$, covers the ground model $\Reals^n$ with countably many rectifiable curves. Hence it is easy to find models where $\cov({\mathcal R}({\Reals^n})) = \aleph_1$ and the continuum is as large as desired. Not so easy is Theorem~\ref{t:shel}, due to S. Shelah, which establishes that it is consistent that $\non({\mathcal Null}) > \aleph_1$ yet $\non({\mathcal R}(\Reals^2)) = \aleph_1$. However, the main results of this paper will be concerned with inequalities relating cardinal invariants of the ideal of meagre sets with invariants of various ideals ${\mathcal R}({\mathfrak Z})$. The motivation for this is to be found in a question due to P. Komjath which can be found in \cite{millerq}: If $\non({\mathcal Null})> \aleph_1$ does it follow that the union of any family of lines in the plane of cardinality $\aleph_1$ is a measure zero set? The analogous question for meagre sets can also be asked: If $\non({\mathcal Meagre})> \aleph_1$ does it follow that the union of any family of lines in the plane of cardinality $\aleph_1$ is meagre? Both of these questions remain open. However, it will be shown that, in the second question, if one relaxes the requirement that the families consist of lines and requires merely that they consist of rectifiable curves (which both topologically and measure theoretically are indistinguishable from line segments) then a negative answer can be obtained. In other words, it will follow from Theorem~\ref{t:main} that it is consistent that $\non({\mathcal Meagre})> \aleph_1$ yet there is some ${\mathcal X}$ which is a family of rectifiable curves in $\Reals^3$ such that $\card{\mathcal X} = \aleph_1$ and $\cup{\mathcal X}$ is not meagre. As well, it will be shown that the cardinal invariant ${\mathcal R}({\mathfrak Z})$ depends, to some extent, on the dimension of ${\mathfrak Z}$. It should be remarked that throughout this paper by a curve is meant a one-to-one function from the unit interval to some metric space. As will be seen in remarks in the last section, the requirement of one-to-oneness can lead to difficulties when trying to prove monotonicity results but it is nevertheless a natural requirement. Not only does it eliminate Peano curves (which, of course, are not rectifiable) but it also guarantees that the images of the curves to be considered are all topologically equivalent to the unit interval. This supports the assertion that the main result obtained in Corollary~\ref{c:main} is a step towards solving the problem of Komjath. \section{Notation and terminology} For the purposes of this paper, by a curve in some metric space ${\mathfrak X} = (X,d)$ will be meant a continuous, one-to-one function $\gamma : [0,1] \to X$. However, this notation will often be abused by statements such as $\gamma \cap X = \emptyset$. In such cases, what is really meant is that the image of $\gamma$ is disjoint from $X$. Similarly, the statement that in a particular metric space there is a family of rectifiable curves whose union is not meagre really means that the union of the images of the curves is not meagre. For any metric space ${\mathfrak X} = (X,d)$ define a length function on curves $\gamma$ in $X$ by \begin{equation}\label{n:Length} \lambda_{\mathfrak X}(\gamma) = \sup_{0 \leq x_0 < x_1 < \ldots < x_n \leq 1}\sum_{i\in n}d(\gamma(x_i),\gamma(x_{i+1})) \end{equation} A curve $\gamma$ in ${\mathfrak X}$ is said to be rectifiable if $\lambda_{\mathfrak X}(\gamma) < \infty$. It will occasionally be useful to have a notion of length defined for sets which are not curves by approximating the set from above by rectifiable curves. So for an arbitrary set $A\subseteq X$ define \begin{equation} \lambda_{\mathfrak X}(A) =\inf\{\lambda_{\mathfrak X}(\gamma) : \gamma\text{ is a curve and }A\subseteq \gamma\}. \end{equation} For any real number $\delta$ such that $0 \leq \delta < \infty$ define \begin{equation}\label{n:DeltaPaths} {\mathbf C}({\mathfrak X},\delta) = \{A\subseteq X : \lambda_{\mathfrak X}(A) \leq \delta\}. \end{equation} The neighbourhood of the point $x \in X$ consisting of the open ball of radius $\epsilon $ around $x$ will be denoted by $B_{{\mathfrak X}}(x,\epsilon)$. The diametre of a subset $X$ of a metric space will be denoted $\diam(X)$ If $\sigma$ and $\tau$ are sequences (in other words, functions whose domains are ordinals) then $\sigma \wedge \tau$ will denote the concatenation of $\sigma$ followed by $\tau$. The sequence $\{(0,x)\}$ of length 1 will be denoted by $\SingleSeq{x}$. All trees will be considered to be nonempty subtrees of some tree of the form $\presup{\stackrel{\omega}{\smile}}{X}$ for some set $X$. If $T$ is a tree and $t\in T$ then \begin{itemize} \item the set of successors of a node $t$ in $T$ is denoted by $\suc_T(t)$ and defined to be the set of all $x$ such that $t\wedge \SingleSeq{x} \in T$ \item $T\angbr{t} = \{s \in T : s\subseteq t \text{ or } t \subseteq s\}$ \item $T\Angbr{t} = \{s : t \wedge s \in T\}$ \item the root of $T$ is denoted by $\sakne(T)$ and is defined to be the maximal member $t$ of $T$ such that $T\angbr{t} = T$ \item the height of a tree $T$ is the supremum of the domains of its elements and is denoted by $\hght(T)$ \item $\Top(T)$ consists of all the maximal elements of $T$ --- in other words, $t\in \Top(T)$ if and only if $\suc_T(t) = \emptyset$ \item $\nonmax(T) = T \setminus \Top(T)$ \item $T\restriction k = \{t\in T : \card{t} \leq k\}$ for any integer $k$ \end{itemize} When dealing with iterations of partial orders involving fusion arguments it will be useful to have notation for trees of trees. By a tree of trees will be understood a finite tree $\mathcal T$ such that for each $t \in \nonmax(\mathcal T)$ the set of successors of $t$ is the set of maximal nodes of some finite tree --- in other words, $\suc_{\mathcal T}(t) = \Top(T)$ where $T$ is some finite tree. The tree $T$ will be denoted by $\tsuc_{\mathcal T}(t)$. \section{The single step partial order} Let ${\mathfrak Y} = (Y,d)$ and ${\mathfrak Z} = (Z,e)$ be metric spaces and suppose that ${\mathbf X} = \{X_n\}_{n\in\omega}$ is a given sequence of finite families of open subsets of $Y$. Define $M_n({\mathbf X}) = \prod_{i\in n}\card{X_i}$, $K_n({\mathbf X}) = 2^{nM_n({\mathbf X})}$ and suppose that $\nu_n:{\mathcal P}(X_n) \to \omega$ are functions --- which, building on ideas first formulated in \cite{MR85g:03004.Shelah}, are called weak measures in \cite{MR94g:03068} --- and $\{\epsilon^i_j\}^{i\in\omega}_{j\in D_i}$ is an infinite triangular matrix satisfying the following conditions: \begin{condi}\label{c:1} If $\bigcup \{A_i : i \in K_n({\mathbf X})\} = A \subseteq X_n $ then there is some $i\in K_n({\mathbf X})$ such that $\nu_n(A_i)\geq \nu_n(A) - 1$. \end{condi} \begin{condi}\label{c:3} If $\gamma$ is a rectifiable curve in $\mathfrak Y$ and $\lambda_{\mathfrak Y}(\gamma) \leq 1$ then $\nu_n(\{a\in X_n: a\cap \gamma\neq \emptyset\}) =0$. \end{condi} \begin{condi}\label{c:2} $\epsilon^i_{D_i} \leq \epsilon^i_{D_i - 1} \leq \ldots \leq \epsilon^i_0 = \epsilon^{i-1}_{D_{i-1}} $ and $\epsilon^i_j < 1$ for all $i$ and $j$. \end{condi} \begin{condi}\label{c:4} For any $n\in\omega$ and any $A\subseteq X_n$ and for any function $F:A \to {\mathbf C}({\mathfrak Z},{\epsilon^n_{k+1}})$ there is some $B\subseteq A$ such that $\nu_n(B) \geq \nu_n(A) - 1 $ and $\lambda_{\mathfrak Z}(\cup F[B]) \leq \diam({\mathfrak Z})\epsilon^n_{k}$. \end{condi} Let $\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ be defined to consist of all trees $T\subseteq \bigcup_{n\in\omega}\prod_{i\in n}X_i$ such that for each $n\in\omega$ there is some $m\in\omega$ such that for all $t\in T$, if $\card{t} > m$ then $\nu_{\card{t}}(\suc_T(t)) > n(M_{\card{t}}({\mathbf X}))^{{\card{t}}^2}$. \begin{defin}\label{d:AxA} If $T \in \Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z}) $ and $t\in T$ define $$\nu^*_T(t) = \nu_{\card{t}}(\suc_T(t))$$ and $$\nu^{**}_T(t) = \frac{\nu_{T}^*(t)}{(M_{\card{t}}({\mathbf X}))^{{\card{t}}^2}}$$ and, for $n\in \omega$, define $\smll_n(T) = \{t\wedge\SingleSeq{x}\in T : \nu^{**}_T(t)\leq n\}$. (With this notation $\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ consists of those trees $T$ such that $\lim_{t\in T}\nu^{**}_T(t) = \infty$.) For $T$ and $T'$ in $\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ define $T \leq_n T'$ if and only if $T\subseteq T'$, $\smll_n(T')\subseteq T$ and $\nu^{**}_T(t) \geq n$ for all $t \in T\setminus \smll_n(T')$. If ${\mathcal T}$ is a tree of trees and $\sigma \in {\mathcal T}$ then define $$\nu^*_{\mathcal T,\sigma}(t) = \nu^{*}_{\tsuc_{\mathcal T}(\sigma)}(t)$$ and define $$\nu^{**}_{\mathcal T,\sigma}(t) = \nu^{**}_{\tsuc_{\mathcal T}(\sigma)}(t) .$$ \end{defin} \begin{lemma}\label{l:AxA} If $T_n \in \Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ and $T_{n+1} \leq _n T_n$ for each $n\in \omega$ then $\cap_{n\in\omega}T_n \in \Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$. \end{lemma} \begin{proof} This is standard. See page 365 of \cite{MR94g:03068} or Lemma 7.3.5 on page 340 of \cite{MR96k:03002} for similar arguments. Note that $\leq_n$ corresponds to $\leq_n^*$ of \cite{MR94g:03068}. \end{proof} \begin{lemma}\label{l:stand} If $T \in \Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ and $T\forces{}{\Dot{x} \in V}$ then for all $n \in \omega$ there is some finite $S\subseteq T$ and $T' \leq_n T$ such that $\Top({S})$ is a maximal antichain in $T$, $\smll_n(T) \subseteq S \subseteq T'$ and a function $\Phi : \Top(S) \to V$ such that $T'\angbr{s}\forces{}{\Dot{x} = \Phi(s)}$ for each $ s\in S$.\end{lemma} \begin{proof} Note that according to Definition~\ref{d:AxA} $T\in\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ if and only if $T\subseteq \prod_{i\in\omega}X_i$ is a tree and for each $n\in\omega$ there are only finitely many $t\in T$ such that $\nu_T^{**}(t) < n$. Moreover, if $t\in T$ and $\suc_T(t) = A_0 \cup A_1$ then there is $i\in 2$ such that $\frac{\nu_{\card{t}}(A_i)}{(M_{\card{t}}({\mathbf X}))^{{\card{t}}^2}} \geq \nu^{**}_T(t) - 1$. This allows the standard Laver argument to be applied. See \cite{MR96k:03002}.\end{proof} \begin{lemma}\label{l:cond:1} If $S \in \Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$, $\card{\sakne(T)} \geq k$ and suppose that $T$ is a finite subtree of $S$. Suppose furthermore, that $\Psi : \Top(T)\to K_k({\mathbf X})$. Then there is $T' \subseteq T$ such that $\Top(T')\subseteq \Top(T)$, $\Psi$ is constant on $\Top(T')$ and $\nu^*_{T'}(t) \geq \nu^*_T(t) - 1$ for all $t \in \nonmax(T')$. \end{lemma} \begin{proof} Proceed by induction on the height of $T$. If $\hght(T) = 1$ then $K_0({\mathbf X}) =1$ so that $\Psi$ is already constant to begin with. So assume that the lemma has been established for trees of height less than or equal to $m$ and let $T$ be a tree of height $m+1$ and let $\Psi :\Top(T) \to M_{k}({\mathbf X})$. Using Condition~\ref{c:1} and the fact that $K_{m}({\mathbf X}) \geq K_k({\mathbf X})$ it follows that for each $s\in \Top(T\restriction m)$ there is $Z_s\subseteq \suc_T(s)$ such that $\nu_m(Z_s) \geq \nu_m(Z) - 1$ and $\Psi$ is constant on $Z_s$ with value $\Psi^*(s)$. Now apply the induction hypothesis to $T\restriction m$ and the mapping $\Psi^*$. \end{proof} \section{Constructing the weak measures} In this section it will be shown that for various metric spaces ${\mathfrak Y} = (Y,d)$ and ${\mathfrak Z} = (Z,e)$ there exist sequences of weak measures ${\mathbf{X}} = \{(X_n,\nu_n)\}_{n\in\omega}$ such that the partial order $\Poset({\mathbf{X}}, {\mathfrak Y},{\mathfrak Z}) $ is not empty. Moreover, it will be shown that for each $D\in \omega$ and open set ${\mathcal U} \subseteq Y$ there is a family $\mathcal X$ of open sub sets of $\mathcal U$ satisfying Conditions~\ref{c:1}, \ref{c:3}, \ref{c:2} and \ref{c:4}. This will guarantee that $\mathcal X$ can be found so that the partial order $\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ forces the union of the ground model rectifiable curves of ${\mathfrak Y}$ to be meagre. \begin{defin}\label{d:tentative} A metric space ${\mathfrak{Y}} = (Y,d)$ will be said to be {\em quasi-infinite dimensional}\/ if for each open set ${\mathcal U} \subseteq Y$ there is $\delta > 0$ such that for all $m \in\omega$ there is $A\in [{\mathcal U}]^m$ such that $d(a,b) > \delta$ for all $\{a,b\} \in [A]^2$. Define a function $\Delta_{\mathfrak{Y}}$ from $(0,1)$ to the cardinals by defining $\Delta_{\mathfrak{Y}}(\epsilon) = \kappa$ if $\kappa$ is the least cardinality of a cover of ${\mathfrak{Y}}$ by open sets of diameter less than $\epsilon$. Recall that the metric space $\mathfrak{Y}$ is said to be {\em totally bounded } \/ if $\Delta_{\mathfrak{Y}}$ takes on only integer values. Finally, define $\mathfrak{Y}$ to be {\em reasonably geodesic} if for every $x$, $y$ and $z$ belonging to $Y$ and every $\epsilon > 0$ such that $x$ and $y$ both belong to $B_{\mathfrak Y}(z, \epsilon)$ there is, for each $\delta > 0$ and each set $W$ such that $\lambda_{\mathfrak Y}(W) < \infty$, a curve $\gamma$ such that $\{x,y\}\subseteq \gamma \subseteq B_{\mathfrak Y}(z, \epsilon)$, $\gamma\cap W \subseteq \{x,y\}$ and $\lambda_{\mathfrak Y}(\gamma) < d(x,y) + \delta$. \end{defin} \begin{propo}\label{p:ERG} $\Reals^3$ is reasonably geodesic. \end{propo} \begin{proof} Let $x$, $y$ and $z$ in $\Reals^3$ and $\epsilon > 0$ be given such that $x$ and $y$ both belong to the ball $B$ of radius $\epsilon$ about $z$. Suppose as well that $\delta > 0$ and $W$ is a set such that $\lambda(W) < \infty$. Let $\rho$ be such that $0 < \rho < \delta/2$ and the spheres $S_x$ and $S_y$ with radii $\rho$ and centres $x$ and $y$ respectively are contained in $B$. Let $\pi_x$ be the mapping which sends a point $a \in \Reals^3$ to the point $\pi_x(a)$ on $S_x$ which lies on the line connecting $x$ and $a$ and let $\pi_y$ be defined similarly for the sphere $S_y$. Note that $\pi_x(W)$ and $\pi_y(W)$ are both the union of countably\footnote{Considering a spiral converging to $x$ shows that the ``countably many'' can not be eliminated here.} many sets of finite length. Moreover, let $D$ be a disk of radius $\rho$ with centre on the line joining $x$ and $y$ and contained in a plane perpendicular to this line. Let $\pi$ be the projection onto this disk parallel to the line connecting $x$ and $y$. Clearly $\pi(W)\cup \pi(\pi_x(W))\cup \pi(\pi_y(W))$ has finite length so it is possible to choose some point $s \in D \setminus\pi(W)\cup \pi(\pi_x(W))\cup \pi(\pi_y(W))$. Now let $s_x$ be the point on $S_x\cap \pi^{-1}\{s\}$ closest to $y$ and let $s_y$ be the point on $S_y\cap \pi^{-1}\{s\}$ closest to $x$. It follows that the three line segments connecting $x$ to $s_x$, $s_x$ to $s_y$ and $s_y$ to $y$ do not intersect $W$ outside of $\{x, y\}$ and have length less than $\delta$.\end{proof} Notice that immediately follows from Proposition~\ref{p:ERG} that any convex subset with nonempty interior of a Banach space whose dimension is at least 3 is reasonably geodesic. \begin{lemma}\label{l:qq} Suppose that $\delta > 0$ and $\{\gamma_i\}_{i\in k}$ are rectifiable curves in a reasonably geodesic metric space $\mathfrak Y$ such that $\lambda_{\mathfrak Y}(\gamma_i) < \delta$ and such that $\gamma_i \subseteq B_{\mathfrak Y}(x, \delta/2)$. Then there is a single rectifiable curve $\gamma$ such that $\bigcup_{i\in k} \subseteq \gamma$ and $\lambda_{\mathfrak Y}(\gamma) < 3k\delta $. \end{lemma} \begin{proof} It is elementary to prove by induction on $k$ that if $\delta > 0$ and $\{\gamma_i\}_{i\in k}$ are rectifiable curves such that $\gamma_i \subseteq B_{\mathfrak Y}(x, \delta/2)$ then, for any $\epsilon > 0$, there is a single rectifiable curve $\gamma$ such that $\bigcup_{i\in k} \subseteq \gamma$ and $\lambda_{\mathfrak Y}(\gamma) < (k-1)\delta + \sum_{i\in k}\lambda_{\mathfrak Y}(\gamma_i) \epsilon$. The lemma follows immediately from this. \end{proof} \begin{lemma}\label{l:comb} Let ${\mathfrak{Y}}$ be a quasi-infinite dimensional metric space and ${\mathfrak{Z}}$ be a totally bounded, reasonably geodesic metric space. Then for all open sets ${\mathcal U} \subseteq Y$, $K\in\omega$, $D\in \omega$ and $\Bar{\epsilon} > 0$ there is a sequence $\{\epsilon_i\}_{i=0}^{D}$ such that $$\epsilon_{D}\leq\epsilon_{D-1} \ldots\epsilon_{D-i} \leq \epsilon_{D-i-1} \leq \ldots \leq \epsilon_0 \leq \Bar{\epsilon}$$ and family ${\mathcal X}$ of open subsets of ${\mathcal Y}$ as well as a weak measure $\nu :{\mathcal P}({\mathcal X}) \to \omega$ such that $\nu({\mathcal X}) = D$ and $\nu$ satisfies Conditions~\ref{c:3}, \ref{c:4} as well as the following version of Condition~\ref{c:1}: If $\bigcup \{A_i\}_{i \in K} = A \subseteq {\mathcal X} $ then there is some $i\in K$ such that $\nu(A_i)\geq \nu(A) - 1$. \end{lemma} \begin{proof} Let $\delta > 0$ be such that for every $m\in\omega$ there is $A\in [{\mathcal U}]^m$ such that $d(a,b) > \delta$ for all $\{a,b\} \in [A]^2$. Construct $\{(L_i,\epsilon_i)\}_{i=0}^{D}$ by induction on $i$. Let $\epsilon_0 = \Bar{\epsilon}$ and $L_0 > K/\delta $. Suppose that $\epsilon_i$ and $L_i$ have been constructed. Let $\epsilon_{i+1} = \frac{\epsilon_i}{3L_i}$ and let $L_{i+1} > K\Delta_{\mathfrak Z}(\epsilon_{i+1})L_i$. Now choose $A\in [{\mathcal U}]^{L_D}$ such that $d(a,b) > \delta$ for all $\{a,b\} \in [A]^2$. Then let $\delta' > 0$ be such that $d(a',b') > \delta$ for all $(a',b')$ for which there exists $\{a,b\} \in [A]^2$ such that $d(a,a') < \delta'$ and $d(b,b') < \delta'$. Let ${\mathcal X} = \{B_{\mathfrak Y}(a,\delta')\}_{a\in A}$. Finally, define $\nu({\mathcal X}') = \max(\{k : \card{{\mathcal X}')} \geq L_{k}\})$ for any ${\mathcal X}'\subseteq {\mathcal X}$. To see that the appropriate version of Condition~\ref{c:1} holds note that if $\nu({\mathcal X}') \geq k+1$ and ${\mathcal X}' = \cup_{i\in K}A_i$ then there is some $i \in K$ such that $\card{A_i} \geq L_{k}$ because $\Delta_{\mathfrak Z}(\epsilon)$ is always at least 1. To see that Condition~\ref{c:3} holds let $\gamma$ be curve such that $\lambda_{\mathfrak Y}(\gamma) \leq 1$ and suppose that $$\nu(\{ B_{\mathfrak Y}(a,\delta') \in {\mathcal X} : B_{\mathfrak Y}(a,\delta') \cap \gamma \neq \emptyset\}) \geq 1.$$ But then choosing $x_a \in B_{\mathfrak Y}(a,\delta') \cap \gamma$ for each $a$ such that $ B_{\mathfrak Y}(a,\delta')\cap \gamma \neq \emptyset$ yields a counterexample to the inequality $\lambda_{\mathfrak Y}(\gamma) \leq 1$ because for any enumeration $\{a_i\}_{i=0}^{L'}$ of $\{a\in A : B_{\mathfrak Y}(a,\delta')\cap \gamma \neq \emptyset\}$ yields that $$\sum_{i=0}^{L'}d(x_{a_i},x_{a_{i+1}}) \geq {L'}\delta \geq L_0\delta \geq (K/\delta )\delta > 1$$ because $L' < L_0$ implies that $\nu(\{a \in A : B_{\mathfrak Y}(a,\delta') \cap \gamma \neq \emptyset\}) = 0$. So all that remains to be checked is Condition~\ref{c:4}. In order to see that this holds, suppose that $\nu({\mathcal X}') \geq k + 1$ for some ${\mathcal X}' \subseteq {\mathcal X}$. Suppose that $F : {\mathcal X}' \to {\mathbf C}({\mathfrak Z},{\epsilon_{k+1}})$ and that, without loss of generality, ${\mathcal X}' \in [{\mathcal X}]^{L_{k+1}}$. Let $\{{\mathcal O}_i\}_{i\in \Delta_{\mathfrak Z}(\epsilon_{k+1})} $ enumerate an open cover of $Z$ with sets of diameter less than $\epsilon_{k+1}$. For each $B_{\mathfrak Y}(a,\delta') \in {\mathcal X}'$ choose $J(a) \in \Delta_{\mathfrak Z}(\epsilon_{k+1})$ such that $F(B_{\mathcal Y}(a,\delta'))\cap {\mathcal O}_{J(a)} \neq \emptyset$. Then choose $j\in \Delta_{\mathfrak Z}(\epsilon_{k+1})$ such that $\card{\{B_{\mathfrak Y}(a,\delta') \in {\mathcal X}': J(a) = j\}} \geq L_{k}$ and let ${\mathcal X}'' \in [\{B_{\mathfrak Y}(a,\delta') \in {\mathcal X}': J(a) = j\}]^{L_k}$. Hence $\nu({\mathcal X}'') \geq k$. Moreover it follows from Lemma~\ref{l:qq} that $\lambda_{\mathfrak Z}(\cup F[{\mathcal X}'']) < 3L_k\epsilon_{k+1} < \epsilon_k$.\end{proof} \begin{corol}\label{cor:norms} Let ${\mathfrak{Y}}$ be a separable, complete, quasi-infinite dimensional metric space and ${\mathfrak{Z}} $ be a totally bounded, reasonably geodesic metric space. Then there is a matrix $\{\epsilon_i\}_{j\in D_i}^{i\in\omega}$ as well as weak norms $\{\nu_i\}_{i\in\omega}$ on $\{X_i\}_{i\in\omega}$ such that Conditions~\ref{c:1}, \ref{c:3}, \ref{c:2} and \ref{c:4} are satisfied by ${\mathbf X} = \{(\nu_i, X_i)\}_{i\in\omega}$, and, furthermore, if $G \subseteq\Poset({\mathbf X}, {\mathfrak Y}, {\mathfrak Z})$ is generic over some model of set theory $V$, then, in $V[G]$ there is a dense $G_\delta$ subset of $\mathfrak Y$ which is disjoint from every rectifiable curve in $\mathfrak Y$ which belongs\footnote{ What is meant here, of course, is that there is a code for the dense $G_\delta$ set and the Borel set which it codes is disjoint from every curve with a code in the ground model. To say that a curve $\gamma$ belongs to $V$ is the same as saying that $\gamma\restriction \Rationals \in V$. } to $V$. \end{corol} \begin{proof} Let $\{{\mathcal U}_n\}_{n\in\omega}$ enumerate a base of open sets for $\mathfrak Y$. Construct $\{\epsilon_i\}_{j\in D_i}^{i\in n}$ and $\nu_i:{\mathcal P}(X_i)\to \omega$ by induction on $n$. Suppose that $\{\epsilon_i\}_{j\in D_i}^{i\in n}$ and $\nu_i:{\mathcal P}(X_i)\to \omega$ have been constructed for $i \in n$. Let $$D_n > n2^{(\prod_{i\in n}\card{X_i})^n}$$ and the use Lemma~\ref{l:comb} to find a sequence $\{\epsilon^n_i\}_{i\in D_n}$ such that $$\epsilon^{n}_{D_n}\leq\ldots\epsilon^{n}_{i+1} \leq \epsilon^{n}_i \leq \ldots \leq \epsilon^{n}_0 \leq \epsilon^{n-1}_{D_{n-1}}$$ as well as a family $X_n$ of open subsets of ${\mathcal U}_n$ and a weak measure $\nu_n$ such that $\nu_n(X_n) = D_n$ and Conditions~\ref{c:1}, \ref{c:3} and \ref{c:4} are all satisfied. It follows that $\Poset({\mathbf X}, {\mathfrak Y}, {\mathfrak Z})$ is not empty. If $G$ is $\Poset({\mathbf X}, {\mathfrak Y}, {\mathfrak Z})$ generic then let $G^*$ denote the sequence defined by $G^*(i) = s(i)$ if and only if there is some $T\in G$ such that $s= \sakne(T)$ and $\card{s} > i$. Then let $${\mathcal V}_G = \bigcap_{i\in\omega}\bigcup_{j \in \omega\setminus i}G^*(j)$$ and note that ${\mathcal V}_G $ is a dense $G_\delta$ in $\mathfrak Z$. To see that ${\mathcal V}_G \cap \gamma = \emptyset$ for every rectifiable curve $\gamma \in V$ let $T \in \Poset({\mathbf X}, {\mathfrak Y}, {\mathfrak Z})$ and let $\gamma$ be a rectifiable curve in $V$. By dividing $\gamma$ into no more than finitely many rectifiable pieces, it may be assumed that $\lambda_{\mathfrak Y}(\gamma) < 1$. Let $r=\card{\sakne(T)}$ and define $$T' = \{t\in T: (\forall i > r)t(i)\cap \gamma = \emptyset\}$$ and $T'$ is a tree. Furthermore, $T'\in \Poset({\mathbf X}, {\mathfrak Y}, {\mathfrak Z})$ because, if $t\in T'$ and $\card{t} > r$ then $\nu_n(\suc_{T'}(t)) \geq \nu_n(\suc_{T}(t)) - 1$ by Conditions~\ref{c:1} and \ref{c:3}. It follows that $$T'\forces{\Poset({\mathbf X}, {\mathfrak Y}, {\mathfrak Z})}{\gamma \cap \bigcup_{j \in \omega\setminus r}G^*(j) = \emptyset}$$ and hence, $1\forces{\Poset({\mathbf X}, {\mathfrak Y}, {\mathfrak Z})}{\gamma \cap{\mathcal V}_G = \emptyset}$. \end{proof} It is worth noting that the hypothesis in Lemma~\ref{l:comb} as well as Corollary~\ref{cor:norms} can be weakened by examining the proof of Lemma~\ref{l:comb}. Given a metric space $\mathfrak Z$, $\epsilon > 0$, integers $K$ and $D$ and $\delta > 0$ it is possible to define a pair of sequences $\{L_i\}_{i=0}^D$ and $\{\epsilon_i\}_{i=0}^D$ as in the proof of Lemma~\ref{l:comb} as follows: $L(K,D,\delta,\epsilon)_0 = K/\delta $, $\epsilon(K,D,\delta,\epsilon)_0 = \epsilon$ and $\epsilon(K,D,\delta,\epsilon)_{i+1} = \epsilon(K,D,\delta,\epsilon)_i/3L_i$ and $L(K,D,\delta,\epsilon)_{i+1} = K\Delta_{\mathfrak Z}(\epsilon(K,D,\delta,\epsilon)_{i+1})L(K,D,\delta,\epsilon)_i$. All that was required of the pair of metric spaces $\mathfrak Y$ and $\mathfrak Z$ in the proof of Lemma~\ref{l:comb} was that in every open subset $\mathcal U$ of $\mathfrak Y$ and for every $\Bar{\epsilon} > 0$ and for all integers $K$ and $D$ there is $\epsilon < \Bar{\epsilon} $ and $\delta > 0$ such that there are $L(K,D,\delta,\epsilon)_D$ points $\mathcal U$ which are pairwise separated by a distance of at least $\delta$ and that $\delta L(K,D,\delta,\epsilon)_0 > 1$. To see an example of such a pair of spaces $\mathfrak Y$ and $\mathfrak Z$ where this occurs yet $\mathfrak Y$ is not an infinite dimensional Banach space, nor is it quasi-infinite dimensional, and $\mathfrak Z = [0,1]^3$ and construct $\mathfrak Y$ to be a compact subspace of $\ell^p$. First choose blocks of independent vectors in $\ell^p$ of cardinality $H_n$ all of which have norm $1/n$. By having $H_n$ grow sufficiently quickly and taking the convex hull of the union of all the blocks it is possible to arrange the desired property. \section{Trees of trees} Trees of trees will arise in the context of of iterations of some version of the partial order $\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$. As a consequence, throughout this section it will be assumed that all trees are finite subtrees of some condition in $\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ and that whenever $\mathcal T$ is a tree of trees and $\sigma \in \mathcal T$ then $\tsuc_{\mathcal T}(\sigma)$ is a finite subtree of some condition in $\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$. \begin{defin}\label{d:<} If $\mathcal T$ and $\mathcal S$ are trees of trees then define ${\mathcal T}\prec{\mathcal S}$ if and only if \begin{gather} {\mathcal T}\subseteq\mathcal S\\ \Top({\mathcal T}) \subseteq\Top(\mathcal S)\\ \label{d:4.5} (\forall \sigma\in \nonmax({\mathcal T})) \Top(\tsuc_{{\mathcal T}}(\sigma))\subseteq\Top(\tsuc_{\mathcal S}(\sigma) )\end{gather}\end{defin} \begin{defin}\label{d:9} Suppose that $\mathcal T$ and $\mathcal S$ are trees of trees and that ${\mathcal T}\prec{\mathcal S}$. For any function $f:\omega\times\omega\times\omega \to \omega$ define ${\mathcal T}\prec_f{\mathcal S}$ if \begin{multline} (\forall \sigma \in \nonmax({\mathcal T})) (\forall t \in \nonmax(\tsuc_{{\mathcal T}}(\sigma))) \nu^*_{{\mathcal T}, \sigma}(t) \geq \nu^*_{{\mathcal S}, \sigma}(t) - f(\hght({\mathcal T}),\card{\sigma},\card{t}) \end{multline} If $f$ is a constant function with value $k$ then $\prec_f$ will be written as $\prec_k$. Notice that this makes sense only when ${\mathcal T} \prec \mathcal S$ because of \eqref{d:4.5}. \end{defin} When dealing with trees of trees in the arguments to follow, it will occasionally turn out to be useful to be able to restrict the tree which determines the successors of a given node to a smaller level. However, it is not possible to do this directly because, for example, it $\mathcal T$ is a tree of trees and $t \in \tsuc_{\mathcal T}(\emptyset)$ comes from the some level below the top then there are many extensions of $t$ to the top level and not all of these will agree on how the tree of trees is to continue. In other words, $t_1$ and $t_2$ could both be extensions of $t$ in $\tsuc_{\mathcal T}(\emptyset)$ yet ${\mathcal T}\angbr{\SingleSeq{t_1}}$ could be different than ${\mathcal T}\angbr{\SingleSeq{t_2}} $. However, if $\mathcal T$ is carefully chosen then the number of extensions of $t$ will be much greater than the number of possibilities for ${\mathcal T}\angbr{\SingleSeq{t'}}$ where $t'$ extends $t$. A pigeonhole argument then makes it possible to find a large set of extensions which all agree on the way the tree of trees should be extended --- at least to small levels. This is the content of the next series of technical definitions and lemmas. \begin{defin}\label{d:10} For a tree of trees $\mathcal T$ and $\sigma \in \mathcal T$ define $R_k(\sigma)$ to be a sequence with the same domain as $\sigma$ and defined by $R_k(\sigma)(i) = \sigma(i)\restriction k$. \end{defin} \begin{defin} For a tree of trees $\mathcal T$ define $\hght^*({\mathcal T})$ to be the maximum of all $\hght(\tsuc_{\mathcal T}(\sigma))$ as $\sigma$ ranges over $\mathcal T$, $\hght_*({\mathcal T})$ to be the minimum of all $\hght(\tsuc_{\mathcal T}(\sigma))$ as $\sigma$ ranges over $\mathcal T$ and define $\sakne_*({\mathcal T})$ to be the minimum of all $\card{\sakne(\tsuc_{\mathcal T}(\sigma)}$ as $\sigma$ ranges over $\mathcal T$. Define $\nu_*({\mathcal T})$ to be the minimum value of $$\{\nu_{{\mathcal T}, \sigma}(t) : \sigma \in {\mathcal T}, t\in \nonmax( \tsuc_{\mathcal T}(\sigma)) \AND t\not\subseteq \sakne(\tsuc_{\mathcal T}(\sigma))\}$$ recalling that $\mathcal T$ arises in the context of $\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ and so it is being assumed that a weak measure $\nu$ is defined on the successors of $t$ for each non-maximal member of $\tsuc_{\mathcal T}(\sigma)$. \end{defin} \begin{defin}\label{d:11} Let $\mathcal T$ be a tree of trees and $k$ an integer. The tree of trees $\mathcal T$ will be said to be $k$-simple if $\tsuc_{\mathcal T}(\sigma) \restriction k = \tsuc_{\mathcal T}(\sigma')\restriction k$ whenever $R_k(\sigma) = R_k(\sigma')$. \end{defin} \begin{lemma}\label{l:30} If $\mathcal T$ is $k$-simple then $${\mathcal R}(k,\mathcal T) = \{R_k(\sigma) : \sigma \in {\mathcal T}\}$$ is a tree of trees such that $\hght^*({\mathcal R}(k,\mathcal T)) \leq k$. \end{lemma} \begin{proof} For any $R_k(\sigma) \in {\mathcal R}(k,\mathcal T)$ define $$\tsuc_{{\mathcal R}(k,\mathcal T)}(R_k( \sigma)) = \tsuc_{\mathcal T}(\sigma)\restriction k$$ and note that, because $\mathcal T$ is $k$-simple, there is no ambiguity in this definition. The fact that $\hght^*({\mathcal R}(k,\mathcal T)) \leq k$ is immediate. \end{proof} \begin{lemma}\label{l:31} If $\mathcal T$ is $k$-simple, $\hght_*({\mathcal T}) \geq k$ and $\sigma \in {\mathcal R}(k,\mathcal T)$ then $${\mathcal R}(k,\mathcal T,\sigma) = \{\tau \in {\mathcal T}:\sigma \restriction \card{\tau} = R_k(\tau)\}$$ is a $k$-simple tree of trees such that $\sakne_*({\mathcal R}(k,\mathcal T,\sigma)) \geq k$. \end{lemma} \begin{proof} For any $\tau \in {\mathcal R}(k,\mathcal T,\sigma)$ define $$\tsuc_{{\mathcal R}(k,\mathcal T,\sigma)}(\tau) = \tsuc_{\mathcal T}(\tau)\angbr{\sigma(\card{\tau}) } .$$ It is immediate that $\sakne^*({\mathcal R}(k,\mathcal T,\sigma)) \geq k$ because $\sakne(\tsuc_{\mathcal T}(\tau)) \supseteq \sigma({\card t})$ for all $\tau \in {\mathcal R}(k,{\mathcal T},\sigma)$ since $\hght_*({\mathcal T}) \geq k$. The $k$-simplicity of ${\mathcal R}(k,{\mathcal T},\sigma)$ follows from observing that if $\tau \in {\mathcal R}(k,{\mathcal T},\sigma)$ then $$\sakne(\tsuc_{{\mathcal R}(k,{\mathcal T},\sigma)}(\tau)) \supseteq \sigma(\card{\tau}).$$ \end{proof} \begin{lemma}\label{l:tt} Let $\mathcal T$ be a tree of trees and let $k$ be an integer greater than $\hght({\mathcal T})$. It is then possible to find a tree of trees ${\mathcal T}^k \prec_{1} \mathcal T$ which is $k$-simple and, furthermore, $$\tsuc_{{\mathcal T}^k}(\sigma)\restriction k = \tsuc_{{\mathcal T}}(\sigma)\restriction k$$ for all $\sigma \in {\mathcal T}^k$. \end{lemma} \begin{proof} Define ${\mathcal T}^k$ by induction on $\hght({\mathcal T})$. If $\hght({\mathcal T}) = 0$ then ${\mathcal T}^k$ is empty. Otherwise, let $\hght({\mathcal T}) = n+1$ and define $\tsuc_{{\mathcal T}^k}(\emptyset)\restriction k$ to be $\tsuc_{{\mathcal T}}(\emptyset)\restriction k$. Now fix $t \in \Top(\tsuc_{{\mathcal T}^k}(\emptyset))$ and use the induction hypothesis to choose $\Tilde{\mathcal T}_s \prec_1 {\mathcal T}\Angbr{\SingleSeq{s}}$ which is $k$-simple. Define a function $\Psi$ by letting $\Psi(s) = {\mathcal R}(k,\Tilde{\mathcal T}_{s})$ for each $s\in \Top(\tsuc_{\mathcal T}(\emptyset)\angbr{t})$. Since the number of possible values for $\Psi(s)$ is not greater than $$(2^{\prod_{i\in k}\card{X_i}})^{\hght({\mathcal T})} \leq 2^{(n+1)M_k({\mathbf X})}\leq 2^{kM_k({\mathbf X})}$$ the last inequality being an immediate consequence of the fact that $k \geq \hght({\mathcal T})$. Consequently, the number of possible values for $\Psi$ is no greater than $K_k({\mathbf X})$ and hence, it follows from Lemma~\ref{l:cond:1} that there is some tree ${\mathcal S}_t \subseteq \tsuc_{\mathcal T}(\emptyset)\angbr{t}$ such that $\Psi$ is constant on $\Top({\mathcal S}_t) \subseteq \Top(\tsuc_{\mathcal T}(\emptyset)\angbr{t})$ and $\nu^*_{{\mathcal S}_t}(r) \geq \nu^*_{{\mathcal T}, \emptyset}(r) - 1$ for all $r \in \nonmax({\mathcal S}_t)$. Now define $$\tsuc_{{\mathcal T}^k}(\emptyset) = \bigcup\{{\mathcal S}_t : {t \in\tsuc_{{\mathcal T}}(\emptyset)\restriction k}\} .$$ Having defined $\tsuc_{{\mathcal T}^k}(\emptyset)$, in order to define ${\mathcal T}^k$ completely, it suffices to define ${\mathcal T}^k\Angbr{\SingleSeq{t}} = \Psi(t)$ for each $t \in \Top(\tsuc_{{\mathcal T}^k}(\emptyset))$ noting that the induction hypothesis implies that ${\mathcal T}^k \prec_1 \mathcal T$. \end{proof} For the next lemma recall that throughout this section, whenever a tree of trees $\mathcal T$ is mentioned it is assumed that $\tsuc_{\mathcal T}(\sigma)$ is a finite subtree of some condition in $\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ for any $\sigma \in \mathcal T$ where ${\mathfrak Z} = (Z,d)$. The integers $ M_k({\mathbf X})$ and the reals $\epsilon^i_j$ mentioned in the statement and proof of the lemma all refer to the parameters of $\Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$. \begin{lemma}\label{l:m} Let $f:\omega^3 \to \omega$ be the function defined by $f(i,j,k) = M_k({\mathbf X})^{i(i-j)}$. Suppose that $\mathcal T$ is a tree of trees and $\Phi:\Top({\mathcal T}) \to {Y}$ is a function. Suppose also that $\sakne_*({\mathcal T}) > \hght({\mathcal T})$ and that $\nu_{*}({\mathcal T}) > \hght({\mathcal T})$. It is then possible to find a tree of trees $\Bar{\mathcal T}$ such that $\Bar{\mathcal T} \prec_f\mathcal T$ and such that $\lambda_{\mathfrak Z}(\Phi[\Top({\Bar{\mathcal T}})]) < \diam({\mathfrak Z})$ \end{lemma} \begin{proof} Induction on the height of the tree of trees $\mathcal T$ will be used to prove the following, stronger, assertion from which the lemma will follow: \begin{claim} Suppose that $\mathcal T$ is a tree of trees of height $h$, $i = \sakne_*({\mathcal T})$, $j = \hght^*({\mathcal T})$, $Q=\nu_*({\mathcal T})$ and $h < i$ and $h < Q$. If, additionally, $\Phi:\Top({\mathcal T}) \to {\mathbf C}({\mathfrak Z},{\epsilon^j_0})$ is a function then it is possible to find a nonempty tree of trees $\Bar{\mathcal T} \prec_f \mathcal T$ such that $\lambda_{\mathfrak Z}(\cup\Phi[\Top({\Bar{\mathcal T}})]) < \diam({\mathfrak Z})\epsilon^i_{Q - h}$. \end{claim} The lemma follows immediately from the claim upon recalling that the $\epsilon^a_b $ have been chosen so that $\epsilon^a_b < 1$ for all integers $a$ and $b$. If the height of ${\mathcal T}$ is zero then there is nothing to prove so suppose that the height of $\mathcal T$ is $n + 1$ and the claim has been established for trees of trees of height $n$ or less. Now proceed by induction on $\hght(\tsuc_{\mathcal T}(\emptyset))= K$. If $K \leq i$ then $\tsuc_{\mathcal T}(\emptyset) $ is either empty or has only a single top node. It is therefore possible to apply the induction hypothesis because either ${\mathcal T}$ has height $n$ or $\suc_{\mathcal T}(\emptyset) = \{\SingleSeq{t}\}$ and so the height of ${\mathcal T}\angbr{\SingleSeq{t}}$ is essentially $ n$. In particular, in the last case it is possible to apply the induction hypothesis to ${\mathcal T}\Angbr{\SingleSeq{t}} $ to get $\Tilde{\mathcal T} \prec_f {\mathcal T}\Angbr{\SingleSeq{t}} $ and then note that $f(n+1,\card{\sigma},\card{t}) \geq f(n,\card{\sigma},\card{t})$ so that it is possible to reattach the node $\SingleSeq{t}$ to the bottom of $\Tilde{\mathcal T}$ to get $\Bar{\mathcal T}$ as desired. Therefore it may be assumed that $K > i$. If $e$ is defined to be $\hght(\tsuc_{\mathcal T}(\emptyset)) - 1$ then $e \geq \hght({\mathcal T})$ and so, it is possible to apply Lemma~\ref{l:tt}. Hence, let ${\mathcal T}^e$ be an $e$-simple tree such that ${\mathcal T}^e \prec_1 {\mathcal T}$ such that \begin{equation}\label{5.45} \tsuc_{{\mathcal T}^e}(\sigma)\restriction k = \tsuc_{{\mathcal T}}(\sigma)\restriction k \end{equation} for all $\sigma \in {\mathcal T}^e$. Now let $\{\tau_w\}_{w\in L}$ enumerate ${\mathcal R}(e,{\mathcal T}^e)$ noting that $L < M_e({\mathbf X})^{h+1}$. Using the induction hypothesis, it is possible to construct, in $L$ steps, a sequence of $e$-simple trees $\{{\mathcal S}_w\}_{w\in L} $ such that \begin{gather} \label{g:0} {\mathcal S}_0 = {\mathcal T}^e\\ \label{g:1} {\mathcal R}(e,{\mathcal S}_w) = {\mathcal R}(e,{\mathcal T}^e)\\ \label{g:2} {\mathcal S}_{w+1} \prec {\mathcal S}_{w}\\ \begin{split} \label{g:3} {\mathcal R}(e,{\mathcal S}_{w+1}, \tau_{w+1}) \prec_{\Bar{f}} {\mathcal R}(e,{\mathcal S}_w,\tau_{w+1}) \text{ where }\Bar{f}\\ \text{ is defined by } \Bar{f}(a, b, c) = f(a-1,b-1,c) \end{split}\\ \label{g:4} (\forall \sigma \in {\mathcal R}(e,{\mathcal S}_{w+1}))\IF \sigma \not\subseteq \tau_{w+1} \text{ then } \tsuc_{{\mathcal S}_{w}}(\sigma) = \tsuc_{{\mathcal S}_{w+1}}(\sigma) \\ \label{g:5} \lambda_{\mathfrak Z}(\cup\Phi[\Top({\mathcal R}(e,{\mathcal S}_w,\tau_w))]) < \diam({\mathfrak Z})\epsilon^{e}_{Q - (n+1)} \end{gather} To see why this can be done, suppose that the ${\mathcal S}_w$ have been constructed satisfying Conditions~\eqref{g:0} to \eqref{g:5} for $ w\in u$. By the choice of $e$, either $\tsuc_{{\mathcal R}(e,{\mathcal S}_{u-1},\tau_u)}(\emptyset)$ has height less than $e$ or $\Top( \tsuc_{{\mathcal R}(e,{\mathcal S}_{u-1},\tau_u)}(\emptyset) ) = \{\tau_u(0)\wedge\SingleSeq{z}\}_{z\in E}$ for some nonempty set $E$. For each $z\in E$ the tree of trees ${\mathcal R}_z = {\mathcal R}(e,{\mathcal S}_{u-1}\Angbr{\SingleSeq{z}},\tau_u)$ has height less than $n+1$ so it is possible to use the induction hypothesis to find a subtree $\Bar{\mathcal R}_z\prec_f {\mathcal R}_z$ such that the conditions of the claim are satisfied. Since $\sakne_*({\mathcal R}_z) = \Bar{e}\geq e$ it follows that $$\lambda_{\mathfrak Y}(\cup\Phi[\Top({\Bar{\mathcal R}_z})]) < \diam({\mathfrak Z})\epsilon^{\Bar{e}}_{Q_z - n } $$ where $Q_z = \nu_*({\mathcal R}_z)$. Notice that the definition of $\nu_*$ guarantees that $Q_z \geq Q$ and hence, $Q_z - n > 0$. Then define $\Phi^* : E \to {\mathbf C}({\mathfrak Z},{\epsilon^{\Bar{e}}_{Q_z - n }}) $ by $\Phi^*(z) = \cup\Phi[\Top({\Bar{\mathcal R}_z})]$ and use the Condition~\ref{c:4} for the weak measure $\nu_{\Bar{e}}$ to find $E^*\subseteq E$ such that $\nu_{\Bar{e}}(E^*) \geq \nu_{\Bar{e}}(E) - 1$ and $\lambda_{\mathfrak Z}(\bigcup_{z\in E^*}\Phi^*(z)) < \epsilon^{\Bar{e}}_{{Q_z - (n +1)}} \leq \epsilon^{{e}}_{{Q_z - (n +1)}} $. Let ${\mathcal S}_u$ be defined by setting declaring $\sigma \in {\mathcal S}_u$ if and only if $\sigma \in {\mathcal S}_{u-1}$ and, if $k \leq \card{\sigma}$ and $ R_e(\sigma\restriction k)\subseteq \tau_u$ and $ \sigma = \SingleSeq{z}\wedge\sigma'$ then $\sigma'\restriction (k-1) \in \Bar{\mathcal R}_z$. It is easy to see that ${\mathcal S}_u$ is indeed a tree of trees. It must be checked that it is also $e$-simple and that conditions~\eqref{g:1} to \eqref{g:5} hold. To see that ${\mathcal S}_u$ is $e$-simple suppose that $\sigma$ and $\sigma'$ belong to ${\mathcal S}_u$ and that $R_e(\sigma) = R_e(\sigma')$. It follows from the induction hypothesis that ${\mathcal S}_{u-1}$ is $e$-simple and that $$\tsuc_{{\mathcal S}_{u-1}}(\sigma)\restriction e = \tsuc_{{\mathcal S}_{u-1}}(\sigma')\restriction e = \tsuc_{{\mathcal T}^e}(\sigma)\restriction e .$$ Let $t \in \Top(\tsuc_{{\mathcal S}_{u-1}}(\sigma)\restriction e)$. It suffices to show that there is some $\tau \in {\mathcal S}_u$ such that $\tau = \sigma \wedge \SingleSeq{t'}$ such that $t \subseteq t'$. There are two cases to consider. First, suppose that $R_e(\sigma)\wedge\SingleSeq{t}\not\subseteq \tau_u$. In this case it is possible to choose $t' \in \tsuc_{{\mathcal T}^e}(\sigma)$ which extends $t$ and to then set $\tau = \sigma \wedge \SingleSeq{t'}$. Otherwise, $R_e(\sigma)\wedge\SingleSeq{t}\subseteq \tau_u$. It should be observed that, in this case, $\sigma = \SingleSeq{z}\wedge \Tilde{\sigma}$ for some $z\in E^*$ and $\Tilde{\sigma} \in \Bar{\mathcal R}_z $. This means that $t\subseteq \sakne(\tsuc_{\Bar{\mathcal R}_z}(\sigma)$ since $\sakne_*(\Bar{\mathcal R}_z) > e$. Hence it suffices to choose any $\tau \in \suc_{\Bar{\mathcal R}_z }(\Tilde{\sigma})$ since it will automatically follow that $R_e(\tau) = R_e(\Tilde{\sigma})\wedge\SingleSeq{t}$ and hence $\SingleSeq{z}\wedge {\tau} \in {\mathcal S}_u$. This same argument establishes condition~\eqref{g:4} as well. Condition~\eqref{g:2} follows from the fact that ${\mathcal R}_z \prec {\mathcal R}(e,{\mathcal S}_{u-1},\tau_u)$ for each $z \in E^*$. In order to verify that condition~\eqref{g:3} holds let $\sigma \in \nonmax({\mathcal R}(e,{\mathcal S}_{u},\tau_u))$ and $t \in \nonmax(\tsuc_{{\mathcal R}(e,{\mathcal S}_{u},\tau_u)}(\sigma))$. If $\sigma = \emptyset$ then either $t \subset \sakne(\tsuc_{{\mathcal R}(e,{\mathcal S}_{u},\tau_u)}(\sigma))$, and in this case $\nu^{*}_{{\mathcal R}(e,{\mathcal S}_{u-1},\tau_u),\sigma}(t) = \nu^{*}_{{\mathcal R}(e,{\mathcal S}_{u},\tau_u),\sigma}(t)$, or else $t = \sakne(\tsuc_{{\mathcal R}(e,{\mathcal S}_{u},\tau_u)}(\sigma))$, and in this case $\nu^{*}_{{\mathcal R}(e,{\mathcal S}_{u-1},\tau_u),\sigma}(t) - 1 = \nu_e(E)- 1\leq \nu(E^*) = \nu^{*}_{{\mathcal R}(e,{\mathcal S}_{u},\tau_u),\sigma}(t)$. On the other hand, if $\sigma \neq \emptyset$ and $\sigma = \SingleSeq{z}\wedge \sigma'$ then $$\nu^{*}_{{\mathcal R}(e,{\mathcal S}_{u-1},\tau_u),\sigma}(t) = \nu^{*}_{{\mathcal R}(e,{\mathcal S}_{u-1}\Angbr{\SingleSeq{z}},\tau_u),\sigma'}(t)$$ and, moreover, the choice of ${\mathcal R}_z$ guarantees that \begin{equation}\begin{split} \ & \ \nu^{*}_{{\mathcal R}(e,{\mathcal S}_{u-1}\Angbr{\SingleSeq{z}},\tau_u),\sigma'}(t) - f({\mathcal R}(e, \hght({\mathcal S}_{u-1}\Angbr{\SingleSeq{z}},\tau_u),\card{\sigma'}, \card{t}) \\ \leq \ & \nu^{*}_{{\mathcal R}_z,\sigma'}(t) - f(\hght({\mathcal R}(e,{\mathcal S}_{u-1}\Angbr{\SingleSeq{z}},\tau_u)),\card{\sigma'}, \card{t}) \\ = \ & \nu^{*}_{{\mathcal R}_z,\sigma'}(t) - f(\hght({\mathcal R}(e, {\mathcal S}_{u}, \tau_u)-1,\card{\sigma} -1, \card{t}) \\ = \ & \nu^{*}_{ {\mathcal R}(e,{\mathcal S}_{u},\tau_u) ,\sigma}(t) - f(\hght({\mathcal R}(e,{\mathcal S}_{u},\tau_u))-1,\card{\sigma} -1, \card{t}) \\ = \ & \nu^{*}_{{\mathcal R}(e,{\mathcal S}_{u},\tau_u),\sigma}(t) - \Bar{f}(\hght({\mathcal R}(e,{\mathcal S}_{u},\tau_u)),\card{\sigma}, \card{t}) \end{split}\end{equation} Condition~\eqref{g:5} follows from the fact that $Q_z \geq Q$ for each $z\in E^*$. Note that $${\mathcal R}(e,{\mathcal S}_{L}) = {\mathcal R}(e,{\mathcal T}^e) $$ and that, furthermore, for each $\sigma \in \Top({\mathcal R}(e,{\mathcal S}_L))$ and $\tau \in {\mathcal R}(e,{\mathcal T}^e,\sigma) $ condition~\eqref{g:3} of the induction has been applied no more than $M_{\card{t}}({\mathbf X})^{n+1 - \card{\tau}}$ times. To be precise, if $\sigma\restriction \card{\tau} \neq\tau_w\restriction \card{\tau}$ then Condition~\eqref{g:4} rather than Condition~\eqref{g:3} applies at stage $w+1$ of the construction. In other words, Condition~\eqref{g:3} is in force only if $\tau \in {\mathcal R}(e,{\mathcal T}^e,\tau_w)$ and $\sigma\restriction \card{\tau} =\tau_w\restriction \card{\tau}$. By counting the number of indices $w$ such that $\sigma\restriction \card{\tau} = \tau_w\restriction\card{\tau}$ is possible to conclude that, for each $\tau \in {\mathcal R}(e,{\mathcal S}_L,\sigma)$ and each $t \in \tsuc_{ {\mathcal R}(e,{\mathcal S}_L,\sigma)}(\tau)$ \begin{equation}\begin{split}\label{m:ff} \nu^*_{{\mathcal R}(e,{\mathcal S}_L,\sigma),\tau}(t) \geq & \nu^*_{{\mathcal R}(e,{\mathcal T}^e,\sigma),\tau}(t) - ( M_{\card{t}}({\mathbf X})^{(n+1 - \card{\tau})} - 1)\Bar{f}(n+1,\card{\tau},\card{t}) \\ \geq & \nu^*_{{\mathcal R}(e,{\mathcal T},\sigma),\tau}(t) - (M_{\card{t}}({\mathbf X})^{(n+1 - \card{\tau})} - 1)\Bar{f}(n+1,\card{\tau},\card{t}) - 1\\ = & \nu^*_{{\mathcal R}(e,{\mathcal T},\sigma),\tau}(t) - (M_{\card{t}}({\mathbf X})^{(n+1 - \card{\tau})} - 1) M_{\card{t}}({\mathbf X})^{n(n- (\card{\tau} - 1))} - 1\\ \geq & \nu^*_{{\mathcal R}(e,{\mathcal T},\sigma),\tau}(t) - M_{\card{t}}({\mathbf X})^{(n+1 - \card{\tau})} M_{\card{t}}({\mathbf X})^{n(n- (\card{\tau} - 1))} \\ = & \nu^*_{{\mathcal R}(e,{\mathcal T},\sigma),\tau}(t) - M_{\card{t}}({\mathbf X})^{(n+1)(n+1 - \card{\tau})}\\ = & \nu^*_{{\mathcal R}(e,{\mathcal T},\sigma),\tau}(t) -f(\card{\mathcal T},\card{\tau},\card{t}) \end{split}\end{equation} %and hence, it follows that %\begin{multline} %\nu^*_{{\mathcal R}(e,{\mathcal S}_L,\sigma),\tau}(t) \geq \\ %\nu^*_{{\mathcal T},\tau}(t) - M_{\card{t}}^{(n+1 - \card{\tau})} %(M_{\card{t}} + g(n,\card{\tau}-1,\card{t}))^{n(n - \card{\tau} +1)} - %g(n+1,\card{\tau},\card{t})\\ %\geq %\nu^*_{{\mathcal T},\tau}(t) - (M_{\card{t}} + %g(n+1,\card{\tau},\card{t}))^{(n+1 - %\card{\tau})} (M_{\card{t}} + g(n,\card{\tau}-1,\card{t}))^{n(n - %\card{\tau} +1)} \\ %\nu^*_{{\mathcal T},\tau}(t) -(M_{\card{t}} %+g(n+1,\card{\tau},\card{t}) %)^{(n+1)(n+1 - \card{\tau})} % \geq \nu^*_{{\mathcal T},\tau}(t) - f(\card{\mathcal %T},\card{\tau},\card{t}) %\end{multline} Because the hypothesis of the claim and the definition of $\nu_*$ guarantee that $Q_z\geq Q > n$, it follows that $\epsilon^e_{Q_z - (n+1)} \leq \epsilon^{e}_{0}$. Hence there is now a natural mapping $\Psi :{\mathcal R}(e, {\mathcal T}^e) \to {\mathbf C}({\mathfrak Z},{\diam({\mathfrak Z})\epsilon^{e}_{0}})$ defined by $\Psi(\sigma) =\cup\Phi[\Top({\mathcal R}(e,{\mathcal S}_{L},\sigma))]$. Since $\hght^*({\mathcal R}(e, {\mathcal T}^e)) = e$, $\sakne_*({\mathcal R}(e, {\mathcal T}^e)) = \sakne_*({\mathcal T}) = i$ and the height of $\tsuc_{{\mathcal R}(e, {\mathcal T}^e)}(\emptyset) $ is not greater than the height of $\tsuc_{{\mathcal T}}(\emptyset)$ it possible to apply both the induction hypothesis again to ${\mathcal R}(e, {\mathcal T}^e)$ and $\Psi$. This yields a tree ${\mathcal T}^* \prec_f {\mathcal R}(e, {\mathcal T}^e)$ such that $\lambda_{\mathfrak Z}(\bigcup_{\sigma \in {\mathcal T}^*}\Psi(\sigma)) < \diam({\mathfrak Z})\epsilon^{i}_{\Bar{Q} - (n+1)}$ where $\Bar{Q} = \nu_*({\mathcal R}(e,{\mathcal T}^e)$ but, note that $\Bar{Q} \geq Q$ so $\epsilon^{i}_{\Bar{Q} - (n+1)} \leq \epsilon^{i}_{{Q} - (n+1)}$. Finally, let $\Bar{\mathcal T}$ be the tree defined by $$\Bar{\mathcal T} = \{\sigma \in {\mathcal S}_L : R_e(\sigma) \in {\mathcal T}^*\} .$$ The inequality~\eqref{m:ff}, equality~\eqref{5.45} and condition~\eqref{g:1} together guarantee that $\Bar{\mathcal T} \prec_f \mathcal T$ because if $\sigma \in \Bar{\mathcal T}$ and $t \in \tsuc_{\Bar{\mathcal T}}(\sigma)$ then, if $\card{t} < e$ it is possible to use equality~\eqref{5.45} and, otherwise, it is possible to use inequality~\eqref{m:ff}. Hence, all the requirements of the claim are satisfied. \end{proof} \section{The iteration} Given metric spaces ${\mathfrak Y}$ and ${\mathfrak Z}$ and a sequence of pairs ${\mathbf X} = \{(X_n,\nu_n)\}_{n\in\omega}$ such for each $n$, $\nu_n$ is a weak measure $\nu_n:{\mathcal P}(X_n) \to \omega$ satisfying Conditions~\ref{c:1} to \ref{c:4}, let $\Poset_\alpha = \Poset_\alpha({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ be the countable support iteration of length $\alpha$ of the partial order $\Poset({\mathbf X},{\mathfrak Y},{\mathfrak Z})$. This notation will be fixed until the end of Theorem~\ref{t:main} and will be used without further explanation. Although these hypotheses will not be needed until Theorem~\ref{t:main}, it may as well be assumed that ${\mathfrak Y}$ and ${\mathfrak Z}$ are both complete, separable and, in addition, ${\mathfrak Y}$ is quasi-infinite dimensional and ${\mathfrak Z}$ is reasonably geodesic and totally bounded since all the results will be used in this context. \begin{defin}\label{d:fusion} If $\Gamma \in [\omega_2]^{< \aleph_0}$, $\psi : \Gamma \to \omega$ and $p$ and $q$ are conditions in $\Poset_{\omega_2}$ then define $p \leq _{\Gamma,\psi} q$ if and only if $p\restriction \gamma \forces{\Poset_{\gamma}}{p(\gamma) \leq_{\psi(\gamma)} q(\gamma)}$ for each $ \gamma \in \Gamma$. If $\psi$ is a constant function with value $n$ then, as usual, $p \leq _{\Gamma,\psi} q$ will be denoted by $p \leq _{\Gamma,n} q$. \end{defin} \begin{lemma}\label{l:fusion} If $\{p_n\}_{n\in\omega}$ is a sequence of conditions in $\Poset_{\omega_2}$ and $\{\Gamma_n\}_{n\in\omega}$ is a sequence of finite subsets of $\omega_2$ such that $\cup_{n\in\omega}\Gamma_n = \cup_{n\in\omega}\dom{p_n}$ and $p_{n+1} \leq_{\Gamma_n,n} p_n$ then there is some $p_{\omega} \in\Poset_{\omega_2}$ such that $p_\omega \leq p_n$ for all $n\in\omega$. \end{lemma} \begin{proof} This follows from Lemma~\ref{l:AxA} and Lemma~\ref{l:stand} since Axiom~A partial orders satisfy the standard fusion lemma. See Definition~7.1.1 and Lemma~7.1.3 on page 326 of \cite{MR96k:03002}. \end{proof} \begin{defin}Let $\Gamma \in [\omega_2]^{< \aleph_0}$ and let $p \in \Poset_{{\omega_2}}$. Define $\mathcal S$ to be good for $p$ and $\Gamma$ by induction on $\card{\Gamma}$. If $\card{\Gamma} = 0$ then $\mathcal S$ is good for $p$ and $\Gamma$ if and only if ${\mathcal S} = \emptyset$. Now suppose that the notion of good for $p$ and $\Gamma$ has been defined for all $p\in \Poset_{\omega_2}$. Now, if $\Gamma \in [\omega_2]^{n+1}$, $\zeta$ is the first element of $\Gamma$ and $p\in \Poset_{\omega_2}$, then define $\mathcal S$ to be good for $p$ and $\Gamma$ if $\mathcal S$ is a tree of trees, \begin{multline}\label{good:1} p\restriction \zeta\forces{\Poset_{\zeta}}{\tsuc_{\mathcal S}(\emptyset) \subseteq p(\zeta)\AND\tsuc_{\mathcal S}(\emptyset) \text{is a maximal antichain in }p(\zeta)} \end{multline} and, for each $t \in \Top(\tsuc_{\mathcal S}(\emptyset))$, \begin{multline} p\restriction\zeta\forces{\Poset_{\zeta}}{p(\zeta) \angbr{t}\forces{\Poset}{{\mathcal S}\angbr{\SingleSeq{t}} \text{ is good for } \Gamma\setminus \{\zeta\} \AND p/ \Poset_{\zeta + 1}}} \end{multline} \end{defin} \begin{defin} Now, if $\mathcal S$ is good for $p$ and $\Gamma$ and $\sigma \in {\mathcal S}$ then define $p[\Gamma,\sigma]$ by induction on $\card{\sigma}$. Define $p[\Gamma,\emptyset] = p$ so assume that $p[\Gamma,\sigma]$ has been defined whenever $\card{\sigma} = n$ and that $\tau \in {\mathcal T}$ and that $\card(\tau) = n+1$. Let $\zeta$ be the minimal element of $\Gamma$ and let $t \in \tsuc({\mathcal T})$ and $\tau'$ be such that $\tau = \SingleSeq{t}\wedge\tau'$. The definition of ${\mathcal T}$ being good for $p$ and $\Gamma$ implies that \begin{multline} p\restriction\zeta \forces{\Poset_{\zeta}}{p(\zeta)\angbr{t}\forces{\Poset} {{\mathcal S}\angbr{\SingleSeq{t}} \text{ is good for } \Gamma\setminus \{\zeta\} \AND p\restriction (\zeta, \omega_2)}} \end{multline} and, hence, \begin{equation}p\restriction\zeta\forces{\Poset_{\zeta}}{ p(\zeta\angbr{t})\forces{\Poset}{p/\Poset_{\zeta+1}[\Gamma\setminus \{\zeta\},\tau'] \text{ is defined}}} \end{equation} by induction. Therefore, it is possible to define \begin{equation} p[\Gamma,\tau](\beta) = \begin{cases} p(\beta) & \IF \beta \in \zeta\\ p(\beta)\angbr{t}& \IF \beta = \zeta\\ (p/\Poset_{\zeta+1})[\Gamma\setminus\{\zeta\},\tau'](\beta') & \IF \beta = \zeta + 1 + \beta' .\end{cases} \end{equation} \end{defin} \begin{lemma}\label{l:maxac} If $p\in \Poset_\beta$ and $\Gamma\in [\beta]^{<\aleph_0}$ and $\mathcal S$ is a tree of trees which is good for $p$ and $\Gamma$ then for any $q \leq p$ there is some $\sigma \in \Top({\mathcal S})$ and $q' \leq q$ such that $q' \leq p[\Gamma,\sigma]$. \end{lemma} \begin{proof} Standard by induction on $\card{\Gamma}$. \end{proof} \begin{defin}Finally, if $p \in \Poset_{\omega_2}$ and $\Gamma \in [\omega_2]^{< \aleph_0}$ and $\mathcal S$ is good for $p$ and $\Gamma$ then $\mathcal S$ will be said to be $n$-sufficient for $p$ and $\Gamma$ if \begin{equation}\label{e:suff} p[\Gamma,\sigma] \restriction \alpha \forces{\Poset_\alpha}{\tsuc_{\mathcal S}(\sigma) \supseteq \smll_{n + \card{{\mathcal S}\Angbr{\sigma}}} (p(\alpha))} \end{equation} for each $\sigma \in \mathcal S$ and $\alpha \in \Gamma$ such that $\card{(\alpha+1)\cap \Gamma} = \card{\sigma}$. \end{defin} \begin{lemma}\label{l:sufffus} Suppose that $p \in \Poset_{\omega_2}$, $\Gamma \in [\omega_2]^{< \aleph_0}$, $ k\in \omega$ and $\mathcal S$ is a tree of trees which is $k$-sufficient for $q$ and $\Gamma$. If $q \leq p$ is such that $\mathcal S$ is good for $q$ and $\Gamma$ and, furthermore, $q[\Gamma,\sigma] \leq_{\Gamma,k} p[\Gamma,\sigma]$ for each $\sigma \in \Top({\mathcal S})$ then $q \leq_{\Gamma,k} p$. \end{lemma} \begin{proof} Once again proceed by induction on $\card{\Gamma}$. In fact, the full strength of $k$-sufficiency is not needed for this lemma. \end{proof} \begin{lemma}\label{l:x} If $\Gamma \in [\omega_2]^{<\aleph_0}$, $k\in \omega$ and $p\in \Poset_{\alpha}$ is such that $p\forces{}{\Dot{x} \in V}$ then there is some $\Tilde{p} \leq_{\Gamma,k} p$ and a tree of trees $\mathcal S$ which is $k$-sufficient for both $\Tilde{p}$ and $\Gamma$ and ${p}$ and $\Gamma$ such that there is some $x_\sigma$ such that $$\Tilde{p}[\Gamma,\sigma] \forces{}{\Dot{x} = \Check{x}_\sigma}$$ for every $\sigma \in \Top({\mathcal S})$. \end{lemma} \begin{proof} Proceed by induction on $\card{\Gamma}$ noting that if $\Gamma = \emptyset$ then there is nothing to prove. Assuming the lemma has been established for all ${\Gamma}$ of cardinality $ n$ let $\card{\Gamma} = n+1$, let $\mu = \min(\Gamma)$ and $\Gamma' = \Gamma\setminus \{\mu\}$. First find $q_1 \leq p \restriction \mu$ such that $q_1 \forces{\Poset_{\mu}}{\smll_k(p(\mu)) = \Top(\Check{S})}$. Using Lemma~\ref{l:stand} and the induction hypothesis it is possible to find $q_2 \in \Poset_{\mu+1}$ as well as $T\supseteq S$ such that \begin{gather} q_2\restriction \mu \leq q_1\\ q_2\restriction \mu\forces{}{q_2(\mu) \leq p(\mu)}\\ q_2\restriction \mu\forces{\Poset_{\mu}}{\Top(\check{T}) \text{ is a maximal antichain in }q_2(\mu) }\\ \begin{split} q_2\restriction \mu\forces{\Poset_{\mu}}{(\forall t \in \Top(\check{T}))(\exists {\mathcal T}_t)(\exists \{x^t_\sigma\}_{\sigma \in \Top({\mathcal T}_t)}) q_2(\mu)\angbr{t}\forces{\Poset}{\check{{\mathcal T}_t}\\ \text{ is $k$-sufficient} \text{for } \Gamma'\AND q^t} \AND q^t \leq_{\Gamma',k} p\restriction (\mu,\omega_2)\\ \AND (\forall \sigma \in \Top({\mathcal T}_t)) q^t[\Gamma,\sigma]\forces{\Poset_{\omega_2}/\Poset_{\mu + 1}}{\Check{x}_\sigma^t = \Dot{x}}}.\end{split} \end{gather} Now let $M = \sum_{t\in \Top(T)}\card{{\mathcal T}_t}$. Let $q_3 \leq q_2 \restriction \mu$ and $T'$ be such that $$q_3\forces{\Poset_{\mu}}{T\cup \smll_M(q_2(\mu)) \subseteq \check{T'} \subseteq q_2(\mu)} .$$ For each $t \in \Top({T'})$ let $\Bar{t}$ be the unique member of $\Top(T)$ such that $\Bar{t} \subseteq t$. Define $\mathcal T$ by setting $\tsuc_{\mathcal T}(\emptyset) = T'$ and ${\mathcal T}\Angbr{\SingleSeq{t}} = {\mathcal T}_{\Bar{t}}$ for each $t \in \Top(T')$. If $\sigma \in \Top({\mathcal T})$ then there are unique $t\in \Top(T')$ and $\sigma' \in \Top({\mathcal T}_{\Bar{t}})$ such that $\sigma = \SingleSeq{t}\wedge \sigma'$ and so define $x_\sigma = x^{\Bar{t}}_{\sigma'}$. Let $$q(\eta) = \begin{cases} q_3(\eta) & \text{if } \eta \in \mu\\ q_2(\eta) & \text{if } \eta = \mu\\ q^t(\eta) & \text{if } t\in T\AND\eta > \mu \AND (\exists r \in \Dot{G})\\ & r\restriction \mu \leq q_3\AND r\restriction \mu\forces{\Poset_\mu}{r(\mu) \leq q_2(\mu)\angbr{t}} \end{cases}$$ where $\Dot{G}$ is a name for the generic filter on $\Poset_{\omega_2}$. It is easy to check that $q \leq_{\Gamma,k} p$ and that $\mathcal T$ is $k$-sufficient for $q$. Moreover $q[\Gamma,\sigma]\forces{\Poset_{\omega_2}}{\Dot{x} = x_\sigma}$ for each $\sigma \in \Top({\mathcal T})$. \end{proof} \begin{corol}\label{l:suff} Given $p \in \Poset_{\omega_2}$, $\Gamma \in [\omega_2]^{< \aleph_0}$ and $ k\in \omega$ there is $q \leq_{\Gamma, k} p$ and a tree of trees $\mathcal S$ which is $k$-sufficient for $q$ and $\Gamma$. \end{corol} \begin{proof} Simply ignore $\Dot{x}$ in Lemma~\ref{l:x}. \end{proof} \begin{defin} Given $p \in \Poset_{\omega_2}$, $\Gamma \in [\omega_2]^{<\aleph_0}$, a tree of trees ${\mathcal S}$ which is $k$-sufficient for $p$ and $\Gamma$ and $\sigma \in \Top({\mathcal S})$ define $p \sqsubseteq_{{\mathcal S},\sigma,M} q$ if and only if \begin{itemize} \item $ p \leq q$ \item for each $\alpha \in \Gamma$ \begin{multline}p\restriction \alpha\forces{\Poset_\alpha}{(\forall t \in p[\Gamma,\sigma](\alpha)) \text{ either } \nu^{**}_{p[\Gamma,\sigma], \alpha}(t) \geq M\\ \OR \nu^{**}_{p[\Gamma,\sigma],\alpha }(t) \geq \nu^{**}_{q[\Gamma,\sigma],\alpha }(t) - 1}\end{multline} \item if $\tau\in \Top {\mathcal S}$ and $\tau\restriction m \neq \sigma\restriction m$ and $\alpha \in \Gamma$ and $\card{\Gamma\cap (\alpha + 1)} = m$ then $$p[\Gamma,\tau]\forces{\Poset_{\alpha}}{p[\Gamma,\tau](\alpha) = q[\Gamma,\tau](\alpha)} .$$ \end{itemize} \end{defin} \begin{lemma}\label{l:newkey} Let $p \in \Poset_{\omega_2}$, $\Gamma \in [\omega_2]^{<\aleph_0}$ and suppose that ${\mathcal S}$ is $k$-sufficient for $p$ and $\Gamma$. There is some enumeration $\{\sigma_i\}_{i\in L}$ of $\Top({\mathcal S})$ such that if $\{p_i\}_{i\in L}$ is a sequence of conditions satisfying the following conditions: \begin{itemize} \item $p_0 = p$ \item $\mathcal S$ is good for each $p_i$ and $\Gamma$ \item $p_{i+1} \sqsubseteq_{{\mathcal S},\sigma_i,L+k} p_i$ \end{itemize} then $p_L \leq_{\Gamma, k} p$. \end{lemma} \begin{proof} Recall that $\Poset = \Poset({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ and that ${\mathbf X} = \{X_i\}_{i\in \omega}$. For each $i\in \omega$ fix a well-ordering $\leq^*$ of $\bigcup_{i\in\omega}X_i$. Given any tree of trees which is good for some $p\in \Poset_{\omega_2}$ the natural enumeration of $\Top({\mathcal S})$ will be the one which corresponds to the lexicographic ordering inherited from the ordering $\leq^*$. It will be shown that the lemma holds if $\{\sigma_i\}_{i\in L}$ is the natural enumeration of $\Top({\mathcal S})$. To establish this, proceed by induction on $\card{\Gamma}$ noting that if $\Gamma = \emptyset$ then there is nothing to do. If the lemma is proved for $\Gamma$ of size less $n$ let $\card{\Gamma} = n$. Let $\{p_i\}_{i\in L}$ be given such that $p_{i+1} \sqsubseteq_{{\mathcal S},\sigma_i,L+k} p_i$. Let $\mu = \min(\Gamma)$ and $\Gamma' = \Gamma \setminus \{\mu\}$ and $\tsuc_{\mathcal S}(\emptyset) = T$. Notice that for each $t \in \Top(T)$ $$\{i \in L : \sigma _i(0) = t\}$$ forms an interval of integers in $L$. Therefore let $\{t_j\}_{j\in J}$ be the enumeration of $\Top(T)$ induced by $\leq^*$. and let $\{L(j)\}_{j\in J+1}$ be integers such that $L_0 = 0$ and $$\{\sigma_{i}\}_{i=L_j}^{L_{j+1}-1} =\{i \in L : \sigma _i(0) = t_j\}.$$ Fix $j \in J$ and let $\sigma_{j,i}\in{\mathcal S}\Angbr{\SingleSeq{t}}$ be such that $\Angbr{\SingleSeq{t_j}}\wedge \sigma_{j,i} = \sigma_{i+L_j}$. Therefore $\{\sigma_{j,i}\}_{i \in L_j}$ is the natural enumeration of $\{\sigma \in \Top({\mathcal S}\Angbr{\SingleSeq{t_j}}) : \SingleSeq{t_j}\wedge\sigma \in \Top({\mathcal S})\}$. Define $p_{i+ L_j}[\Gamma,\SingleSeq{t_j}] = p_{j,i}$. The definition of $k$-sufficient entails that ${\mathcal S}\Angbr{\SingleSeq{t}}$ is $k$-sufficient for each $p_{j,i}$ and $\Gamma'$. Moreover, it is easy to see that $$p_{j,i+1} \sqsubseteq_{{\mathcal S}\Angbr{\SingleSeq{t_j}},\sigma_{j,i},L+k} p_{j,i}$$ because this simply requires restricting the domain of the universal quantifier in the definition of $\sqsubseteq_{{\mathcal S}\Angbr{\SingleSeq{t_j}},\sigma_{j,i},L+k}$. From the induction hypothesis it therefore follows that $p_{j,L_{j+1}} \leq_{\Gamma',k} p_{j,0}[\Gamma,\SingleSeq{t_j}]$. Since $\mathcal S$ is $k$-sufficient it follows that $$p\restriction \mu\forces{\Poset_\mu}{(\forall s \in p(\mu)\angbr{t_j}) \IF \card{s} \geq \card{t_j}\text{ then }\nu_{p(\mu)}(s) \geq L_{j+1} + k}$$ and hence, the definition of $\sqsubseteq_{{\mathcal S}\Angbr{\SingleSeq{t_j}},\sigma_{j,i},L+k}$ and $L \geq L_{j+1}$ guarantees that $$p\restriction \mu\forces{\Poset_\mu}{(\forall s \in p_{j,L_{j+1}}(\mu)\angbr{t_j}) \IF \card{s} \geq \card{t_j}\text{ then }\nu_{p_{j,L_{j+1}}(\mu)}(s) \geq k}$$ and, moreover, if $j < k$ then $p_{L_k + i}[\Gamma,\SingleSeq{t_j}] = p_{L_{j+1} - 1}[\Gamma,\SingleSeq{t_j}] $ for any $i < L_{k+1}$. It therefore follows that $p_L$ is equivalent to the condition $p^*$ defined by $$p^*(\alpha) = \begin{cases} p_L(\alpha) & \IF \alpha \in \mu\\ \bigcup_{j\in J}p_{j,L_j}(\mu) &\IF \alpha = \mu\\ p_{j,L_j}(\alpha) &\IF \alpha > \mu \AND p_{j,L_j}\restriction \mu +1 \in G . \end{cases}$$ Then, since ${\mathcal S}$ is good for $\Gamma$ and $p_L$, it follows that $$p_L\forces{\Poset_\mu}{ T \text{ is a maximal antichain in }p_L(\mu)}$$ and hence $p^* \leq_{\Gamma,k} p$. \end{proof} \begin{lemma}\label{l:nk2} Let $p \in \Poset_{\omega_2}$, $\Gamma \in [\omega_2]^{<\aleph_0}$ and suppose that ${\mathcal S}$ is good for $p$ and $\Gamma$. Suppose further that $\sigma \in \Top({\mathcal S})$ and that $q \leq p[\Gamma,\sigma]$ satisfies, for each $\alpha\in \Gamma$, that \begin{equation}\label{e:c4-1} {q}\forces{\Poset_\alpha}{(\forall t \in {q}(\alpha)) \text{ either } \nu^{**}_{q(\alpha)}(t) \geq M\OR \nu^{**}_{q(\alpha)}(t) \geq \nu^{**}_{p[\Gamma,\sigma],\alpha}(t) - 1}\end{equation} Then there exists some $q' \in \Poset_{\omega_2}$ such that \begin{itemize} \item $q'\sqsubseteq_{{\mathcal S},\sigma,M} p$ \item $\mathcal S$ is good for $q'$ and $\Gamma$ \item $q'[\Gamma,\sigma] \leq q$.\end{itemize} \end{lemma} \begin{proof} Proceed by induction on $\card{\Gamma}$ noting that if $\Gamma = \emptyset$ there is nothing to do. Let $\mu = \min(\Gamma)$ and $T = \tsuc_{\mathcal S}(\emptyset)$. Let $\sigma(0) = t$ and $\sigma = \SingleSeq{t} \wedge \sigma'$. Using the induction hypothesis it is possible to find a $\Poset_{\mu+1}$ name $q^*$ for a condition in $\Poset_{\omega_2}/\Poset_{\mu+1}$ such that \begin{itemize} \item $1\forces{\Poset_{\mu + 1}}{q^*\sqsubseteq_{{\mathcal S}\Angbr{\SingleSeq{t}},\sigma',M} p[\Gamma,\SingleSeq{t}]/\Poset{\mu + 1}}$ \item $1\forces{\Poset_{\mu + 1}}{\mathcal S\Angbr{\SingleSeq{t}} \text{ is good for }q^* \AND\Gamma' \ }$ \item $1\forces{\Poset_{\mu + 1}}{q^*[\Gamma,\sigma'] \leq q/\Poset_{\mu + 1}}$. \end{itemize} Now define $$q'(\eta) = \begin{cases} q(\eta) & \IF \eta\in \mu \\ (p[\Gamma,\sigma](\eta) \setminus \left(p(\eta)\Angbr{t})\right) \cup {q}(\eta) & \IF \eta = \mu \\ p(\eta) & \IF \mu\in\eta \AND q\restriction (\mu + 1) \notin G_{\mu + 1}\\ q^*(\eta) & \IF \mu\in\eta \AND q\restriction (\mu + 1) \in G_{\mu + 1} \end{cases}$$ where $G_{\mu + 1}$ is a name for the generic filter on $\Poset_{\mu + 1}$. The fact that $q'$ is the desired condition follows from the induction hypothesis. \end{proof} \begin{lemma}\label{new:tech} Suppose that ${\mathcal S}$ is good for $\Gamma \in [\omega_2]^{<\aleph_0}$ and $p\in \Poset_{\omega_2}$ and that $\Bar{\mathcal S} \prec \mathcal S$. If ${p_{\Bar{\mathcal S}}}$ is defined by $${p_{\Bar{\mathcal S}}}(\alpha) = \begin{cases} p(\alpha) & \IF \alpha \notin \Gamma\\ \bigcup\{p(\alpha)\angbr{s} : s\in \tsuc_{\Bar{\mathcal S}}(\tau) \AND & \IF \alpha \in \Gamma\\ \card{\tau} = \card{(\alpha + 1) \cap \Gamma} \AND {p_{\Bar{\mathcal S}}}[\Gamma,\tau]\restriction \alpha \in \Dot{G}\cap \Poset_\alpha\} & \end{cases}$$ where $\Dot{G}$ is a name for the generic filter on $\Poset_{\omega_2}$ then $\Bar{\mathcal S}$ is good for $\Gamma$ and ${p_{\Bar{\mathcal S}}}$ and ${p_{\Bar{\mathcal S}}} \leq p$. \end{lemma} \begin{proof} Once again, proceed by induction on $\card{\Gamma}$. Let $\card{\Gamma} = n$ and assume the lemma proved if $\card{\Gamma} < n$. Let $\mu = \min(\Gamma)$ and $\Gamma' = \Gamma \setminus \{\mu\}$ and $\tsuc_{\Bar{\mathcal S}}(\emptyset) = T$. For each $t \in \Top(T)$ the induction hypothesis gives that $\Bar{\mathcal S}\Angbr{\SingleSeq{t}}$ is good for ${p[\Gamma,\SingleSeq{t}]_{\Bar{\mathcal S}}}$ and $\Gamma'$. Since $\Top(T) \subseteq \Top(\tsuc_{\mathcal S}(\emptyset))$ it follows that $$ {p_{\Bar{\mathcal S}}}\restriction \mu \forces{}{\Top(T)\text{ is a maximal antichain in }{p_{\Bar{\mathcal S}}}(\mu)}$$ and hence $\Bar{\mathcal S}$ is good for $\Gamma$ and ${p_{\Bar{\mathcal S}}}$. The fact that ${p_{\Bar{\mathcal S}}} \leq p$ is obvious. \end{proof} \begin{theor}\label{t:main} If $\mathfrak Z$ is totally geodesic and $p \in \Poset_\alpha = \Poset_\alpha({\mathbf X},{\mathfrak Y}, {\mathfrak Z})$ is such that $$p \forces{\Poset_\alpha}{{\mathcal U}\text{ is a dense } G_\delta\text{ in }{\mathfrak Z}} $$ then there is some $q \leq p$ as well as a rectifiable curve $\gamma$ in the metric space ${\mathfrak Z}$ such that $q\forces{\Poset_\alpha}{\gamma\cap {\mathcal U}\text{ is dense in } \gamma}$. \end{theor} \begin{proof} In this theorem the hypothesis that $\mathfrak Z = (Z,e)$ is reasonably geodesic will play an important role. For later reference, notice that one of the consequences of this hypothesis is the following \begin{multline}\label{h:3} (\forall \text{ open } {\mathcal V}\subseteq Z)(\forall m\in \omega)(\forall \delta > 0)(\exists \{x_i\}_{i\in m})(\exists \theta > 0) (\forall i \in m)\\ B_{\mathfrak Z}(x_i,\theta)\subseteq {\mathcal V} \AND (\forall (\gamma_1,\gamma_2,\ldots,\gamma_m) \in \prod_{i = 1}^m{\mathbf C}(B_{\mathfrak Z}(x_i,\theta/2),\theta))\\ (\exists \text{ a curve } \gamma)\lambda_{\mathfrak Z}(\gamma) < \delta \AND\gamma \supseteq \cup_{i\in m}\gamma_i \end{multline} This follows from Lemma~\ref{l:qq} by choosing $\theta < \delta/3m$ such that $$B_{\mathfrak Z}(x,\theta/2)\subseteq \mathcal V$$ for some point $x$. Then, noticing that any reasonably geodesic space with more than two points can have no isolated points, simply choose $\{x_i\}_{i\in m}$ so that $\diam(\{x_i\}_{i\in m}) < \theta$. A standard fusion argument using Lemma~\ref{l:fusion} shows that in order to prove the theorem, it suffices to prove the following preservation property: \begin{condi}\label{c:pres} Suppose that \begin{itemize} \item $p \forces{\Poset_\alpha}{{\mathcal U} \subseteq Z \text{ is dense open}}$ \item $\delta $ and $\epsilon$ are real numbers greater than 0 \item $\Bar{\gamma} : [a,b] \to Z$ is a rectifiable curve \item $k\in\omega$ \item $\Gamma \in [\omega]^{\leq k}$ \item $A$ is a finite subset of $[a,b]$. \end{itemize} Then there is some $q \leq_{\Gamma,k} p$ as well as a rectifiable curve ${\gamma}: [a,b] \to Z$ and a finite $B\subseteq [a,b]$ such that \begin{itemize} \item $q\forces{\Poset_\alpha}{B \cap {\gamma}^{-1} ({\mathcal U}) \neq\emptyset}$ \item $\lambda_{\mathfrak Z}({\gamma}) < \lambda_{\mathfrak Z}(\Bar{\gamma}) + \delta$ \item $\norm{\gamma - \Bar{\gamma}} < \epsilon$ \item $\Bar{\gamma}(z) = \gamma(z)$ for each $ z\in A$. \end{itemize} \end{condi} To see that establishing Condition~\ref{c:pres} suffices it may, without loss of generality, be assumed that $$p \forces{\Poset_\alpha}{{\mathcal U} = \bigcap_{n\in\omega}{\mathcal U}_n \text{ and each }{\mathcal U}_n \text{ is dense open}}$$ and that $\{(a_n,b_n,e_n)\}_{n\in\omega}$ is an enumeration of $\Rationals\times\Rationals\times \omega$. Suppose that sequences $\{p_n\}_{n\in k}$, $\{\Gamma_n\}_{n\in k}$, $\{A_n\}_{n\in\omega}$ and $\{\gamma_n\}_{n\in k}$ have been constructed. Let $\Gamma_k\supseteq \Gamma_{k-1}$ be chosen by some scheme that will ensure that $\bigcup_{n\in\omega}\Gamma_n =\bigcup_{n\in\omega}\dom(p_n)$ and $\card{\Gamma_k} \leq k$ and let $A = \bigcup_{n\in k}A_n\cap [a_n,b_n]$. Using Condition~\ref{c:pres} it is possible to construct $p_k$, $A_k$ and $\gamma_k$ such that \begin{itemize} \item if $(a_k,b_k)$ is a nonempty interval in $[a,b]$ then $$p_k\forces{\Poset_\alpha}{{\gamma}_k^{-1} ({\mathcal U}_{e_k}) \cap A_k \neq \emptyset}$$ \item $\lambda_{\mathfrak Z}(\gamma_k) < \lambda_{\mathfrak Z}({\gamma_{k-1}}) + 2^{-k}$ \item $\norm{\gamma_k - {\gamma}_{k-1}} < \theta_k$ \item $\gamma_k(z) = \gamma_{k-1}(z)$ for each $z\in A$ \end{itemize} where $\theta_k$ has been chosen so that $\theta_k < 2^{-k}$ and, using compactness and the fact that $\gamma_{k-1}$ is one-to-one, such that if $\card{x - y} > 1/k$ and $\{x,y\} \subseteq (a_k, b_k)$ then $d(\gamma_{k_1}(x),\gamma_{k_1}(y)) > \theta_k$. Hence, $\gamma_\omega = \lim_{n\to\infty}\gamma_n$ is a rectifiable curve and, from the choice of the $\theta_k$, it follows that $\gamma_\omega$ is also one-to-one. Moreover, by Lemma~\ref{l:fusion} it follows that there is some condition $p_\omega$ such that $p_\omega \leq p_n$ for all $n \in \omega$. Hence $p_\omega\forces{\Poset_\alpha}{{\gamma}_k^{-1} ({\mathcal U}_{e_k}) \cap A_k \neq \emptyset}$ for each $k \in \omega$. Since the construction guarantees that $\gamma_\omega\restriction A_k = \gamma_k\restriction A_k $ it follows that $p_\omega\forces{\Poset_\alpha}{{\gamma}_\omega^{-1} ({\mathcal U}_{e_k}) \cap A_k \neq \emptyset}$ for each $k \in \omega$. Hence $p_\omega\forces{\Poset_\alpha}{{\gamma}_\omega^{-1} ({\mathcal U}_n) \cap [a,b] \text{ is dense in }[a,b]}$ for each $n\in \omega$ and so $p_\omega\forces{\Poset_\alpha({\mathbf X},{\mathfrak Z})}{{\gamma}_\omega^{-1} ({\mathcal U}) \cap [a,b] \text{ is dense in }[a,b]}$ . In order to show that it is possible to satisfy Condition~\ref{c:pres} it suffices to show that it is possible, for any reasonable geodesic ${\mathfrak Z} = (Z,d)$, to satisfy the following condition: \begin{condi}\label{c:pres2} Suppose that \begin{itemize} \item $p \forces{\Poset_\alpha}{{\mathcal W} \subseteq Z \text{ is dense open}}$ \item $\delta > 0 $ \item $k\in\omega$ \item $\Gamma \in [\omega]^{\leq k}$ \end{itemize} Then there is some $q \leq_{\Gamma,k} p$ as well as a rectifiable curve ${\gamma}: [0,1] \to Z$ and a finite $B\subseteq [0,1]$ such that \begin{itemize} \item $q\forces{\Poset_\alpha}{{\gamma}^{-1} ({\mathcal W}) \cap B \neq\emptyset}$ \item $\lambda_{\mathfrak Z}({\gamma}) < \delta$ \end{itemize} \end{condi} The reason this suffices it that, given $\Bar{\gamma}:[a,b] \to Z$, $A$, $k$ and $\epsilon > 0$ as in Condition~\ref{c:pres} it is possible to choose a point $w\in [a,b]$ and $\beta' > 0$ such that $\beta' < \min(\epsilon, \delta/6)$ and such that $B_{{\mathfrak Z}}(\Bar{\gamma}(w),\beta)$ is disjoint from $\{\Bar{\gamma}(z) : z\in A\}$. Let $\emptyset \neq [a',b'] \subseteq [a,b]$ and $\beta \leq \beta'$ be such that $B_{{\mathfrak Z}}(\Bar{\gamma}(w),\beta)\cap \Bar{\gamma}[a,b]\subseteq\Bar{\gamma}[a',b'] \subseteq B_{{\mathfrak Z}}(\Bar{\gamma}(w),\beta)$ and define ${\mathfrak Z}' = (Z\cap B_{{\mathfrak Z}}(\Bar{\gamma}(w),\beta),d)$ and note that ${\mathfrak Z}'$ is reasonably geodesic. Then apply Condition~\ref{c:pres2} to find $\gamma:[0,1] \to B_{{\mathfrak Z}}(\Bar{\gamma}(w),\beta) $ and $B$ such that $\lambda_{{\mathfrak Z}'}(\gamma) < \delta/3$ and note that, since $\gamma$ is a curve and not just a set, it follows that $\lambda_{\mathfrak Z}(\gamma) = \lambda_{{\mathfrak Z}'}(\gamma) < \delta/3$. It is then an easy matter to reparametrize $\gamma$ so that it can be assumed to have domain $[a'',b'']$ where $a''$ and $b''$ are chosen so that $a' < a'' < b'' < b'$. Then use the hypothesis that $\mathfrak Z$ is reasonably geodesic to find curves $\gamma_{a}:[a',a''] \to B_{{\mathfrak Z}}(\Bar{\gamma}(w),\beta)$ and $\gamma_{b}[b'',b'] \to B_{{\mathfrak Z}}(\Bar{\gamma}(w),\beta)$ such that \begin{align} \gamma_{a}(a') = &\Bar{\gamma}(a') &\AND& & \gamma_{a}(a'')& = {\gamma}(a'')\\ \gamma_{b}(b'') =& {\gamma}(b'') &\AND& & \gamma_{b}(b') &= \Bar{\gamma}(b')\\ \lambda_{{\mathfrak Z}}(\gamma_{a}) <& \delta/3 &\AND& & \lambda_{{\mathfrak Z}}(\gamma_{b}) &< \delta/3 \end{align} and $\gamma_a \cap \gamma_b = \emptyset$, $\gamma_a \cap \gamma = \{a''\}$, $ \gamma_b \cap \gamma = \{b''\}$, $\gamma_a \cap \Bar{ \gamma} = \{a'\}$, $\gamma_b \cap \Bar{\gamma} = \{b'\}$. Note also that $$\gamma \cap \Bar{\gamma}\restriction ([a,a']\cup [b',b]) = \emptyset$$ because $\gamma[a'',b''] \subseteq B_{{\mathfrak Z}}(\Bar{\gamma}(w),\beta)$. Hence $\Bar{\gamma}\restriction ([a,a']\cup [b',b])\cup \gamma\cup \gamma_{a'}\cup \gamma_{b'}$ is the desired curve and, after reparametrizing, $B$ is the desired finite set. To verify that Condition~\ref{c:pres2} can be satisfied, let $$p \forces{\Poset_\alpha}{{\mathcal W} \subseteq Z \text{ is dense open}}$$ $\delta > 0 $, $\Gamma \in [\omega]^{<\aleph_0}$ and $k\in\omega$. Using Corollary~\ref{l:suff} choose $q' \leq_{\Gamma,k} p$ and a tree of trees $\mathcal S$ which is $k$-sufficient for $q'$ and $\Gamma$. Let $M = \card{\Top({\mathcal S})}$ and let $\{\sigma_i\}_{i\in M}$ enumerate $\Top({\mathcal S})$. Without loss of generality, it may be assumed that $k\geq \card{\Gamma}$ and so $\hght_*({\mathcal S}) > \card{\Gamma}$. Using \ref{h:3} choose $\{x_i\}_{i\in M}$ and and let $\theta >0$ be such that for any choice of $$(\gamma_1,\gamma_2\dots,\gamma_{M}) \in \prod_{i\in M}{\mathbf C}(B_{\mathfrak Z}(x_i,\theta/2),\theta)$$ there exists a single curve $\gamma$ such that $\lambda_{\mathfrak Z}(\gamma) < \delta$ and $\gamma \supseteq \bigcup_{i= 1}^M\gamma_i$. By shrinking $\theta$ if necessary, it may also be assumed that $B_{\mathfrak Z}(x_i,\theta/2)\cap B_{\mathfrak Z}(x_i,\theta/2)= \emptyset$. Now construct, in $M$ steps, conditions $\{q_i\}_{i\in M+1}$ as well as finite sets $\{A_i\}_{i\in M+1}$ such that: \begin{gather} \label{ih:1} q_0 = q'\\ \label{ih:9} {\mathcal S}\text{ is good for } q_i \AND \Gamma\\ \label{ih:2} q_{i+1} \sqsubseteq_{{\mathcal S},\sigma_i,M+k} q_i\\ \label{ih:5} q_{i+1}[\Gamma,\sigma_i] \forces{\Poset_\alpha({\mathbf X},{\mathfrak Z})}{{\mathcal W} \cap A_i \neq \emptyset}\\ \label{ih:6} \lambda_{{\mathfrak Z}_i}(A_i) < \theta\\ \label{ih:7} A_i \subseteq { Z}_i \end{gather} Before proceeding it is worth pausing to see what would be accomplished by this construction. It follows from Lemma~\ref{l:newkey} that $q_M\leq_{\Gamma,k} q' $. Using the properties of the family $\{{B_{{\mathfrak Z}}(x_i,\theta/2)}_{i}\}_{i\in M}$ obtained from hypothesis~\ref{h:3} it is possible to find a curve $\gamma$ such that $\lambda_{\mathfrak Z}(\gamma) < \delta$ and $\cup_{i\in M}A_i \subseteq \gamma$. Let $B = \gamma^{-1}(\cup_{i\in M}A_i)$ and note that $q_M\forces{}{B\cap \gamma^{-1}({\mathcal W})) \neq \emptyset}$ because, by Lemma~\ref{l:maxac}, if $q \leq q_M$ then there is some $q' \leq q$ and $i\in M$ such that $q' \leq q_M[\Gamma,\sigma_i] \leq q_{i+1}[\Gamma,\sigma_i]$ and $ q_{i+1}[\Gamma,\sigma_i]\forces{}{A_i\cap {\mathcal W} \neq \emptyset}$. Hence, all that has to be shown is how to construct $q_{i+1}$ given $q_i$. In order to do this, it suffices to find $\Tilde{q} \leq q_i[\Gamma,\sigma_i]$ and $A_i\subseteq Z_i$ such that for each $\alpha \in \Gamma$ \begin{equation}\label{e:c4} \Tilde{q}\forces{\Poset_\alpha}{(\forall t \in \Tilde{q}(\alpha)) \text{ either } \nu^{**}_{\Tilde{q}}(t) \geq M +k\OR \nu^{**}_{\Tilde{q}}(t) \geq \nu^{**}_{q_i[\Gamma,\sigma_i]}(t) - 1}\end{equation} and $\Tilde{ q} \forces{}{{\mathcal W} \cap A_i \neq \emptyset}$ and condition~\eqref{ih:6} is satisfied since, once this has been accomplished, all that remains to be done is to choose $q_{i+1}$ using Lemma~\ref{l:nk2} such that that $$q_{i+1}\sqsubseteq_{{\mathcal S},\sigma_i,M+k} q_i$$ $\mathcal S$ is good for $q_{i+1}$ and $\Gamma$ and $q_{i+1}[\Gamma,\sigma_i] \leq \Tilde{q}$. Note that $1\forces{\Poset_{\omega_2}}{\Dot{x} \in V\cap {\mathcal W} \cap Z_i}$ for some name $\Dot{x}$. Then use Lemma~\ref{l:x} to find $\Tilde{\Tilde{q}} \leq_{\Gamma,M+k} q_i[\Gamma,\sigma_i]$ and a tree of trees $\Tilde{\mathcal S}$ which is $(M+k)$-sufficient for both $\Tilde{\Tilde{q}}$ and $\Gamma$ and $q_{i}[\Gamma,\sigma_i]$ and $\Gamma$ such that $$\Tilde{\Tilde{q}}[\Gamma,\tau] \forces{\Poset_{\omega_2}}{\Dot{x} = \Check{x}_\tau}$$ for each $\tau \in \Top(\Tilde{\mathcal S})$. Note that \begin{equation}\label{e:rr} \sakne_*(\Tilde{\mathcal S}) \geq \hght_*({\mathcal S}) \geq \card{\Gamma} = \hght({\Tilde{\mathcal S}}) \end{equation} and, moreover, since $\mathcal S$ is $k$-sufficient, it follows that if $\tau \in \Tilde{\mathcal S}$ and $t \not\subseteq\sakne(\tsuc_{\Tilde{\mathcal S}}(\tau))$ then $\nu^{**}_{\Tilde{\mathcal S},\tau}(t) \geq k \geq \card{\Gamma} = \hght({\Tilde{\mathcal S}})$ and hence $\nu_*(\Tilde{{\mathcal S}}) > \hght(\Tilde{{\mathcal S}})$. Let the function $\Phi$ be defined on $\Top({\Tilde{\mathcal S}})$ by $\Phi(\tau) = x_\tau$. It is therefore possible to apply Lemma~\ref{l:m} to $\Tilde{\mathcal S}$, $\Phi$ and the metric space ${\mathfrak Z}_i = (Z_i,d)$ to find a nonempty tree of trees $\Bar{\mathcal S} \prec_f \Tilde{\mathcal S}$ such that \begin{equation}\label{ih:6f} \lambda_{{\mathfrak Z}_i}(\{x_\tau\}_{\tau\in\Top({\Bar{\mathcal S}})})) < \diam({\mathfrak Z}_i) \leq \theta . \end{equation} Then define $\Tilde{q}$ by $$\Tilde{q}(\alpha) = \begin{cases} \Tilde{\Tilde{q}}(\alpha) & \IF \alpha \notin \Gamma\\ \bigcup\{\Tilde{\Tilde{q}}(\alpha)\angbr{s} : s\in \tsuc_{\Bar{\mathcal S}}(\tau) \AND & \IF \alpha \in \Gamma\\ \card{\tau} = \card{(\alpha + 1) \cap \Gamma} \AND \Tilde{q}[\Gamma,\tau]\restriction \alpha \in \Dot{G}\cap \Poset_\alpha\} & \end{cases}$$ where $\Dot{G}$ is a name for the generic filter on $\Poset_{\omega_2}$. It will be shown that $\Tilde{q}$ and $A_i = \{x_\tau\}_{\tau\in\Top({\Bar{\mathcal S}})}$ are as desired. First note that $\Tilde{q} \leq \Tilde{\Tilde{q}} \leq q_i[\Gamma, \sigma_i]$ by Lemma~\ref{new:tech}. Condition~\eqref{ih:6} follows immediately from \eqref{ih:6f}. To see that $\Tilde{q}\forces{\Poset_\alpha}{{\mathcal W}\cap A_i \neq \emptyset}$ note that Lemma~\ref{new:tech} implies that $\Bar{\mathcal S}$ is good for $\Tilde{q}$ and $\Gamma$. Hence, by Lemma~\ref{l:maxac}, if $r \leq \Tilde{q}$ then there is $ \sigma \in \Top(\Bar{\mathcal S})$ and $r' \leq r$ such that $r' \leq \Tilde{q}[\Gamma,\sigma] \leq \Tilde{\Tilde{q}}[\Gamma,\sigma]$ and $\Tilde{\Tilde{q}}[\Gamma,\sigma]\forces{}{\Dot{x} = \Check{x}_\sigma \in A_i\cap {\mathcal W}}$. Therefore all that has to be checked is that \eqref{e:c4} holds for all $\alpha \in \Gamma$. This follows from the fact that $\Bar{\mathcal S} \prec_f \Tilde{\mathcal S}$. To see that this is so notice that, since $\Bar{\mathcal S}$ is good for $\Tilde{q}$ and $\Gamma$, by Lemma~\ref{l:maxac}, it suffices to show that for any $\alpha\in \Gamma$ and any $\tau \in \Bar{\mathcal S}$ such that $\card{\tau} = \card{\Gamma \cap (\alpha + 1)}$ $$ \Tilde{q}[\Gamma,\tau]\restriction \alpha\forces{\Poset_\alpha}{(\forall t \in \Tilde{q}(\alpha)) \text{ either } \nu^{**}_{\Tilde{q}}(t) \geq M +k\OR \nu^{**}_{\Tilde{q}}(t) \geq \nu^{**}_{q_i[\Gamma,\sigma_i]}(t) - 1} .$$ Let such a $\tau$ and $\alpha \in \Gamma$ be fixed. Since $\Tilde{\mathcal S}$ is $(M+k)$-sufficient for $\Tilde{\Tilde{q}}$ and $\Gamma$ and, furthermore, $\Bar{\mathcal S} \prec \Tilde{\mathcal S}$ it follows that $$\Tilde{q}[\Gamma,\tau]\restriction \alpha\forces{\Poset_\alpha}{(\forall t \in \Tilde{q} \setminus \nonmax(\tsuc_{\Tilde{\mathcal S}}(\tau) )) \nu_{\Tilde{q}(\alpha)}^{**}(t) = \nu_{\Tilde{\Tilde{q}}(\alpha)}^{**}(t) \geq M+k} .$$ Therefore it suffices to show that $$ \Tilde{q}[\Gamma,\tau]\restriction \alpha\forces{\Poset_\alpha}{(\forall t \in \nonmax(\tsuc_{\Tilde{\mathcal S}}(\tau))) \nu^{**}_{\Tilde{q}}(t) \geq \nu^{**}_{q_i[\Gamma,\sigma_i]}(t) - 1}$$ and, since $\Tilde{\mathcal S}$ is good for $q_i[\Gamma,\sigma_i]$ and $\Gamma$, this is the same as showing that $$\nu^{**}_{\Bar{\mathcal S}, \tau}(t) \geq \nu^{**}_{\Tilde{\mathcal S}, \tau}(t) -1$$ for each $ t \in \nonmax(\tsuc_{\Bar{\mathcal S}}(\tau))$. For a fixed $ t \in \nonmax(\tsuc_{\Bar{\mathcal S}}(\tau))$ the definition of $\Bar{\mathcal S}\prec_f \Tilde{\mathcal S}$ guarantees that \begin{equation}\begin{split} \nu^{**}_{\Bar{\mathcal S}, \tau}(t) = & \frac{\nu^{*}_{\Bar{\mathcal S}, \tau}(t)}{(M_{\card{t}}({\mathbf X}))^{{\card{t}}^2}} \geq \frac{\nu^{*}_{\Tilde{\mathcal S}, \tau}(t) - f(\hght({\Bar{\mathcal S}}),\card{\tau},\card{t})} {(M_{\card{t}}({\mathbf X}))^{{\card{t}}^2}}\\ = & \frac{\nu^{*}_{\Tilde{\mathcal S}, \tau}(t) - M_{\card{t}}({\mathbf X})^{\hght({\Bar{\mathcal S}}) (\hght({\Bar{\mathcal S}}) - \card{\tau})}} {(M_{\card{t}}({\mathbf X}))^{{\card{t}}^2}} \end{split} \end{equation} Recalling that $\card{t} \geq \hght({\mathcal S}) = \hght(\Bar{\mathcal S})$ from \eqref{e:rr} yields that $$\nu^{**}_{\Bar{\mathcal S}, \tau}(t) \geq \frac{\nu^{*}_{\Tilde{\mathcal S}, \tau}(t)} {(M_{\card{t}}({\mathbf X}))^{{\card{t}}^2}} - 1 = \nu^{**}_{\Tilde{\mathcal S}, \tau}(t) - 1 $$ which establishes \eqref{e:c4}. \end{proof} \begin{corol}\label{c:main} It is consistent with set theory that there are two complete, separable metric spaces $\mathfrak Y$ and $\mathfrak Z$ such that the union of any $\aleph_1$ rectifiable curves in $\mathfrak Y$ is meagre yet there are $\aleph_1$ rectifiable curves in $\mathfrak Z$ is whose union is not meagre. \end{corol} \begin{proof} Let ${\mathfrak Y} = \ell^2$ and ${\mathfrak Z} = [0,1]^3$. Let $\mathbf X$ be the sequence of weak measures constructed in Corollary~\ref{cor:norms}. From Lemma~\ref{cor:norms} it follows that if $\Poset_{\omega_2}$ is the countable support iteration of $\Poset({\mathbf X}, {\mathfrak Y}, {\mathfrak Z})$ to $\omega_2$ then, if $G$ is $\Poset_{\omega_2}$ generic over $V$ then the union of any $\aleph_1$ rectifiable curves in $\mathfrak Y$ is meagre. On the other hand, it follows from Theorem~\ref{t:main} that the union of the ground model rectifiable curves in $\mathfrak Z$ is not meagre. \end{proof} \section{Measure and rectifiable curves} This section contains a result due to S. Shelah\footnote{The result was obtained by S.~Shelah in October of 1994 and is included with his permission.} which is somewhat, but not quite, the measure analogue of Theorem~\ref{t:main}. The measure analogue would assert the consistency of all sets of reals of size $\aleph_1$ having measure zero yet there existing a family of rectifiable curves whose union is not of measure zero. The following theorem produces, not a family of of rectifiable curves, but simply points which can not be covered by countably many rectifiable curves. \begin{theor}[S. Shelah]\label{t:shel} It is consistent with set theory that\/ $\non(Null) > \aleph_1$ yet\/ $\non({\mathcal R}(\Reals^2)) = \aleph_1$. \end{theor} \begin{proof} For certain pairs of functions $(g,g^*) \in \fomom\times\fomom$ a partial order $\Sacks_{g,g^*} $ is defined on pages 348-349 of \cite{MR96k:03002}. It is shown that $\Sacks_{g,g^*} $ is proper and $(g^*,g)$-bounding\footnote{The careful reader will notice that, in fact, Lemma~7.3.23 claims that $\Sacks_{g,g^*} $ is $(g,g^*)$-bounding but this is clearly a misprint.} and hence, from Lemma~7.2.19, which establishes that the countable support iteration of proper $(f,h)$-bounding partial orders is $(f,h)$-bounding, it follows that the countable support iteration of length $\omega_2$ of $\Sacks_{g,g^*} $ is also $(g^*,g) $-bounding. Furthermore, it follows directly from Theorem~7.3.21 that if $G$ is generic for this iteration over a model $V$ then $V[G]$ is a model of $\non(Null) > \aleph_1$. Hence it suffices to show that if $g$ and $g^*$ are chosen appropriately then $V[G]$ will also be a model of $\non({\mathcal R}(\Reals^2)) = \aleph_1$. In particular, it will be shown that if $\{\gamma_i\}_{i\in\omega}$ is any family of rectifiable curves in $\Reals^2$ in the generic extension then there is a point $x\in V\cap \Reals^2$ such that $x \notin \bigcup_{i\in\omega}\gamma_i$. Let $f$, $g$ and $g^*$ be chosen so that $g$ and $g^*$ satisfy (1) and (2) on page 348 of \cite{MR96k:03002} and, in addition, so that $$g^*(n) = \card{[f(n)^2]^{<5f(n)}}$$ $f(0) = 5$, $g(0) = 1$ and $$\frac{f(n+1)}{f(n)^2} > 5g(n+1)$$ as well. To see that this works, let $\{\gamma_i\}_{i\in\omega}$ be a family of rectifiable curves in the generic extension and, without loss of generality, suppose that none of the curves has length greater than 1. Attention will be focussed on the unit square $[0,1]^2$. Before continuing, observe that if the unit square is covered by $m^2$ congruent closed squares with sides of length $1/m$ and $\gamma$ is any rectifiable curve of length less than $1$ then there are at most ${5m}$ squares which the curve intersects. The way to see this is to choose a point on the curve from each square which the curve intersects. Let these points be $\{\gamma(x_i)\}_{i=0}^k$ where $x_i \leq x_{i+1}$. Since no 5 of the squares can have a common point of intersection, it follows that for any $i\in k-5$ there must be some $\{i,j\} \in [5]^2$ such that the ${\lVert {\gamma(x_i) - \gamma(x_j)} \rVert} \geq 1/m$. Since $\sum_{i\in k}{\lVert {\gamma(x_i) - \gamma(x_{i+1})} \rVert} < 1$ it follows that $k < 5m$. Now let ${\mathcal L}(n)$ be the natural family of size $f(n)^2$ consisting of congruent closed squares of length $1/f(n)$ whose union covers the unit square. For each $n \in \omega$ let $S_n$ be the set of squares in ${\mathcal L}(n)$ which the curve $\gamma_n$ intersects. Since $S_n \in [{\mathcal L}(n)]^{<5f(n)}$ it follows that there is $H: \omega \to [[{\mathcal L}(n)]^{<5f(n)}]^{g(n)}$ in $V$ such that $S_n \in H(n)$ for all $n\in\omega$. Let $S^*_n = \cup H(n)$. It follows that $\card{S_0^*} < 5\cdot 5$ and so there is some square $Q_0 \in {\mathcal L}(n)$ such that $Q_0\notin S^*_0$. Suppose that squares $Q_0 \supseteq Q_1 \supseteq \ldots \supseteq Q_n$ have been chosen inductively so that $Q_i\notin S^*_i$. The number of squares in ${\mathcal L}(n+1)$ which are subsets of $Q_n$ is $\frac{f(n+1)^2}{f(n)^2}$ and $\card{S^*_{n+1}} \leq 5f(n+1)g(n+1)$. Since $$\frac{f(n+1)}{f(n)^2} > 5g(n+1)$$ it follows that it is possible to find $Q_{n+1}\subseteq Q_n$ which does not belong to $S^*_{n+1}$. Letting $x \in \bigcap_{n\in\omega}Q_n$ yields that $x\notin \bigcup_{n\in\omega}S^*_n$ and hence $x\notin \bigcup_{n\in\omega}\gamma_n$ \end{proof} \section{Remarks and open questions} The proof of Theorem~\ref{t:main} relied heavily on the hypothesis that $\mathfrak Z$ is reasonably geodesic and this, in turn, implies some form of dimensionality greater than 2. This raises the obvious question of whether this is only a technical problem or if there is some intrinsic geometric reason for the hypothesis. Answering the following questions may shed some light on this. \begin{quest} Is it consistent that $\non(Meagre) > \aleph_1$ yet there is a family of $\aleph_1$ rectifiable curves in $\Reals^2$ whose union is not meagre? \end{quest} \begin{quest} Is it consistent that that the union of any $\aleph_1$ rectifiable curves in $\ell^2$ is meagre yet there is a family of $\aleph_1$ rectifiable curves in $\Reals^2$ whose union is not meagre? \end{quest} \begin{quest}\label{3} Are there integers $n$ and $k$ such that it consistent that the union of any $\aleph_1$ rectifiable curves in $\Reals^n$ is meagre yet there is a family of $\aleph_1$ rectifiable curves in $\Reals^k$ whose union is not meagre? \end{quest} At this point it should be pointed out that there is an obvious strategy for proving some monotonicity results by taking a family of rectifiable curves in $\Reals^{n+1}$ whose unions is not meagre and projecting them down to $\Reals^n$. The curves will still be rectifiable and their union will not be meagre. However the projected curves may not be one-to-one. Indeed the following observation shows that they can not even be covered by a small number of one-to-one curves. \begin{propo}\label{prop:rem} There is a one-to-one, rectifiable curve in $\Reals^3$ which, when projected to the plane along any axis cannot be covered by less than $\cov(Meaagre)$ one-to-one curves in $\Reals^2$. \end{propo} \begin{proof} The curve $\gamma$ is constructed as the uniform limit of one-to-one, rectifiable curves --- in other words, $\gamma - \lim_{n\to\infty}\gamma_n$. Let $\{(a_n,b_n)\}_{n\in\omega}$ enumerate all rational intervals in $[0,1]$ infinitely often. The curves $\gamma_n$ are constructed so that $\Gamma_0$ is arbitrary and, if $\gamma_n$ is given then $\gamma_{n+1}$ is constructed by tying a knot in $\Gamma_n$ on the copy of the interval $(a_n,b_n)$ of diameter less than $2^{-2n-1}$ around a core of diameter $2^{-2n-2}$. It follows that $\Gamma$ has the property that any projection $\pi(\gamma)$ of $\gamma$ onto the plane has a dense set of points $x\in \pi(\gamma)$ such that there there are intervals $I_x\subseteq \pi(\gamma)$ and $J_x\subseteq \pi(\gamma)$ such that $I_x \cap J_x = \{x\}$. Now note that if $\eta$ is any one-to-one curve then $\eta \cap \pi(\gamma)$ is nowhere in $ \pi(\gamma)$ because if ${\mathcal U}$ is open then let $x\in \pi(\gamma)\cap \mathcal U$ be such that $I_x$ and $J_x$ are defined. Since $\eta$ is one-to-one it follows that at most one of $I_x\cap \mathcal U$ and $J_x\cap \mathcal U$ are contained in $\eta$ --- so say that $I_x\cap {\mathcal U}\not\subseteq \eta$. Since $\eta $ is closed, there must be an open set ${\mathcal V}\subseteq \mathcal U$ such that ${\mathcal V}\cap I_x\neq \emptyset$ and ${\mathcal V}\cap \eta = \emptyset$. The result now follows.\end{proof} In light of Proposition~\ref{prop:rem} it is reasonable to ask the following variant of Question~\ref{3}. \begin{quest}\label{3.1} Are there integers $n$ and $k$ such that it consistent that $\cov({\mathcal R}(\Reals^n)) > \aleph_1$ yet $\cov({\mathcal R}(\Reals^k)) = \aleph_1$? \end{quest} It would be interesting to see if Theorem~\ref{t:main} can be improved so that, in the case ${\mathfrak Z} = \Reals^n$, the rectifiable curve mentioned in the conclusion is nicely embedded. There are various interpretations what nicely embedded might mean here. The best would be a linear embedding of the unit interval since this would solve Komjath's question mentioned in the introduction. A less ambitious alternative is to see if it is possible to avoid the difficulties of Proposition~\ref{prop:rem} by answering the following. \begin{quest} Is it consistent that $\non(Meagre) > \aleph_1$ yet there is a family of $\aleph_1$ rectifiable curves in $\Reals^n$ whose union is not meagre and all of which are homotopic to the unit interval? \end{quest} In this same spirit, one can ask for smoothness properties. \begin{quest} Is it consistent that $\non(Meagre) > \aleph_1$ yet there is a family of $\aleph_1$ differentiable curves in $\Reals^n$ whose union is not meagre.? \end{quest} Of course, one can ask for higher orders of smoothness as well. Nothing is known in any of these cases either. The proof of Theorem~\ref{t:main} runs into the usual difficulties when try to obtain larger values of the continuum. The following question points this out. \begin{quest} Is it consistent that there are complete separable metric spaces ${\mathfrak Z}_0$, ${\mathfrak Z}_1$ and ${\mathfrak Z}_2$ such that for each $i\in 3$ there are $\aleph_{i+1}$ rectifiable curves in ${\mathfrak Z}_i$ whose union is not meagre yet the union of fewer such curves is meagre?\end{quest} Note that there is a certain monotonicity for $\non({\mathcal R}(\Reals^n))$ in that $\non({\mathcal R}(\Reals^n)) \leq \non({\mathcal R}(\Reals^{n+1}))$ is immediate. This begs the following question. \begin{quest} Is it consistent for some integer $n$ that $\non({\mathcal R}(\Reals^n)) \neq \non({\mathcal R}(\Reals^{n+1}))$? \end{quest} Finally, it should be mentioned that all of the questions considered here have their measure analogues. \begin{quest} Is there some integer $n$ such that it consistent that $\non(Null) > \aleph_1$ yet the union of any $\aleph_1$ rectifiable curves in $\Reals^n$ has measure zero? \end{quest} Indeed, rectifiable curves are special examples of sets of Hausdorff dimension~1 in $\Reals^2$. One can also consider the $\sigma$-ideal generated by Borel sets of Hausdorff dimension~$p$ in $\Reals^n$ and ask similar questions in this context. 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