% LaTeX file % Linear subspace of $\reals^\lambda$ without dense % totally disconnected subset \documentstyle[12pt]{article} \pagestyle{myheadings} % Use blackboard bold for the normal real subsets \newfont{\amssymBig}{msbm10 scaled\magstep4} \newfont{\amssymten}{msbm10 scaled\magstep1} % real numbers \newcommand{\bigreals}{\mbox{{\amssymBig R}}} \newcommand{\reals}{\mbox{{\amssymten R}}} %closure of a set \newcommand{\closure}[1]{\mbox{$\mbox{\rm cl}\left(#1\right)$}} %domain of a function \newcommand{\dom}[1]{\mbox{$\mbox{\rm dom}\left(#1\right)$}} %Macro to include theorem labels in the margins. \newcommand{\marginlabels}[1]{\def\margintags{#1}} \marginlabels{n} \def\nolabels{n} \newcommand{\Label}[1] {\if\margintags\nolabels \label{#1} \else \label{#1}\marginpar{{\tiny #1}} \fi} \newtheorem{Theorem}{Theorem} \newtheorem{Lemma}{Lemma} \newtheorem{Problem}{Problem} \newcommand{\theorem}[2]{\begin{Theorem}\Label{#1}$\!\!\!${\bf.} #2\end{Theorem}} \newcommand{\lemma}[2]{\begin{Lemma}\Label{#1}$\!\!\!${\bf.} #2\end{Lemma}} \newcommand{\problem}[2]{\begin{Problem}\Label{#1}$\!\!\!${\bf.} #2\end{Problem}} \newcommand{\e}{\mbox{$\varepsilon$}} \newcommand{\E}{\mbox{$\cal E$}} \newcommand{\G}{\mbox{$\cal G$}} \newcommand{\F}{\mbox{$\cal F$}} \newcommand{\D}{\mbox{$\cal D$}} \newcommand{\B}{\mbox{$\cal B$}} \markright{Linear subspace of $\reals^\lambda$} \title{Linear subspace of $\bigreals^\lambda$ without dense totally disconnected subset} \author{} \date{} \begin{document} \maketitle \begin{flushleft} {\small \noindent Krzysztof Ciesielski, Department of Mathematics, West Virginia University, Morgantown, WV 26506 } \end{flushleft} \begin{abstract} In \cite{KC:Lspace} the author showed that if there is a cardinal $\kappa$ such that $2^\kappa=\kappa^+$ then there exists a completely regular space without dense 0-dimensional subspace. This was a solution of a problem of Arhangiel'skii. Recently Arhangiel'skii asked the author (private communication) whether we can generalize this result by constructing a completely regular space without dense totally disconnected subspace, and whether such a space can have a structure of a linear space. The purpose of this paper is to show that indeed such a space can be constructed under the additional assumption that there exists a cardinal $\kappa$ such that $2^\kappa=\kappa^+$ and $2^{\kappa^+}=\kappa^{++}$. \end{abstract} \section{Notation and lemmas} The topological terminology used in this paper is standard and follows \cite{Engelking:GenTop} with the exception that we will use term {\em totally disconnected} for the topological spaces which do not contain connected subsets with more than one point. (In \cite{Kunen,Engelking:GenTop} such spaces are called hereditary disconnected.) The set theoretical terminology and notation used in this paper is standard and follows \cite{Kunen}. In particular, ordinals are identified with their sets of predecessors and cardinals with the initial ordinals. Symbol $\omega$ denotes the first infinite ordinal as well as first infinite cardinal. ${\cal P}(X)$ will denote the power set of $X$ and $|X|$ the cardinality of $X$. If $\kappa$ is a cardinal number than $\kappa^+$ denotes the cardinal successor of $\kappa$ and $2^\kappa=|{\cal P}(\kappa)|$. $[X]^{\leq\kappa}$ will denote $\{Y\subset X: |Y|\leq\kappa\}$. Similarly we define $[X]^{<\kappa}$. The functions will be identified with their graphs. The class of all functions $f: X\to Y$ from a set $X$ to a set $Y$ is denoted by $Y^X$. The space $\reals^\lambda$ will be always considered as a linear topological space over \reals\ with the standard operations and the product topology. For a cardinal number $\kappa$ a topological space $X$ is said to be $\kappa$-Lindel\"of provided every open cover of $X$ has a subcover of cardinality $\leq\kappa$. We will also need the following notation. Let $\B_0$ denote a fixed countable base for \reals, let $\B(A)=\{\e\colon D\to\B_0\colon D\in[A]^{<\omega}\}$ and for $\e\in\B(A)$ let $[\e]=\{f\in\reals^A\colon (\forall a\in\dom{\e})(f(a)\in \e(a)\}$ be a basic open set for $\reals^A$. For a cardinal number $\kappa$ let $H_\kappa(A)$ denote the set of all functions $g\colon D\to\reals$ such that $D\in[A]^{\leq\kappa}$, and let $\F_\kappa(A)$ the class of all functions $f\colon H_\kappa(A)\to H_\kappa(A)$ such that $f(g)\supset g$ for all $g\in H_\kappa(A)$. For $g\in H_\kappa(A)$ let $[g]=\{f\in\reals^A\colon g\subset f\}$. Moreover, for an ordinal number $\xi$, $\kappa^+\leq\xi\leq\kappa^{++}$, let $\D_\kappa(\xi)$ be the family of all sets of the form \[ D_f=\left( \reals^\zeta\setminus\bigcup_{g\in H_\kappa(\zeta)}[f(g)]\right) \times\reals^{\kappa^{++}\setminus\zeta} \] where $\kappa^+\leq\zeta\leq\xi$ and $f\in\F_\kappa(\zeta)$. Finally, let us define \[ \D_\kappa=\bigcup_{\kappa^+<\xi<\kappa^{++}}\D_\kappa(\xi). \] In what follows we will use the following well known fact. For completeness sake, we will sketch here its proof. \lemma{lemknown}{For any disconnected set $S\subset\reals^X$ there exists $\delta\in H_\omega(X)$ such that $[\delta]\cap S=\emptyset$. } Proof. If $S$ is not dense in $\reals^X$ then we can easily find appropriate $\delta$. Assume that $S$ is dense in $\reals^X$ and let $U,V\subset\reals^X$ be non-empty open sets separating such that $S\subset U\cap V$. Let $\{[\e_n]: n<\omega\}$ be a maximal family of non-empty disjoint basic open sets $[\e]$ such that either $[\e]\subset U$ or $[\e]\subset V$. It is countable since $\reals^X$ has the Suslin property. Now, if $D=\bigcup_{n<\omega}\dom{\e_n}$, $U_0=U\cap\bigcup_{n<\omega}[\e_n]$, $V_0=V\cap\bigcup_{n<\omega}[\e_n]$ and $U_1$ and $V_1$ are projections of $U_0$ and $V_0$ into $\reals^D$, then the sets $U_1$ and $V_1$ are non-empty, open, disjoint and, by connectedness of $\reals^D$, there is $\delta\in\closure{U_1}\cap\closure{V_1}$. Then, $[\delta]\subset\closure{U_0}\cap\closure{V_0} \subset\closure{U}\cap\closure{V}$ and indeed $[\delta]\cap S\subset[\delta]\cap(U\cup V)=\emptyset$. \bigskip Now we are ready for our main lemma. \lemma{lem}{Let us assume that $2^\kappa=\kappa^+$ and $2^{\kappa^+}=\kappa^{++}$. Then, {\rm (1)} $|\D_\kappa|=\kappa^{++}$; {\rm (2)} for every totally disconnected set $S\subset\reals^{\kappa^{++}}$ there is $D\in\D_\kappa$ such that $S\subset D$; {\rm (3)} if $\kappa^+<\xi<\kappa^{++}$, $\D_0\subset \D_\kappa(\xi)$, and $|\D_0|\leq\kappa^+$ then there is a $g\in H_{\kappa^+}(\kappa^{++})$ such that $[g]\cap\bigcup\D_0=\emptyset$; {\rm (4)} $r(h+D)\in\D_\kappa$ for any $h\in\reals^{\kappa^{++}}$, $r\in\reals\setminus\{0\}$ and $D\in\D_\kappa$. } Proof. (1) To see it notice that for $\kappa^+<\xi<\kappa^{++}$ the cardinality $|H_\kappa(\xi)|=|\xi|^\kappa |\reals|^\kappa=\kappa^\kappa 2^\kappa=\kappa^+$ and so, $|\D_\kappa(\xi)|= |\bigcup_{\kappa^+\leq\zeta\leq\xi}\F_\kappa(\zeta)|=\kappa^+$. Hence, $|\D_\kappa|\leq\kappa^{++} |\D_\kappa(\xi)|=\kappa^{++}$. (2) Since $S$ is totally disconnected, for every $g\in H_\kappa(\kappa^{++})$ the set $S\cap [g]$ must be disconnected or have at most one point. Since $[g]$ is connected and homeomorphic to $\reals^{\kappa^{++}}$, by Lemma \ref{lemknown} we can find a countable extension $f(g)\in H_\kappa(\kappa^{++})$ of $g$ such that $S\cap [f(g)]=\emptyset$. Thus, we defined $f\in \F_\kappa(\kappa^{++})$ such that $S\subset\reals^{\kappa^{++}}\setminus \bigcup\{[f(g)]\colon g\in H_\kappa(\kappa^{++})\}$. Now, define $\{\xi_\eta: \eta<\kappa^+\}$ by induction on $\eta<\kappa^+$ by putting $\xi_0=\kappa^+$, $\xi_\lambda=\bigcup_{\eta<\lambda}\xi_\eta$ for limit ordinals $\lambda<\kappa^+$ and choose $\xi_{\eta+1}<\kappa^{++}$ such that $f(H_\kappa(\xi_\eta))\subset H_\kappa(\xi_{\eta+1})$. This can be done, since $|H_\kappa(\xi_\eta)|\leq\kappa^+$. Define $\xi=\bigcup_{\eta<\kappa^+}\xi_\eta<\kappa^{++}$. Then, $f|_{H_\kappa(\xi)}\in\F_\kappa(\xi)$ since for every $g\in H_\kappa(\xi)$ there is $\eta<\kappa^+$ such that $g\in H_\kappa(\xi_\eta)$. So, \[ S\subset\reals^{\kappa^{++}}\setminus \bigcup\{[f(g)]\colon g\in H_\kappa(\kappa^{++})\} \subset \reals^{\kappa^{++}}\setminus \bigcup\{[f(g)]\colon g\in H_\kappa(\xi)\}\in\D_\kappa. \] (3) Let $\{D_{f_\eta}\colon\eta<\kappa^+\}$ be an enumeration of $\D_0$ where $f_\eta\in\F_\kappa(\zeta_\eta)$. Put $\zeta=\sup\{\zeta_\eta\colon \eta<\kappa^+\}$ and construct, by induction on $\eta<\kappa^+$, an increasing (in sense of inclusion) sequence of functions $\{g_\eta\in H_\kappa(\zeta)\colon\eta<\kappa^+\}$ by taking $g_\eta=f_\eta\left(\bigcup_{\gamma<\eta}g_\gamma\right)$. Thus, $[g_\eta]\cap D_{f_\eta}=\emptyset$. It is easy to see that $g=\bigcup_{\eta<\kappa^+}g_\eta \in H_{\kappa^+}(\kappa^{++})$ satisfies the requirements. (4) It is easy to check that for $f\in\F_\kappa(\zeta)$ we have $r(h+D_f)=D_{f^\prime}\in\D_\kappa$, where $f^\prime\in\F_\kappa(\zeta)$ is defined for every $g\in H_\kappa(\zeta)$ and $\xi\in\dom{f(g)}$ by formula $f^\prime\left(r\left[g+h|_{\dom{g}}\right]\right)(\xi)= r[f(g)(\xi)+h(\xi)]$. Function $f^\prime$ is indeed defined on $H_\kappa(\zeta)$ since for every $g^\prime\in H_\kappa(\zeta)$ there is $g\in H_\kappa(\zeta)$ such that $g^\prime=r\left[g+h|_{\dom{g}}\right]$. \section{The example} Now we are ready to prove our main theorem. \theorem{Example}{Assume that there exists an infinite cardinal $\kappa$ such that $2^\kappa=\kappa^+$ and $2^{\kappa^+}=\kappa^{++}$. Then, there exists a linear subspace $L\subset\reals^{\kappa^{++}}$ which does not contain any dense totally disconnected subset. } Proof. Let $\{D_\eta\colon\eta<\kappa^{++}\}$ be an enumeration of $\D_\kappa$. We will define an increasing sequence $\{\alpha_\eta<\kappa^{++}\colon\eta<\kappa^{++}\}$ of ordinals and a sequence $\{g_\eta\in\reals^{\kappa^{++}}\colon\eta<\kappa^{++}\}$ by induction on $\eta<\kappa^{++}$. So, let us assume that for some $\eta<\kappa^{++}$ our construction is done for all $\zeta<\eta$. Let $L_\eta$ be a linear subspace of $\reals^{\kappa^{++}}$ generated by $\{g_\zeta\colon\zeta<\eta\}$ and define \[ \E_\eta=\{r(h+D_\zeta): r\in\reals\setminus\{0\}, h\in L_\eta, \zeta<\eta\}. \] By Lemma \ref{lem}(4), $\E_\eta\subset\D_\kappa$ and it is easy to see that $|\E_\eta|\leq\kappa^+$. Hence, by Lemma \ref{lem}(3), there exists $g\in H_{\kappa^+}(\kappa^{++})$ such that $[g]\cap\bigcup\E_\eta=\emptyset$. We can also find $\alpha_\eta<\kappa^{++}$ such that $g\in H_{\kappa^+}(\alpha_\eta)$ and $\alpha_\zeta<\alpha_\eta$ for all $\zeta<\eta$. Define $g_\eta\in\reals^{\kappa^{++}}$ by taking ${g_\eta}\supset g$, $g_\eta(\alpha_\eta)=1$, and $g_\eta(\xi)=0$ for $\xi>\alpha_\eta$ and notice that $g_\eta\not\in\bigcup\E_\eta$, since $g_\eta\in[g]$. Define $L$ as a linear subspace of $\reals^{\kappa^{++}}$ generated by $\{g_\eta\colon\eta<\kappa^{++}\}$. To see that $L$ satisfies the theorem first notice that $L=\bigcup_{\eta<\kappa^{++}}L_\eta$. If $g\in L$ then there are $\eta_1<\ldots<\eta_n<\kappa^{++}$ and non-zero real numbers $r_1,\ldots,r_n$ such that $g=r_1g_{\eta_1}+\ldots+r_ng_{\eta_n}$. Then, $g(\alpha_{\eta_n})=r_n\neq 0$ while $g(\xi)=0$ for $\xi>\alpha_{\eta_n}$. Hence, for a function $z_\eta$ defined by $z_\eta(\xi)=0$ for all $\alpha_{\eta}\leq\xi<\kappa^{++}$ we have $L\cap[z_\eta]=L_\eta\neq L$. But for every $D\subset [L]^{\leq\kappa^+}$, there is $\eta<\kappa^{++}$ such that $D\subset L_\eta$. Since every set $[z_\eta]$ is closed in $\reals^{\kappa^{++}}$, we conclude that $L$ does not have a dense subset of cardinality $\kappa^+$. On the other hand, we will show that for every $\xi<\kappa^{++}$ we have $L\cap D_\xi\subset L_\xi$. This will finish the proof, since $|L_\xi|\leq\kappa^+$ and, by Lemma \ref{lem}(2), every totally disconnected set in $\reals^{\kappa^{++}}$ is subset of some $D_\xi$. So, let $g=h+rg_\eta\in L\setminus L_\xi$, where $h\in L_\eta$, $\eta\geq\xi$ and $r\in\reals\setminus\{0\}$. Then, $r^{-1}(-h+D_\xi)\in\E_\eta$ and so, $g_\eta\not\in r^{-1}(-h+D_\xi)$. Hence, indeed, $g=h+rg_\eta\not\in D_\xi$. This finishes the proof of Theorem \ref{Example}. \section{Remarks} The example from \cite{KC:Lspace} mentioned in the abstract is hereditary $\kappa$-Lindel\"of if the assumption $2^\kappa=\kappa^+$ is used in the construction. In particular, under the Continuum Hypothesis the space is hereditary Lindel\"of, and so, also normal. By the similar method we can generalize the example from Theorem \ref{Example} to be hereditary $\kappa^+$-Lindel\"of. However, the following problem remains open. \problem{pr1}{ Does it exists (at least consistently with ZFC) a linear topological space without dense totally disconnected subspace which is normal? Lindel\"of? hereditarily Lindel\"of?} Let us also mentioned that set theoretical assumption in Theorem \ref{Example} can be weaken to the following: there exists an infinite cardinal $\lambda$ such that $2^{<\lambda}=\lambda$ and $2^\lambda=\lambda^+$. The proof remains essentially the same. We will finish the paper quoting another problem of Arhangiel'skii (private communication) concerning the same subject. \problem{pr2}{ Does it exists a completely regular topological space such that $C_p(X)$ does not have dense 0-dimensional (or totally disconnected) subspace, where $C_p(X)$ stands for the space of all continuous functions $f\colon X\to\reals$ considered with the topology of pointwise convergence?} \begin{thebibliography}{22} \bibitem{KC:Lspace} K. Ciesielski, L-space without any uncountable 0-dimensional subspace, {\sl Fund. Math. 125} (1985), 231-235. \bibitem{Engelking:GenTop} R. Engelking, {\sl General Topology}, Polish Scientific Publishers, Warsaw, 1977. \bibitem{Kunen} K. Kunen, {\sl Set Theory}, North-Holland, 1983. \end{thebibliography} \end{document}