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\markright{Compact sets ...
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\title{Ideals of compact sets associated with Borel functions}

%\MathReviews{Primary:  26A15; Secondary: 03E75, 54A25.}
%\keywords{cardinal functions; extendable, Darboux, almost continuous
%and peripherially continuous functions; functions with perfect road. }


\author{{\small Francis Jordan}%
\thanks{AMS classification numbers: Primary 26A15, 26A21
\endgraf  Key words and phrases: Descriptive set theory, Borel functions, compact sets.
\endgraf  },
\small  Department of Mathematics, Loyola University,\\
New Orleans, La 70118\\ (fejord@hotmail.com) }

\date{}


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%\title{The minimum number of Darboux functions needed to cover Baire
%class 1}
%\author{Francis Jordan\\
%\small Dept. of Mathematics, Univ., Morgantown, WV
%26506-6310}

\date{}


\pagestyle{myheadings}

\begin{document}\maketitle

\begin{abstract}

We investigate the connection between the Borel class of a function $f$ and
the Borel complexity of the set 
$\T(f)=\{C\in\comp(X)\colon f|_C\text{ is continuous}\}$ where $\comp(X)$ denotes the compact subsets of $X$ with the Hausdorff metric.  For example, we show that for a
function $f\colon X\to Y$ between Polish spaces; if $\T(f)$ is $F_{\sigma\delta}$ in $\comp(X)$, then $f$ is Borel class one. 
\end{abstract}
\section{Introduction}
Given a Polish space $X$ let $\comp(X)$ denote the collection of nonempty compact subsets of $X$ with the Hausdorff metric.  
We investigate the connection between the Borel class a function $f$ and
the Borel complexity of the set 
$\T(f)=\{C\in\comp(X)\colon f|_C\text{ is continuous}\}$.  Generally, the set $\T(f)$ is an ideal.  One can see the subject of this paper from at least two directions.  First, one can see the complexity of $\T(f)$ as a measure of how discontinuous $f$ is, since for $f$ continuous $\T(f)=\comp(X)$ which is a very simple set.  Secondly, descriptive set theorist have an interest in finding natural examples of objects such as ideals of compact sets which are complex.  

\section{Preliminaries}
Let $X$ be a set.  We let $|X|$ denote the cardinality of $X$.  Given a cardinal $\kappa$ we let $[X]^{<\kappa}$, $[X]^{\leq\kappa}$ and, $[X]^{\kappa}$ denote the subsets of $X$ of cardinality strickly less than $\kappa$, less or equal to $\kappa$, and equal to $\kappa$, respectively.  Given a function $f\colon X\to Y$ and $A\subseteq X$ we let $f|_A$ denote the restriction of $f$ to $A$.  Given a product of two sets $X\times Y$ we let $\pi_X$ and $\pi_Y$ denote the usual projections onto $X$ or $Y$, respectively.

Suppose $X$ is a Polish space with metric $d$.  For a set $A\subseteq X$ we write
$\cl_X(A)$, $\interior_X(A)$, $\bndy_X(A)$ for the closure, interior, and boundary of $A$ in
$X$, respectively.   When it is understood what space we are referring to the subscript
will be dropped.  Given sets $A,B\subseteq X$ we define $\dis(A,B)=\inf(\{d(x,y)\colon x\in A\ \&\ y\in B\})$.  Given sets $A,B\subseteq X$ we define the Hausdorff distance between $A$ and $B$ to be $\dist(A,B)=\max(\sup(\{\dis(\{x\},B)\colon x\in A\}),\sup(\{\dis(A,\{y\})\colon y\in B\}))$.  When $\dist$ is restricted to the compact subsets of $X$ it is a metric known as the Hausdorff metric.  The diameter of a nonempty set $A\subseteq X$ is defined by $\diam(A)=\sup\{d(x,y)\colon x,y\in A\}$, if $A=\emptyset$ we let $\diam(A)=0$.  It is known that if $X$ is Polish, then $\comp(X)$ is Polish as well \cite[4.25]{kech}.  


By a Cantor set we mean a compact totally disconnected metric space with no isolated points.  


Let $X$ be Polish.  By $\bor(X)$ we denote the
Borel subsets of $X$ as defined in \cite[11.A]{kech}.  For
$0<\alpha<\omega_1$ let $\Sigma_{\alpha}^0(X)$,
$\Pi_{\alpha}^0(X)$, $\Delta_{\alpha}^0(X)$ stand for the subclasses of
$\bor(X)$
defined as in \cite[11.B]{kech} (e.g., $\Pi_2^0$ is $G_{\delta}$ and $\Sigma_2^0$ is $F_{\sigma}$).  The analytic subsets of $X$ and the coanalytic
subsets of $X$ as defined in \cite{kech} will be denoted by $\Sigma_1^1(X)$ and $\Pi_1^1(X)$, respectively.  A set $A\subseteq X$ is said to be coanalytic hard provided that for any zero-dimensional Polish space $Y$ and coanalytic $B\subseteq Y$ there is a continuous function $f\colon Y\to X$ such that $f^{-1}(A)=B$.  To say that $A$ is coanalytic hard is essentially saying that $A$ is at least as complex as any coanalytic set.  In particular, if $A$ is coanalytic hard, then $A$ is neither Borel nor analytic. 

If a function $f\colon X\to Y$ has the property that for every
open set
$U\subseteq Y$ the set $f^{-1}(U)\in\Sigma_2^0(X)$, then we say
$f$ is a Borel class one function.  Let $\baire_1$ denote the
Borel class one functions.  If a function $f\colon X\to Y$ has the property that for every
open set
$U\subseteq Y$ the set $f^{-1}(U)\in\bor(X)$, then we say
$f$ is a Borel function.  We let $\baire$ denote the Borel functions.
For a function $f\colon X\to Y$ and $S\subseteq X$ we let
$\osc(f,S)=\sup\{\dis(f(x),f(y))\colon x,y\in S\}$.  For a function $f\colon X\to Y$ we let $\disc(f)$ denote the set of discontinuity points of $f$.  



We say $f\colon X\to Y$ is a discrete limit of a sequence of functions $\{f_n\}_{n\in\omega}$ provided that for every $x\in X$ there is an $n_{x}\in\omega$ such that $f_k(x)=f(x)$ for all $k\geq n_x$.  For more facts about discrete limits see \cite{disconv}.

\section{Results}
If a function $f\colon X\to Y$ has
the property that for every $x\in X$ there exist open sets $U\subseteq X$ and $V\subseteq Y$ such that $x\in U$, $f(x)\in V$, and  $f|_{\cl(f^{-1}(V)\cap U)}$ is continuous then we say
$f\in\type_0$.  If a function $f\colon X\to Y$ has
the property that for every $x\in X$ there exist open sets $U\subseteq X$ and $V\subseteq Y$ such that $f(x)\in V$, $x\in U$, and $f|_{f^{-1}(V)\cap U}$ is continuous, then we say
$f\in\type_1$.

We may now state our theorems.
\thm{thm:last}{If $X$ and $Y$ are Polish spaces and $f\colon X\to Y$ is a function, then $f$ is continuous if and only if $\T(f)\in\Pi_2^0(\comp(X))$.}

\thm{thm:0}{If $X$ and $Y$ are Polish spaces and $f\colon X\to Y$, then the following are equivalent:
\begin{description}
\item[(i)] $f\in\type_0$
\item[(ii)] $\T(f)\in\Sigma_2^0(\comp(X))$
\item[(iii)] there is a $\subseteq$-increasing sequence $\{T_n\}_{n\in\omega}$ of closed subsets of $X$ such that $\T(f)=\bigcup_{n\in\omega}\comp(T_n)$

\item[(iv)] $\T(f)\in\Delta_3^0(\comp(X))$.
\end{description}
Moreover, if $Y=\real$ the conditions (i)-(iv) are equivalent to :
\begin{description}
\item[(v)] $f$ is open in $\cl(f)$
\item[(vi)] $f$ is the discrete limit of continuous functions $\{f_n\}_{n\in\omega}$ such that $\T(f)=\{C\in \comp(X)\colon \{f_n|_C\}_{n\in\omega}\text{ is eventually constant}\}$.
\end{description}}

\thm{thm:0000}{If $X$ and $Y$ are Polish spaces and $f\colon X\to Y$, then $(i)\Rightarrow (ii) \Rightarrow (iii)$ where :
\begin{description}
\item[(i)] $\T(f)\in\Pi_{3}^0(\comp(X))$
\item[(ii)] $f\in\baire_{1}$

\item[(iii)] $\T(f)\in\Pi_{4}^0(\comp(X))$
\end{description} and none of the implications may be reversed.  Moreover, there is a $\baire_1$ function $f$ such that $\T(f)\notin\Sigma_4^0(\comp(X))$.}

\thm{thm:000}{If $X$ and $Y$ are Polish spaces, and $f\colon X\to Y$, then the following are equivalent:
\begin{description}
\item[(i)] $\T(f)\in\Sigma_{3}^0(\comp(X))$
\item[(ii)] $f\in\type_{1}$ and $f$ has $G_{\delta}$-graph.
\end{description}}

The following theorem shows the importance of the assumption in Theorem~\ref{thm:000} (ii) that $f$ has $G_{\delta}$-graph:

\thm{thm:dich}{If $X$ and $Y$ are Polish and $f\colon X\to Y$ is Borel, then the following are equivalent:
\begin{description}
\item[(i)] $\T(f)$ is Borel,
\item[(ii)] $f$ has $G_{\delta}$ graph, and
\item[(iii)] $\T(f|_A)$ is coanalytic hard for no $A\in\comp(X)$.
\end{description}}

In particular, let $g$ be the characteristic function of the rationals.  Clearly, $g\in\type_1$ but does not have $G_{\delta}$-graph, so $\T(g)\notin\Sigma_3^0(X)$.

We note the following propositions which will be used repeatedly:

\prop{prop:02}{(\cite[23.1]{kech}) The set \[\{\sigma\in 2^{\omega\times\omega}\colon
(\forall m\in\omega)(\exists k\in\omega)(\forall n\geq k)
(\sigma(\langle m,n\rangle))=0\}.\]
is in $\Pi_3^0(2^{\omega\times\omega})\setminus\Sigma_3^0(2^{\omega\times\omega})$.}
We will let $H$ denote the subset of $2^{\omega\times\omega}$ described in Proposition~\ref{prop:02}.


\prop{newt1}{\cite[23.6]{kech} The set 

\[\{\sigma\in 2^{\omega\times\omega}\colon(\exists l\in\omega)(\forall m\geq l)(\exists k\in\omega)(\forall n\geq k)(\sigma(\langle m,n\rangle)=0)\}.\]

is in $\Sigma_4^0(2^{\omega\times\omega})\setminus\Pi_4^0(2^{\omega\times\omega})$.}

We will let $I$ denote the subset of $2^{\omega\times\omega}$ described in Proposition~\ref{newt1}.


\section{Proof of Theorem~\ref{thm:last}}
If $f\colon X\to Y$ is continuous, then $\T(f)=\comp(X)\in\Pi_2^0(\comp(X))$.  Suppose now that $f\colon X\to Y$ is not continuous.  There exist $x\in X$ and $x_n\in X$ such that $\lim_{n\to\infty} x_n=x$ and no subsequence of $\{f(x_n)\}_{n\in\omega}$ converges to $f(x)$.  Let $A=\{x_n\colon n\in\omega\}\cup\{x\}$.  Notice that $B=\{Y\in\comp(A)\colon x\in Y\}$ is compact in $\comp(X)$ and has no isolated points.  Clearly, a compact set $K\in\T(f)\cap B$ if and only if $K$ is finite.  Since the finite members of $B$ form a countable dense subset of $B$, we have $B\cap\T(f)\in\Sigma^0_2(\comp(X))\setminus\Pi^0_2(\comp(X))$.  Thus, $\T(f)\notin\Pi_2^0(\comp(X))$.\qed

\section{Proof of Theorem~\ref{thm:dich}}
We begin with two lemmas the first being a version of the Blumberg Theorem \cite{blum} the proof of which is similar to the method used in \cite{stewart}. 

\lem{lem:toot01}{Let $X$ and $Y$ be separable metric spaces with $|X|>1$.  If $f\colon X\to Y$ has no isolated points, then there is a nonempty set $D\subseteq X$ such that $D$ has no isolated points and $f|_D$ is continuous.}
\proof 
Let ${\cal U}$ and ${\cal V}$ be countable bases for $X$ and $Y$, respectively.  We may assume that both bases are closed under the operation $W_1\setminus\cl(W_2)$ where $W_1,W_2\in {\cal U}$ or $W_1,W_2\in{\cal V}$.  Let ${\cal R}$ denote the rational rectangles, i.e., sets of the form $U\times V$ where $U\in{\cal U}$ and $V\in{\cal V}$.  Let $X_1\subseteq X$ be a countable set such that $f|_{X_1}$ is dense in $f$.  Notice that $f|_{X_1}$ has no isolated points.  Let $A\subseteq X_1$ and $|A|>1$.  We define the mesh of $A$ to be $\mesh(A)=\sup\{\dis(x,A\setminus\{x\})\colon x\in A\}$.  
Let $P$ be the collection of all pairs $(A,S)\in [X_1]^{<\omega}\times [{\cal R}]^{<\omega}$ such that 
\begin{description}
\item[(0)] $|A|>1$,
\item[(1)] $\pi_X[R_1]\cap\pi_X[R_2]=\emptyset$ for all distinct $R_1,R_2\in S$, and 
\item[(2)]$f|_A\subseteq\cup S$.
\end{description}
We say $(A_1,S_1)\leq (A_2,S_2)$ provided $A_2\subseteq A_1$, $\mesh(A_1)\leq \mesh(A_2)$, and $\cup S_1\subseteq\cup S_2$.   
Now $(P,\leq)$ is a reflexive and transitive ordering.  

For each $x\in X_1$ and $n>0$, let \[E^x_n=\{(A,S)\in P\colon \text{if $\langle x,f(x)\rangle\in T\in S$, then $\diam(\pi_Y[T])<1/n$}\}.\]
We show $E^x_n$ is dense in $P$.  Let $(A,S)\in P$.  If $\langle x,f(x)\rangle\notin\cup S$, then $(A,S)\in E^x_n$ by failure of hypothesis.  So we may asume that $\langle x,f(x)\rangle\in T$ for some $T\in S$.  Pick $V\in{\cal V}$ so that $\diam(V)<1/n$ and $f(x)\in V$.  If $x\in A$, then pick $U$ open such that $\cl(U)\subseteq\pi_X(T)$ and $\{x\}=A\cap U=A\cap\cl(U)$.  If $x\notin A$, then pick an open set $U$ such that $\cl(U)\subseteq\pi_X[T]$ and $A\cap U=\emptyset$.  Let $S^*=(S\setminus\{T\})\cup\{U\times V, (\pi_X[T]\setminus\cl(U))\times(\pi_Y[T]\setminus\cl(V))\}$.  Now $(A,S^*)\leq (A,S)$ and $(A,S^*)\in E^x_n$.  So, $E^x_n$ is dense for all $x\in X_1$ and $n\in\omega$. 

For each $n>0$, let \[F_n=\{(A,S)\colon\text{$\dis(\{x\},A\setminus\{x\})<1/n$\ for all $x\in A$}\}.\]
We show $F_n$ is dense in $P$.  Let $(A,S)\in P$.  Fix $x\in A$.  Since $(A,S)\in P$, there is a $T\in S$ such that $\langle x,f(x)\rangle\in T$.  Since $T$ is open and $f|_{X_1}$ has no isolated points we can find an $x^*\in X_1\setminus A$ such that $\langle x^*,f(x^*)\rangle\in T$ and $\dis(x,x^*)<\min\{\mesh(A),1/n\}$.  Let $A^*=A\cup\{x^*\colon x\in A\}$.  Now $(A^*,S)\leq (A,S)$ and $(A^*,S)\in F_n$.  So, $F_n$ is dense in $P$ for all $n>0$.  

Since $|\{E^x_n\colon x\in X_1\ \&\ n>0\}\cup \{F_n\colon n>0\}|\leq\omega$ we may find a filter $G\subseteq P$ such that $G$ has nonempty intersection with each of the dense sets defined.  Let $D=\bigcup\{A\colon (A,S)\in G\}$.  For every $(A,S)\in G$ we have $f|_D\subseteq\cup S$.  To see it let $x\in D$ and $(A,S)\in G$.  By definition of $D$, there is an $(A_1,S_1)\in G$ such that $x\in A_1$.  Pick $(A_2,S_2)\in G$ such that $(A_2,S_2)\leq (A,S)$ and $(A_2,S_2)\leq(A_1,S_1)$.  Since $x\in A_2$ there is a $T\in S_2$ such that $\langle x,f(x)\rangle\in T$.  Thus, $\langle x,f(x)\rangle\in T\subseteq\cup S_2\subseteq\cup S$.    

We show that $f|_D$ is continuous.  Let $x\in D$ and $\epsilon>0$.  Pick $n>0$ such that $1/n<\epsilon$ and pick $(A,S)\in G\cap E^x_n$.  Since $x\in D$, there is an $(A_1, S_1)\in G$ such that $x\in A_1$.  Pick $(B,M)\in G$ such that $(B,M)\leq(A_1,S_1)$ and $(B,M)\leq(A,S)$.  Now $x\in B$ so there is a $N\in M$ such that $\langle x,f(x)\rangle\in N$.  Since $(B,M)\leq (A,S)$, we have $N\subseteq\cup M\subseteq\cup S$.  So, $\langle x,f(x)\rangle\in\cup S$.  Hence, there is a $T\in S$ such that $\langle x,f(x)\rangle\in T$ and $\pi_Y[T]<1/n$.  Since $f|_{D\cap\pi_X[T]}\subseteq T$, there is an open neighborhood $U$ of $x$ such that $\dis(f(x),f(w))<1/n<\epsilon$ for all $w\in U\cap D$.  Therefore, $F|_D$ is continuous.

We now show that $D$ has no isolated points.  Let $x\in D$ and $\epsilon>0$. Pick $n>0$ such that $1/n<\epsilon$.  There is an $(A,S)\in G$ such that $x\in A$. Pick $(A_1,S_1)\in G\cap F_n$.  Pick $(A_2,S_2)\in G$ such that $(A_2,S_2)\leq (A,S)$ and $(A_2,S_2)\leq(A_1,S_1)$.  Now $x\in A_2$.  Since $(A_2,S_2)\leq(A_1,S_1)$ and $\mesh(A_1)<1/n$, we have $\mesh(A_2)<1/n$.  Thus, there is a $w\in A_2\subseteq D$ such that $\dis(x,w)<1/n<\epsilon$.  Thus, $D$ has no isolated points.\qed    


\lem{lem:toot1}{Let $C$ be a Cantor set and $D\subseteq C$ be countable and dense.  If ${\cal S}$ is the collection of all $K\in\comp(C)$ such that $K\cap D$ and $K\cap (C\setminus D)$ are both compact, then ${\cal S}$ is coanalytic hard.}
\proof
Let $N\subseteq 2^{\omega}$ be the set all binary sequences $\tau$
such that $\tau^{-1}(1)$ is finite.  Notice that $N$ is countable
and dense in $2^{\omega}$.  It is well known \cite[33.B]{kech} that
$I=\{K\in\comp(2^{\omega})\colon K\subseteq N\}$
is a coanalytic hard set.  For $K\subseteq 2^{\omega}$ and
$n\in\omega$ let $K|_n=\{\sigma|_n\colon\sigma\in K\}$.

Define $\Theta\colon \comp(2^{\omega})\to\comp(2^{\omega})$ by
\[\Theta(K)=\cl(\bigcup_{n\in\omega}
\{\sigma\in
2^{\omega}\colon\sigma|_n\in K|_n\ \&\
(\forall k\geq n)(\sigma(k)=0)\})\]
for every $K\in\comp(2^{\omega})$.  It is easily seen that $\Theta$ is continuous.  It should be clear that $\Theta(K)\subseteq N$ if $K\subseteq N$.  On the other hand, suppose $K\setminus N\neq\emptyset$.
Let $k\in K\setminus N$.  Now $k\in\Theta(K)$ and there exist $\{n_l\in N\}_{l\in\omega}$ such that $\lim_{l\to\infty}n_l=k$.  So, in this case $\Theta(K)\cap N$ is not compact.  Thus, $\Theta^{-1}({\cal S})= I$.  Therefore, ${\cal S}$ is coanalytic hard.\qed


\lem{lem:star}{Let $X$ be Polish.  If $G\in\Pi_2^0(X)$ is dense and $D$ is a dense set disjoint from $G$, then there is a countable $E\subseteq D$ such that $E$ is dense in $X$ and $G\cup E\in\Pi_2^0(X)$.}

\proof

Let $X\setminus G=\bigcup_{n\in\omega}F_n$ where each $F_n$ is closed.  We may assume that $\lim_{n\to\infty}\diam(F_n)=0$ and that $F_m\setminus\bigcup_{n<m}F_n\neq\emptyset$ for every $m\in\omega$.  Fix $m\in\omega$.  If $(F_m\setminus\bigcup_{n<m}F_n)\cap D\neq\emptyset$, pick $e_m\in(F_m\setminus\bigcup_{n<m}F_n)\cap D$.  Let $E=\{e_{m}\colon D\cap (F_m\setminus\bigcup_{n<m}F_n)\neq\emptyset\}$.  Clearly, $E\in\Sigma_2^0(X)$ and $E\subseteq D$.



Let $U\subseteq X$ be a nonempty open set.  Since $F_n$ is nowhere dense for every $n\in\omega$ and $D$ is dense, there is no $N\in\omega$ such that $D\cap U\subseteq\bigcup_{n\leq N}F_n$.  Thus, there is a $d\in D$ and a $k\in\omega$ such that $(F_k\setminus\bigcup_{n<k}F_n)\cap D\neq\emptyset$ and $F_k\subseteq U$.  Now $e_{k}\in U$.  Thus, $E$ is dense in $X$.



Since $E\cup G=X\setminus(\bigcup_{m\in\omega}F_m\setminus E)$ and $F_m\cap E$ is finite for every $m\in\omega$, we have $E\cup G\in\Pi_2^0(X)$.\qed



{\sc Proof of Theorem~\ref{thm:dich}}
Clearly, (i) implies (iii).  

We show that (ii) implies (i).  First notice that the set $\comp(f)$ is a $G_{\delta}$-subset of $\comp(X\times Y)$ and that $\T(f)$ is an injective continuous image of $\comp(f)$ by the function $\Theta\colon\comp(f)\to\comp(X)$ defined by $\Theta(K)=\pi_X[K]$.  Since $\T(f)$ is an injective image of a Borel set, $\T(f)$ is Borel by \cite[15.1]{kech}.

We now show that (iii) implies (ii).  Suppose $f\colon X\to Y$ is Borel and $f$ does not have $G_{\delta}$-graph.  Since $f$ is Borel we have that $f$ is a $\Pi^1_1$-subset of $X\times Y$.  By a theorem of Hurewicz \cite[21.18]{kech}, there is a relatively closed subset $B$ of $f$ such that $B$ is homeomorphic to the rational numbers.  Let $B_1=\pi_X(B)$.  Since $f|_{B_1}$ has no isolated points, by  Lemma~\ref{lem:toot01}, there is a $B_2\subseteq B_1$ such that $f|_{B_2}$ is continuous and $B_2$ has no isolated points.  Let $C\subseteq X$ be a Cantor set such that $B_2$ is dense in $C$.  Since $f$ is Borel, there is a dense $G_{\delta}$-subset $D$ of $C$ such that $f|_D$ is continuous.   By Lemma~\ref{lem:star} there is a dense subset $B_3$ of $B_2$ such that $D\cup B_3\in\Pi_2^0(C)$.  Notice $B_3$ has no isolated points.  Let $d\in D$.  Since $B$ is relatively closed in $f$, there exist $\epsilon,\delta>0$ such that for any $x\in\ball_{\delta}(d)\cap D$ and $w\in\ball_{\delta}(d)\cap B_3$ we have $|f(x)-f(w)|>\epsilon$.  Let $C_1\subseteq\ball_{\delta}(d)\cap (D\cup B_3)$ be a Cantor set such
that $B_3\cap C_1$ is countable and dense in $C_1$.  It is clear that $\T(f|_{C_1})$ is exactly the compact subsets $P$ of $C_1$ with the property that both $B_3\cap P$ and $(C_1\setminus B_3)\cap P$ are both compact.  By Lemma~\ref{lem:toot1}, $\T(f|_{C_1})$ is coanalytic hard.\qed

\section{Proof of Theorem~\ref{thm:000}}
Suppose $X$ and $Y$ are Polish spaces and $f\colon X\to Y$.  

We show that (i) implies (ii).
\lem{lem:12}{If $\T(f)\in\Sigma_3^0(\comp(X))$, then $f\in\type_1$.}
\proof
Suppose $f\notin\type_1$.  Let $x\in X$ be such that for every pair of open sets $U\subseteq X$ and $V\subseteq Y$ with $x\in U$ and $f(x)\in V$ we have $f|_{f^{-1}(V)\cap U}$ not continuous.
Let $\{V_n\}_{n\in\omega}$ be a decreasing sequence of open
subsets of $Y$ such that $\dist(V_n,f(x))<1/2^n$ and $f(x)\in V_n$
for every $n\in\omega$.  Let $\{U_n\}_{n\in\omega}$ be a
decreasing sequence of open subsets of $X$ such that
$\dist(U_n,x)<1/2^n$ and $x\in U_n$ for every $n\in\omega$.
For each $n\in\omega$ pick $x_n\in\disc(f|_{f^{-1}(V_n)\cap U_n})$.  For each $n\in\omega$
we may find $\{w_{n,k}\in f^{-1}(V_n)\cap U_n\}_{k\in\omega}$ such
that $\lim_{k\to\infty} w_{n,k}=x_n$ and no subsequence of
$\{f(w_{n,k})\}_{k\in\omega}$ converges to $f(x_n)$, we may also
assume that $\cl(\{\langle w_{n,k},f(w_{n,k})\rangle\colon k\in
\omega\})\cap\cl(\{\langle w_{m,k},f(w_{m,k})\rangle
\colon k\in\omega\})=\emptyset$ for all $n,m\in\omega$ such that $n\neq m$.
Let $C=\cl(\{w_{n,k}\colon n,k\in\omega\})$.
Define $h\colon 2^{\omega\times\omega}\to\comp(C)$ by
$h(\sigma)=\{w_{n,k}\colon\sigma(\langle n,k\rangle)=1\}\cup\{x_n\colon n\in\omega\}\cup\{x\}$.
Notice that $h$ is continuous.  It is straight forward to check that
$f|_{h(\sigma)}$ is continuous if and only if \[
(\forall m\in\omega)(\exists k\in\omega)(\forall n\geq k)(\sigma(\langle m,n\rangle))=0.\]  Thus, $h^{-1}(\T(f)\cap \comp(C))=H$.  By Proposition~\ref{prop:02} and the continuity of $h$, 
we have $\T(f)\cap\comp(C)\notin\Sigma_3^0(\comp(X))$.  Since $\comp(C)$ is closed, $\T(f)\notin\Sigma_3^0(\comp(X))$. \qed

We now show that (ii) implies (i).  We first define an operation $M$ on collections of subsets of product spaces.
Given a collection ${\cal A}$ of subsets of $X\times Y$.  Define 
\[M({\cal A})=\bigcup_{x\in X}\left(\pi_{X}^{-1}(\{x\})\cap\bigcap\{A\in{\cal A}\colon x\in\pi_X[A]\}\right).\]

\lem{lem:new0}{If $f\colon X\to Y$ is a function and ${\cal A}$ is a finite collection of subsets of $X\times Y$ such that $\pi_X[A]$ is closed for every $A\in{\cal A}$ and $f|_{\pi_X[A\cap f]}$ is continuous for each $A\in {\cal A}$, then $f|_{\pi_{X}[M({\cal A})\cap f]}$ is continuous.}
\proof 
Let $\{x_n\}_{n\in\omega}$ be a sequence of points in $\pi_X[M({\cal A})\cap f]$ which converges to some $x\in\pi_X[M({\cal A})\cap f]$.  Since ${\cal A}$ is finite, we may assume that there is an ${A\in\cal A}$ such that $x_n\in\pi_X[A\cap f]$ for every $n\in\omega$.  Since $A$ has closed $X$-projection, $x\in\pi_X[A]$.  Since $x\in\pi_X[M({\cal A})\cap f]$ and $x\in\pi_X[A]$,  we have $\langle x,f(x)\rangle\in A$.  In particular, $\{x_n\colon n\in\omega\}\cup\{x\}\subseteq \pi_X[A\cap f]$.  Thus, 
$\lim_{n\in\omega}f(x_n)=f(x)$.  Therefore, $f|_{\pi_{X}[M({\cal A})\cap f]}$ is continuous.\qed      

\lem{lem:new22}{If ${\cal A}$ is a finite collection of closed subsets of $X\times Y$ such that $\pi_X[A]$ is closed for every $A\in{\cal A}$, then $M({\cal A})\in\Pi_2^0(X\times Y)$.}
\proof
Notice that for every $A\in{\cal A}$ we have 
\[A\cup\left(\left(\pi_X[\cup{\cal A}]\setminus\pi_X[A]\right)\times Y\right)\in\Pi_2^0(X\times Y).\]
It is easily checked that \[M({\cal A})=\bigcap_{A\in{\cal A}}\left(A\cup\left(\left(\pi_X[\cup{\cal A}]\setminus\pi_X[A]\right)\times Y\right)\right).\]
Thus, $M({\cal A})\in\Pi_2^0(X\times Y)$.\qed 

\lem{lem:new2}{Let $f\in\type_1$ and ${\cal B}_1$ and ${\cal B}_2$ be countable bases for $X$ and $Y$, respectively.  If $x\in A\subseteq X$ and $f|_{A}$ is continuous, then there exist $B_1\in{\cal B}_1$ and $B_2\in{\cal B}_2$ such that  
$f|_{f^{-1}(\cl(B_2))\cap\cl(B_1)}$ is continuous and $f[A\cap\cl(B_1)]\subseteq\cl(B_2)$.}
\proof
Since $f\in\type_1$, there exist open sets $U\subseteq X$ and, $V\subseteq Y$ such that $x\in U$, $f(x)\in V$, and
$f|_{f^{-1}(V)\cap U}$ is continuous.  Pick 
$B_1\in{\cal B}_1$ and $B_2\in{\cal B}_2$ so that $\cl(B_1)\subseteq U$, 
$\cl(B_2)\subseteq V$, $x\in B_1$, and $f(x)\in B_2$.  Since
$f^{-1}(\cl(B_2))\cap\cl(B_1)\subseteq f^{-1}(V)\cap U$, we have that
$f|_{f^{-1}(\cl(B_2))\cap\cl(B_1)}$ is continuous.  Since $f|_A$ is continuous we may assume $B_1$ is small enough that $f[A\cap\cl(B_1)]\subseteq\cl(B_2)$.\qed

\lem{lem:new3}{If $f\in\type_1$ and $f$ has $G_{\delta}$-graph, then $\T(f)\in\Sigma_3^0(X)$.}
\proof
Let ${\cal B}_1$ and ${\cal B}_2$ be countable bases for $X$ and $Y$ respectively.  Let ${\cal W}$ be the collection of all finite collections $Z=\{W_{0},\ldots W_{n}\}$ of sets of the form $W_i=\cl(B_1)\times\cl(B_2)$ (where $B_1\in{\cal B}_1$ and $B_2\in{\cal B}_2$) such that $f|_{\pi_X[M(Z)\cap f]}$ is continuous.  Let $Z\in{\cal W}$.  By Lemma~\ref{lem:new22} and the assumption that $f$ has $G_{\delta}$-graph, $M(Z)\cap f\in\Pi_2^0(X\times Y)$.  Since $f|_{\pi_X[M(Z)\cap f]}$ is continuous, $\pi_X[M(Z)\cap f]\in\Pi_2^0(X)$.  Thus, ${\cal T}=\bigcup\{\comp(\pi_X[M(Z)\cap f])\colon Z\in{\cal W}\}\in\Sigma_3^0(X)$.  



The proof will be complete if we show that $\T(f)={\cal T}$.  The containment ${\cal T}\subseteq\T(f)$ is obvious.  We work for the opposite containment.  Let $C\in\T(f)$.  
We will construct a finite collection $W=\{W_1, W_2,...W_n\}$ of sets of the form $W_i=\cl(B_1)\times\cl(B_2)$ where $B_1\in{\cal B}_1$ and $B_2\in{\cal B}_2$ such that 
\begin{description}
\item[(a)]$f|_C\subseteq\bigcup W$, 
\item[(b)]$f|_{\pi_X[f\cap W_i]}$ is
continuous for every $1\leq i\leq n$, and
\item[(c)] $f|_{C\cap\pi_X(W_i)}\subseteq W_i$ for every $1\leq i\leq n$.  
\end{description} 
By Lemma~\ref{lem:new2}, for every $x\in C$ there exist $B_1^x\in{\cal B}_1$ and $B_2^x\in{\cal B}_2$ such that $x\in B_1^x$, $f(x)\in B^x_2$, $f|_{f^{-1}(\cl(B_2^x))\cap\cl(B^x_2)}$ is continuous, and $f[\cl(B^x_1)\cap C]\subseteq\cl(B^x_2)$.  Since $f|_C$ is compact, we we may find a finite subcover $W^*=\{W^*_1,\ldots W^*_n\}$ of $\{B^x_1\times B_2^x\colon x\in C\}$.  For each $1\leq i\leq n$ let $W_i=\cl(W^*_i)$.  The collection $W=\{W_1\ldots W_n\}$ clearly satisfies conditions (a), (b), and (c).  

By (b), and Lemma~\ref{lem:new0}, $f|_{\pi_X[f\cap M(W)]}$ is continuous.  So $W\in{\cal W}$.  We will be done if we show that $C\subseteq \pi_X[M(W)\cap f]$.  Let $x\in C$.  By (a), there is some $W_i\in W$ such that $\langle x,f(x)\rangle\in W_i$.  By (c), for any $W_k\in W$ if $x\in\pi_X(W_k)$, then $\langle x,f(x)\rangle\in W_k$.  Thus, $\langle x,f(x)\rangle\in M(W)\cap f$.  So, $x\in\pi_X[M(W)\cap f]$.  Therefore, $C\subseteq\pi_X[M(W)\cap f]$. \qed

\section{Proof of Theorem~\ref{thm:0}}
Suppose $X$ and $Y$ are Polish spaces and $f\colon X\to Y$.

\lem{lem:11}{If $\T(f)\in\Delta_3^0(\comp(X))$, then $f\in\type_0$.}
\proof
By way of contradiction assume $f\notin\type_0$.  Let $x\in X$ be such that for every pair of open sets $U\subseteq X$ and $V\subseteq Y$ with $x\in U$ and $f(x)\in V$ we have $f|_{\cl(f^{-1}(V)\cap U)}$ not continuous.
Let $\{V_n\}_{n\in\omega}$ be a decreasing sequence of open
subsets of $Y$ such that $\dist(V_n,f(x))<1/2^n$ and $f(x)\in V_n$
for every $n\in\omega$.  Let $\{U_n\}_{n\in\omega}$ be a
decreasing sequence of open subsets of $X$ such that
$\dist(U_n,x)<1/2^n$ and $x\in U_n$ for every $n\in\omega$.  By Lemma~\ref{lem:12}, we may assume that $f|_{f^{-1}(V_0)\cap U_0}$ is continuous.

Fix $n>0$.  Since $f|_{\cl(f^{-1}(V_n)\cap U_n)}$ is not continuous and $f|_{f^{-1}(V_0)\cap U_0}$ is continuous,
we may find an $x_n\in\cl(f^{-1}(V_n)\cap U_n)\setminus (f^{-1}(V_0)\cap U_0)$.  There exist $\{w_{n,m}\in f^{-1}(V_n)\cap U_n\}_{m\in\omega}$ such
that $\lim_{m\to\infty} w_{n,m}=x_n$.  Since $x_n\notin f^{-1}(V_0)$, $\lim_{m\to\infty}f(w_{n,m})\neq f(x_n)$.  

Since $x_n\neq x$ for all $n\in\omega$, we may assume that
$\cl(\{w_{n+1,m}\colon m\in\omega\})
\cap\cl(\{w_{l+1,m}
\colon m\in\omega\})=\emptyset$ for distinct $n,l\in\omega$.
Let $C=\cl(\{w_{n+1,m}\colon n,m\in\omega\})$.  We will have a contradiction if we show that $\T(f)\cap\comp(C)\notin\Delta_3^0(\comp(C))$.  Define $h\colon 2^{\omega\times\omega}\to\comp(C)$ by the formula $h(\sigma)=\{x\}\cup(\bigcup_{n\in\omega}L_n)$ where 

\begin{equation}\notag L_n=
\begin{cases} 
\emptyset& \text{if $\{m\colon\sigma(\langle n+1,m\rangle)=1\}=\emptyset$}\\ \{x_{n+1}\} & \text{if $\{m\colon\sigma(\langle n+1,m\rangle)=1\}$ is infinite;}\\
\{w_{n+1,\max\{m\colon\sigma(\langle n,m\rangle)=1\}}\}& \text{otherwise.}
\end{cases}
\end{equation}
We claim that $h\in\baire_1$.  For each $l\in\omega$ define $h_l\colon 2^{\omega\times\omega}\to\comp$ by the formula
$h_l(\sigma)=\{x\}\cup(\bigcup_{n\in\omega}L_{n,l})$ where 
\begin{equation}\notag L_{n,l}=
\begin{cases} \{w_{n+1,\max\{m\leq l\colon\sigma(\langle n,m\rangle)=1\}}\}& \text{if $\{m\leq l\colon\sigma(\langle n+1,m\rangle)=1\}\neq\emptyset$;}\\
\emptyset& \text{if $\{m\leq l\colon\sigma(\langle n+1,m\rangle)=1\}=\emptyset$.}
\end{cases}
\end{equation}



Notice that $h_l$ is continuous for all $l\in\omega$ and that $h_l(\sigma)\to h(\sigma)$ for every $\sigma\in 2^{\omega\times\omega}$.  So $h\in\baire_1$.  Notice that $h(\sigma)\in\T(f)$ if and only if $\sigma\in I$ where $I$ is the set from Proposition~\ref{newt1}.  In particular, $h^{-1}(\T(f))\notin\Pi_4^0(2^{\omega\times\omega})$.  Since $h\in\baire_1$, we must have $\T(f)\notin\Pi_3^0(\comp(C))$.  Hence, $\T(f)\notin\Delta_3^0(\comp(C))$.   Thus, we have the desired contradiction.\qed

\lem{lem:3}{If $f\in\type_0$, then
there exist $\{\langle U_n,V_n\rangle\}_{n\in\omega}$ such that  for every
$n\in\omega$ we have:
$\langle U_n,V_n\rangle\in\Sigma_1^0(X)\times\Sigma_1^0(Y)$,
$f|_{\cl(f^{-1}(\cl(V_n))\cap\cl(U_n))}$ is continuous, and
$f\subseteq\bigcup_{n\in\omega}U_n\times V_n$.}
\proof
Let $x\in X$.  Let $U^*_x\in\Sigma_1^0(X)$ and $V^*_x\in\Sigma_1^0(Y)$
be such that $x\in U^*_x$, $f(x)\in V^*_x$, and
$f|_{\cl(f^{-1}(V^*_x)\cap U^*_x)}$ is continuous.  Pick
$U_x\in\Sigma_1^0(X)$ and $V_x\in\Sigma_1^0(Y)$ such that $\cl(U_x)\subseteq U^*_x$ and
$\cl(V_x)\subseteq V^*_x$ and $x\in U_x$ and $f(x)\in V_x$.  Since
$\cl(f^{-1}(\cl(V_x))\cap\cl(U_x))\subseteq\cl(f^{-1}(V^*_x)\cap U^*_x)$, we have that
$f|_{\cl(f^{-1}(\cl(V_x)\cap\cl(U_x))}$ is continuous.   Since the graph of $f$
is second countable and $f\subseteq\bigcup_{x\in X}U_x\times V_x$,
we may find the desired countable collection.\qed

\lem{lem:4}{If $f\in\type_0$, then there exists
a $\subseteq$-increasing sequence $\{W_n\}_{n\in\omega}$ of closed subsets of $X$ such that
$\T(f)=\bigcup_{n\in\omega}\comp(W_n)$.  In particular,  $\T(f)\in\Sigma_2^0(\comp(X))$.}
\proof
Let ${\cal U}=\{U_n\times V_n\}_{n\in\omega}$
be as in Lemma~\ref{lem:3}.  For each $n\in\omega$ let
$W_n=\bigcup_{k\leq n}\cl(f^{-1}(\cl(V_k))\cap\cl(U_k))$.  We show that
$\T(f)=\bigcup_{n\in\omega}\comp(W_n)$.  Fix $n\in\omega$.  Since $f|_{\cl(f^{-1}(\cl(V_k))\cap\cl(U_k))}$ is continuous for every $k\leq n$, we have that $f|_{W_n}$ is continuous.  Thus, $\bigcup_{n\in\omega}\comp(W_n)\subseteq\T(f)$.  We now show the reverse inequality.  Suppose $C$ is compact and $f|_C$ is continuous.  Since $f|_C$ is compact, $f|_C$ is contained in a finite number of members of ${\cal U}$.  So $C\subseteq W_n$ for some $n\in\omega$.  Thus, $\T(f)\subseteq\bigcup_{n\in\omega}\comp(W_n)$. \qed     

Lemma~\ref{lem:11} and Lemma~\ref{lem:4} show that (i) (ii) (iii), and (iv) of Theorem~\ref{thm:0} are equivalent when $X$ and $Y$ are  Polish spaces.  

We now assume that $Y=\real$ and $X$ is Polish.  

\lem{lem:5}{If for a function $f\colon X\to\real$ there exists
a $\subseteq$-increasing sequence $\{W_n\}_{n\in\omega}$ of closed subsets
of $X$ such that
$\T(f)=\bigcup_{n\in\omega}\comp(W_n)$, then $f$ is a discrete limit of continuous functions $\{f_n\}_{n\in\omega}$ such that \[\T(f)=\{C\in\comp(X)\colon \{f_n|_{C}\}_{n\in\omega}\text{ is eventually constant }\}.\]}
\proof
Fix $n\in\omega$.  Since $\comp(W_n)\subseteq\T(f)$, we have that $f|_{W_n}$ is continuous. 
By the Tietze Extension Theorem there is a continuous $f_n\colon X\to\real$ such that $f_n|_{W_n}=f|_{W_n}$.   Clearly, $\{f_n\}_{n\in\omega}$ converges discretely to $f$.  We show $\{f_n\}_{n\in\omega}$ is as desired.

Suppose $C\in\T(f)$.  By assumption $C\subseteq W_n$ for some
$n\in\omega$.  In particular, $f_m|_{C}\subseteq f_n|_{W_n}$
for all $m\geq n$.   Thus, $\{f_n|_{C}\}_{n\in\omega}$ is
eventually constant.

Suppose $C\in\comp(X)$ and $\{f_n|_{C}\}_{n\in\omega}$ is eventually constant.  There is an $n\in\omega$ such that $f|_C=f_n|_C$.  Thus,
$C\in\T(f)$.\qed

\lem{lem:6}{If $f\colon X\to\real$ is a discrete limit of continuous functions $\{f_n\}_{n\in\omega}$ such that $\T(f)=\{C\in\comp(X)\colon \{f_n|_{C}\}_{n\in\omega}\text{ is eventually constant }\}$, then $\T(f)\in\Sigma_2^0(\comp(X))$.}
\proof
For each $n\in\omega$ let $Z_n=\{x\in X\colon (\forall m\geq n)(f_n(x)=f_m(x))\}$.  It is easily checked that $Z_n$ is closed for every $n\in\omega$.  Now for every $n\in\omega$ we have that
$\comp(Z_n)=\{C\in\comp(X)\colon (\forall m\geq n)(f_n|_C=f_m|_C)\}$ is closed
in $\comp(X)$.  Therefore, $\T(f)=\{C\in 2^{X}\colon \{f_n|_{C}\}_{n\in\omega}\text{ is eventually constant }\}\in\Sigma_2^0(\comp(X)$.\qed

Lemma~\ref{lem:5} and Lemma~\ref{lem:6} show that (v) is equivalent to (i), (ii), and (iii) when $Y=\real$.

\lem{lem:288}{Let $f\colon X\to Y$.  If $\T(f)\in\Sigma_2^0(\comp(X))$, then $f$ is open in $\cl(f)$.}
\proof
Suppose $f$ is not open in $\cl(f)$.  There exists an $x\in X$ and $\{\langle x_n,y_n\rangle\}_{n\in\omega}$ such that $\langle x_n,y_n\rangle\in\cl(f)\setminus f$ for every $n\in\omega$ and 
$\lim_{n\to\infty}\langle x_n,y_n\rangle=\langle x,f(x)\rangle$.
For each $n\in\omega$ we may find a sequence $\{w_{n,k}\}_{k\in\omega}$ of points in $X$ 
such that $\lim_{n\to\infty}\langle w_{n,k},f(w_{n,k})\rangle=
\langle x_n,y_n\rangle$.  Notice that $\langle x_n,y_n\rangle\neq\langle x_n,f(x_n)\rangle$.  We may assume that
$\cl(\{\langle w_{n,k},f(w_{n,k})\rangle\colon k\in\omega\})
\cap\cl(\{\langle w_{m,k},f(w_{m,k})\rangle
\colon k\in\omega\})=\emptyset$ for all distinct $n,m\in\omega$.
Let $C=\cl(\{w_{n,k}\colon n,k\in\omega\})$.  We will have a contradiction if we show that $\T(f)\cap\comp(C)\notin\Sigma_2^0(\comp(C))$.  Define $g\colon 2^{\omega\times\omega}\to\comp(C)$ by $h(\sigma)=\cl(\{w_{n,k}\colon\sigma(n,k)=1\})\cup\{x\}$.  We claim that $h\in\baire_1$.  For each $m\in\omega$ define $h_m\colon 2^{\omega\times\omega}\to\comp(C)$ by $h_m(\sigma)=\{w_{n,k}\colon\sigma(n,k)=1\text{ and }k\leq m\}\cup\{x\}$.  Notice that $h_m$ is continuous for all $m\in\omega$ and that $h_m(\sigma)\to h(\sigma)$ for every $\sigma\in 2^{\omega\times\omega}$.  So $h\in\baire_1$.  It is also easy to see that $h(\sigma)\in\T(f)$ if and only if $\sigma\in X_0$.  In particular, $h^{-1}(\T(f))\notin\Sigma_3^0(2^{\omega\times\omega})$.  Since $h\in\baire_1$, we must have $\T(f)\notin\Sigma_2^0(\comp(C))$.  Thus, we have the desired contradiction.\qed

\lem{lem:2}{Let $f\colon X\to\real$.  If $f$ is open in $\cl(f)$, then $f\in\type_0$.}
\proof
Let $x\in X$.  Since $f$ is open in $\cl(f)$, we may find an open set $U\subseteq X$ and a bounded open interval $V\subseteq\real$ such that $x\in U$ and $f(x)\in V$ and
$\cl(f)\cap (U\times V)=f\cap (U\times V)$.  Pick open sets $U_1\subseteq U$ and $V_1\subseteq V$ contianing $x$ and $f(x)$, respectively such that $\cl(U_1)\subseteq U_1$ and $\cl(V_1)\subseteq V_1$.  Now $f\cap (\cl(U_1)\times\cl(V_1))=\cl(f)\cap(\cl(U_1)\times\cl(V_1))$.  By way of contradiction, assume that $f|_{\cl(f^{-1}(V_1)\cap U_1)}$ is not continuous.  Let $\{w_n\}_{n_\in\omega}$ be a sequence of points in $\cl(f^{-1}(V_1)\cap U_1)$ and $w\in\cl(f^{-1}(V_1)\cap U_1)$ be such that $\lim_{n\in\omega}w_n=w$ and $\lim_{n\in\omega}f(w_n)\neq f(w)$.  Without loss of generality, we may assume that no subsequence $\{f(w_n)\}_{n\in\omega}$ converges to $f(w)$.  Since $\cl(V_1)$ is compact, there is a $r\in\cl(V_1)$ such that $\lim_{n\in\omega}f(w_n)=r$.  However, $f\cap (\cl(U_1)\times\cl(V_1))$ is closed so $f(w)=r$ which contradicts our choice of $\{w_n\}_{n\in\omega}$.\qed 

Lemma~\ref{lem:288} and Lemma~\ref{lem:2} show that (vi) is equivalent to (i), (ii), and (iii) when $Y=\real$.  Which completes the proof of Theorem~\ref{thm:0}. 

\section{Proof of Theorem~\ref{thm:0000}}
We show that (i) implies (ii).  

\lem{lem:worm}{Let $K$ be a Cantor set with a countable dense subset
$D$.  If ${\cal S}\subseteq\comp(K)$ is the collection of compact sets $C$ with the property that $C\cap D$ is finite and $C\setminus D$ is compact, then ${\cal S}\in\Sigma_3^0(\comp(K))\setminus\Pi_3^0(\comp(K))$.}
\proof
First we show that ${\cal S}\in\Sigma_3^0(\comp(K))$.  Let $D=\{d_n\colon
n\in\omega\}$ be an enumeration of $D$.  Define $f\colon K\to\real$ so that $f(d_n)=n+1$ for every $n\in\omega$ and $f(x)=0$ for $x\in K\setminus D$.  Notice that $\T(f)={\cal S}$.  Since $f$ has $G_{\delta}$-graph and $f\in\type_1$, Theorem~\ref{thm:000} guarantees that ${\cal S}=\T(f)\in\Sigma_3^0(\comp(K))$.

We now work to show that ${\cal S}\notin\Pi_3^0(\comp(K))$.  In what follows we let $\omega+1$ denote $\omega\cup\{\omega\}$ topologized to be a convergent sequence of isolated points with limit point $\omega$.  Let $L=\{\tau\in(\comp(\omega+1))^{\omega}\colon(\forall n\in\omega)(\omega\in\tau(n))\}$ and $E\subseteq L$ be the collection of all $\tau\in L$ such that for some $n\in\omega$ we have $|\tau(k)|<\omega$ for all $k<n$ and $\tau(k)=\omega+1$ for all $k\geq n$.  Since $L$ is a Cantor set and $E$ is countable and dense in $L$, we may assume that $K=L$ and $D=E$.

Define $\Theta\colon 2^{\omega\times\omega}\to L$ by setting $\Theta(\sigma)(n)=\{k\in\omega\colon \sigma(\langle n,k\rangle)=1\}\cup\{\omega\}$ for every $\sigma\in 2^{\omega\times\omega}$ and $n\in\omega$.  Notice that $\Theta$ is continuous.  

Define $\Psi\colon L\to\comp(L)$ by letting $\Psi(\tau)$ be the closure of the collection of all $\rho\in L$ such that for some $n\in\omega$ we have  $\rho|_n=\tau|_n$ and $\rho(k)=\omega+1$ for all $k\geq n$.  If for infinitely many $n\in\omega$ we have $\tau(n)\neq\omega+1$, then $\Psi(\tau)$ is a convergent of sequence points in $L$ with limit point $\tau$.  If there is an  $n\in\omega$ such that for all $k\geq n$ we have $\tau(k)=\omega+1$, then $\Psi(\tau)$ is a finite subset of $L$ containing $\tau$.  

We claim that $\Psi$ is continuous.  Suppose $\{\tau_k\}_{n\in\omega}$ is a sequence points in $L$ converging to some $\tau\in L$.  We show that $\lim_{k\in\omega}\Psi(\tau_k)=\Psi(\tau)$. 

Suppose there exist an infinite $A\subseteq\omega$ such that $\rho_k\in\Psi(\tau_k)$ for every $k\in A$ and $\lim_{k\in A}\rho_k=\rho$.  
We claim that $\rho\in\Psi(\tau)$.  We will consider two exhaustive cases.  First, suppose that there is an $N\in\omega$ such that for infinitely many $k\in A$ we have 
$\rho_k(l)=\omega+1$ for all $l\geq N$.  We may assume that $N$ is minimal with respect to this property.  Let $A^*$ be the set of all $k\in A$ such that $\rho_k(l)=\omega+1$ for all $l\geq N$.  By minimality, there are only finitely many $k\in A^*$ such that $\rho_k(N-1)=\omega+1$.  So, for almost all $k\in A^*$ we have $\rho_k|_N=\tau_k|_N$.  Thus, we have $\rho|_N=\tau|_N$ and $\rho(l)=\omega+1$ for all $l\geq N$, so $\rho\in\Psi(\tau)$.  For the second case, suppose that for every $N\in\omega$ there are only finitely many $k\in A$ such that  
$\rho_k(l)=\omega+1$ for all $l\geq N$.  In this case we have $\lim_{k\in A}\rho_k(j)=\lim_{k\in A}\tau_k(j)=\tau(j)$ for every $j\in\omega$.  Thus, $\rho=\tau\in\Psi(\tau)$.  By cases, we have the claim.

We show that for every $\rho\in\Psi(\tau)$ there is a sequence $\{\rho_k\}_{k\in\omega}$ such that $\rho_k\in\Psi(\tau_k)$ and $\lim_{k\to\infty}\rho_k=\rho$.  If $\rho=\tau$, then we can let $\rho_k=\tau_k$ for every $k\in\omega$ and have $\lim_{k\to\infty}\rho_k=\rho$.  If there is an $n\in\omega$ such that $\rho|_n=\tau|_n$ and $\rho(l)=\omega+1$ for all $l\geq n$, then we pick $\rho_k\in\Psi(\tau_k)$ such that $\rho_k|_n=\tau_k|_n$ and $\rho_{k}(l)=\omega+1$ for all $l\geq n$ to get $\lim_{k\to\infty}\rho_k=\rho$.

By the proceeding two paragraphs, $\lim_{n\in\omega}\Psi(\tau_k)=\Psi(\tau)$.  Thus, $\Psi$ is continuous.

Let $\Gamma\colon 2^{\omega\times\omega}\to\comp(L)$ be defined by $\Gamma(\sigma)=\Psi(\Theta(\sigma))$.  Clearly, $\Gamma$ is continuous.  We claim $\Gamma^{-1}({\cal S})=2^{\omega\times\omega}\setminus H$ where $H$ is the set from Proposition~\ref{prop:02}.  Suppose $\sigma\in 2^{\omega\times\omega}\setminus H$.  By definition of $H$ there is a smallest $n\in\omega$ such that $|\Theta(\sigma)(n)|=\omega$.  It follows that at most $n$ elements of $\Psi(\Theta(\sigma))$ are in $E$.  We will show that $\Psi(\Theta(\sigma))\setminus E$ is compact.  If $\Theta(\sigma)\notin E$, then either $\Psi(\Theta(\sigma))$ is finite or $\Psi(\Theta(\sigma))$ is a convergent sequence with limit point not in $E$.  If $\Theta(\sigma)\in E$, then $\Psi(\Theta(\sigma))$ is finite.  In any of the three cases above $\Psi(\Theta(\sigma))\setminus E$ is compact.  Thus, $2^{\omega\times\omega}\setminus H\subseteq\Gamma^{-1}({\cal S})$.  Suppose $\sigma\in H$.  By definition of $H$,  $|\Theta(\sigma)(n)|<\omega$ for every $n\in\omega$.  Thus,  $\Psi(\Theta(\sigma))$ is a convergent sequence of elements of $E$ with limit point $\Theta(\sigma)\notin E$.  So, $\Psi(\Theta(\sigma))\notin{\cal S}$.  Hence, $\Gamma^{-1}({\cal S})\subseteq 2^{\omega\times\omega}\setminus H$.  Since $\Gamma^{-1}({\cal S})=2^{\omega\times\omega}\setminus H$ and  $H\notin\Sigma_3^0(2^{\omega\times\omega})$, we have ${\cal S}\notin \Pi_3^0(\comp(K))$.\qed

\lem{lem:star1}{If $X$ is Polish and $G\in\Pi_2^0(X)$ is countable, the the set $I$ of isolated points of $G$ is a dense open subset of $G$.}
\proof
Clearly, $G$ is a countable dense $G_{\delta}$-subset of $\cl(G)$.  Since $\cl(G)$ is countable and closed, the set $J$ of isolated points of $\cl(G)$ form a dense open subset of $\cl(G)$.  Clearly, $I=J$. So, $I=G\cap J$ is dense and open in $G$.\qed
 
\medskip

{\sc Proof that $\T(f)\in\Pi_3^0(\comp(X))$ implies $f\in\baire_1$.} Let $f\notin\baire_1$.  If $f$ does not have $G_{\delta}$-graph, then, by Theorem~\ref{thm:dich}, $\T(f)$ is not Borel and so $\T(f)\notin\Pi_3^0(\comp(X))$.  So, we may assume that $f$ has $G_{\delta}$-graph.

Since $f\notin\baire_1$, there is a Cantor set $C$ such that $f|_C$ is nowhere continuous.  We may assume that there is a $K>0$ such that  
\begin{equation}\label{star}
\osc(f|_C,x)>3K
\end{equation} 
for every $x\in C$.  Since $f$ is Borel, there is a $G_{\delta}$-set $G$ such that $G$ is a dense subset of $C$ and $f|_G$ is continuous.

Let $U$ be a nonempty open subset of $C$.  Let $x\in G\cap U$.  There is an open set $V\subseteq U$ such that $x\in V$ and $\diam(f[V\cap G])<K$.  By (\ref{star}) there is a $d\in V$ such that $\dist(f(d),f[V\cap G])>K$.
Since $U$ was arbitrary we may find a countable dense subset $D$ of $C$ such that for every $d\in D$ there is an open set $V_d$ such that $d\in V_d$ and $\dist(f(d),f[V_d\cap G])>K$.  By Lemma~\ref{lem:star}, there is a countable dense subset $E$ of $D$ such that $G_1=G\cup E$ is a $G_{\delta}$-subset of $C$.  Since $f|_{G}$ and $f|_E$ are disjoint open subsets of $f|_{G_1}$ which is a $G_{\delta}$-subset of $X\times Y$, we have that $f|_E$ is a countable $G_{\delta}$-set.  By Lemma~\ref{lem:star1}, the collection $J$ of isolated points of $f|_E$ is dense in $f|_E$.  So, we may find a countable dense $E_1\subseteq E$ such that $f|_{E_1}=J$ is the collection of isolated points in $f|_{G_1}$.  Find a compact perfect set $K\subseteq G\cup E_1$ such that $E_1\cap K$ is dense in $K$.  Letting $Q=E_1\cap K$ and $H=K\setminus Q$ it should be clear that $\T(f|_K)$ is the collection of compact sets $L\in\comp(K)$ with the property that $L\cap Q$ is finite and $L\cap H$ is compact.  By Lemma~\ref{lem:worm}, $\T(f|_K)\notin\Pi_3^0(\comp(K))$.  Since $\T(f)\cap\comp(K)=\T(f|_K)$, we have that $\T(f)\notin\Pi_3^0(J(X))$.\qed
 
\medskip 

{\sc Proof that $f\in\baire_1$ implies $\T(f)\in\Pi^0_4(\comp(X))$.}  Since $Y$ is Polish, we can consider $Y$ as subset of $[0,1]^{\omega}$ with the usual product topology and $f$ to be a function from $X$ into $[0,1]^{\omega}$.  Since every $\baire_1$ function into $[0,1]$ is a pointwise limit of continuous functions, we have that $f\colon X\to[0,1]^{\omega}$ is a pointwise limit of continuous functions $f_i\colon X\to [0,1]^{\omega}$.  



For each $n,k,l\in\omega$ let \[A_{k,l,n}=\{P\in\comp(X)\colon(\exists x,w\in P)(\forall i\geq n)(d(x,w)\leq \frac{1}{2^l})(d(f_i(x),f_i(w))\geq\frac{1}{2^k})\}.\] 

We show that $A_{k,l,n}$ is closed.  Let $P_j\in A_{k,l,n}$ and $P_j\to P$.  For each $j\in\omega$ there are $x_j,w_j\in P_j$ such that $d(x_j,w_j)\leq 1/2^l$ and for all $i\geq n$ we have $d(f_i(x_j),f_i(w_j))\geq 1/2^k$.  Taking a subsequence if necessary we may assume that there exist $x,w\in P$ such that $\lim_{j\in\omega}\{x_j,w_j\}=\{x,w\}$ in $\comp(X)$.  Clearly, $d(x,w)\leq 1/2^l$.  For $i\geq n$ fixed the continuity of $f_i$ implies that $d(f_i(x),f_i(w))\geq 1/2^k$.  Hence, $P\in A_{k,l,n}$.  So, $A_{k,l,n}$ is closed.   



Let $E=\bigcup_{k\in\omega}\bigcap_{l\in\omega}\bigcup_{n\in\omega}A_{k,l,n}$.  Clearly, $E\in\Sigma_4^0(\comp(X))$.  We will be done if we show that $\T(f)=\comp(X)\setminus E$.  



Suppose $P\in E$.  There is a $k\in\omega$ such that  $P\in\bigcap_{l\in\omega}\bigcup_{n\in\omega}A_{k,l,n}$.  So for every $l\in\omega$ there exist $x_l,w_l\in P$ such that $d(x_l,w_l)\leq 1/2^l$ and $d(f_i(x_l),f_i(w_l))\geq 1/2^k$ for all sufficently large $i\in\omega$.  It follows that $d(f(x_l),f(w_l))\geq 1/2^k$.  Since $P$ is compact, there is a $p\in P$ such that $\lim_{l\in\omega}\{x_l,w_l\}=\{p\}$ in $\comp(X)$.  Clearly, the oscillation of $f|_P$ at $x$ is at least $1/2^k$.  Hence, $P\notin\T(f)$.



Suppose $P\notin\T(f)$.  There is a $p\in P$ and a $k\in\omega$ and a sequence $(p_l)_{l\in\omega}$ of elements of $P$ such that $d(p_l,p)\leq 1/2^l$ for every $l\in\omega$ and 

\begin{equation}\label{tent}
d(f(p_l),f(p))> 1/2^k.  
\end{equation}

Since $(f_i)_{i\in\omega}$ converges to $f$ pointwise, (\ref{tent}) implies we may find for each $l\in\omega$ a $n_l\in\omega$ such that for all $i\geq n_l$ we have 
$d(f_i(p_l),f_i(p))\geq 1/2^k$.  Thus, $P\in E$. \qed



We now show that none the implications of Theorem~\ref{thm:0000} may be reversed.



Let $\{q_n\colon n\in\omega\}$ be an enumeration of the rational numbers in $\real$.  Define $f\colon\real\to\omega$ by 
\begin{equation}\notag f(x)=
\begin{cases} n+1& \text{if $x=q_n$;}\\
0& \text{otherwise.}
\end{cases}
\end{equation}

Now $f\notin\baire_1$ since it has no point of continuity.  However, $\T(f)\in\Sigma_3^0(\comp(X))\subseteq\Pi_4^0(\comp(X))$ since $f$ is $T_1$ and has $G_{\delta}$-graph.  So, the implication $(ii)\Rightarrow (iii)$ may not be reversed.  



Let $f\colon\real\to\real$ be the characteristic function of a convergent sequence without its limit point.  Clearly, $f\in\baire_1$.  However, $f$ is not $T_0$ so $\T(f)\notin\Delta_3^0(\comp(X))$.  Since $f$ is $T_1$ and has $G_{\delta}$-graph, we have $\T(f)\in\Sigma_3^0(\comp(X))$.  Thus, $\T(f)\notin\Pi_3^0(\comp(X))$.  So, the implication $(i)\Rightarrow (ii)$ may not be reversed. 


We construct a $\baire_1$ function $f\colon\real\to\real$ such that $\T(f)\in\Pi_4^0(\comp(X))\setminus\Sigma_4^0(\comp(X))$.  For each $n\in\omega$ pick an increasing sequence $(w_{n,m})_{m\in\omega}$ in $(1-1/2^{n},1-1/2^{n+1}]$ which converges to $1-1/2^{n+1}$.  Define $f\colon\real\to\real$ by 

\begin{equation}\notag f(x)=
\begin{cases} \frac{1}{2^{n}}& \text{if $x=w_{n,m}$;}\\
0& \text{otherwise.}
\end{cases}
\end{equation}

It is easily seen that $f\in\baire_1$.  For each $n,m\in\omega$ let $(z_{n,m,l})_{l\in\omega}$ be an increasing sequence in $(1-1/2^n,w_{n,0}]$ if $m=0$ or $(w_{n,m-1},w_{n,m}]$ if $m\neq 0$, in either case let $\lim_{l\in\omega}z_{n,m,l}=w_{n,m}$.  Let $I$ be the set from Proposition~\ref{newt1}.  Define $J=\Pi_{i\in\omega}I$.  By \cite[23.3]{kech}, $J\in\Pi_5^0((2^{\omega\times\omega})^{\omega})\setminus\Sigma_5^0((2^{\omega\times\omega})^{\omega})$.  Define $h\colon (2^{\omega\times\omega})^{\omega}\to\comp(\real)$ by 

\[h(\sigma)=\{1\}\cup\left\{1-\frac{1}{2^{n+1}}\colon n\in\omega\right\}\bigcup_{n,m\in\omega}L_{n,m},\]  

where $L_{n,m}$ is defined by 

\begin{equation}\notag L_{n,m}=
\begin{cases} 
\emptyset& \text{if $\{l\colon\sigma(n)(m,l)=1\}=\emptyset$}\\ \{w_{m,n}\} & \text{if $\{l\colon\sigma(n)(m,l)=1\}$ is infinite;}\\
\{z_{n,m,\max\{l\colon\sigma(n)(m,l)=1\}}\}& \text{otherwise.}
\end{cases}
\end{equation}
By an argument similar to the one used in the proof of Lemma~\ref{lem:12}, one can show that $h$ is in $\baire_1$.  It is easy to verify that $h^{-1}(\T(f))=J$.  Since
$h\in\baire_1$, we have $\T(f)\notin\Sigma_4^0(\comp(X))$.


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\end{thebibliography}
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