I am preparing for my Analysis qualifier. This problem is not from a past exam but from the text book Bruckner, A. W., et. al., Real Analysis (2nd edition), page 459.
A set in $A$ a topological space $X$ has the Baire property if $A=G\triangle P$ where $G$ is open, and $P$ is a set of first category (meager). The problem asks to prove that
- $A$ has the Baire property if and only if $A=F\triangle P$ where $F$ is closed and $P$ is of first category.
- The collection of all sets that have the Baire property is a $\sigma$-algbera.
- If $X$ is a complete separable metric space, and $A$ is an analytic set (the continuous image of $\mathbb{N}^\mathbb{N}$), $A$ has the Baire property.
- Give an example of a set in a complete separable metric space that is not analytic, but has the Baire property.
With my fellow students we worked out parts 1 and 2. This involves the fact that the boundary $\partial G$ of an open (or closed set) $G$ is nowhere dense, and bunch of properties of the symmetric property.
Edit: I would like to ask for some hints, solution or references for parts 3 and 4. We have tried to use the fact, proved in the textbook, that in Polish spces, a set is analytic if and only if $A$ is the Souslin set of of a family of closed sets.
Thank you!
Best Answer
Here is a solution that uses the fact that analytic sets are the Souslin schemes of closed sets (see Theorem 11.19 of the book referenced in the OP).
Let $\mathcal{N}:=\mathbb{N}^{\mathbb{N}}$ denote the space of sequences of natural numbers, and $\mathbb{N}^{<\omega}=\bigcup_{n\in\mathbb{Z}_+}\mathbb{N}^n$, where $\mathbb{N}^0:=\{\emptyset\}$. For any $f\in\mathcal{N}$, let $f|_\ell:=(f(1),\ldots,f(\ell))\in\mathbb{N}^\ell$ if $\ell\geq1$, and $f|_\ell=\emptyset$ if $\ell=0$. This is the restriction of $f$ to $\{1,\ldots,\ell\}$. Elements of $\mathbb{N}^{<\omega}$ can be thought as strings of integers of finite length, $\emptyset$ being the string of length $0$.
Given a class $\mathcal{E}$ of subsets of $X$, a Souslin scheme $E$ is function from $\mathbb{N}^{<\omega}$ into $\mathcal{E}$ with $E_\emptyset:=X$. For each Souslin scheme $E$ on $\mathcal{E}$, we define \begin{align} S(E):=\bigcup_{f\in\mathcal{N}}\bigcap^\infty_{\ell=0} E_{f|_\ell}=\bigcup_{f\in\mathcal{N}}\bigcap^\infty_{\ell=1} E_{(f(1),\ldots,f(\ell))}.\tag{0}\label{zero} \end{align} When $X$ is a Polish space (separable and complete metric space) and $\mathcal{E}$ is closed under finite intersections, then we can (and will) assume that $E_{f|_k}\subset E_{f|_m}$ for $m<k$, for we can define the scheme $E'$ as $E'_{(f(1),\ldots,f(k))}=\bigcap^k_{j=1}E_{(f(1),\ldots,f(j))}$ to get $S(E)=S(E')$.
The big result about analytic sets as defined in Bruckner is summarized in the following result:
This is by no means a trivial result (it takes 15 pages of prepping material to get to it and prove it). I don't know if a more direct solution exists. Part (3) will follow if we show that the collection $\mathcal{B}$ of sets with the Baire property is closed under the Souslin operation $S$, that is, if $E$ is a Souslin scheme on $\mathcal{B}$, then $S(E)\in\mathcal{B}$.
The following auxiliary result will be very useful in the remainder of this post.
A proof of this result is given at the end.
We now prove the following result from which part (3) of the problem follows:
Proof: Suppose $E$ is a scheme on $\mathcal{B}$. Since $\mathcal{B}$ is a $\sigma$-algebra, we may assume without loss of generality that $E_{f|_k}\subset E_{f|_m}$ for all $f\in\mathcal{N}$ and $m\leq k$. Let $\phi=(k_1,\ldots,k_n)\in \mathbb{N}^{<\omega}$ be a string of integers of length $n$, $n\in\mathbb{Z}_+=\{0\}\cup\mathbb{N}$ (recall that when $n=0$, $\phi=\emptyset$, $f|_0=\emptyset$, and $E_\phi=X$). Let $\mathcal{N}_\phi=\{f\in\mathcal{N}:f|_n=\phi\}$, and define \begin{align} E^\phi=\bigcup_{f\in\mathcal{N}^\phi}\bigcap^\infty_{\ell=1}E_{(f(1),\ldots,f(\ell))}.\tag{1}\label{one} \end{align} Observe that $E^\emptyset=S(E)$, and that $E^\phi\subset E_\phi$ for all $\phi\in\mathbb{N}^{<\omega}$. Furthermore, if $\psi=(k_1,\ldots,k_m)\subset \phi=(k_1,\ldots,k_m,k_{m+1},\ldots,k_n)$, then $E^\phi\subset E^\psi$.
By Proposition B, for each $\phi\in \mathbb{N}^{<\omega}$ there is a set $B^\phi\in \mathcal{B}$ such that $E^\phi\subset B^\phi$ and such that for any other $B'\in \mathcal{B}$ that contains $E^\phi$, $B^\phi\setminus B'$ is of first category. By taking $B^\phi\cap E_\phi$ if necessary, we may assume without loss of generality that $$E^\phi\subset B^\phi\subset E_\phi.$$ Furthermore, since $E^\phi\subset E^\psi$ whenever $\psi\subset \phi$, by taking $\bigcap_{\psi\subset \phi}B^\psi$ in place of $B^\phi$, we may assume that $$ B^\phi\subset B^\psi\qquad \text{whenever}\quad\psi\subset \phi.$$ Define \begin{align} C^\phi:=B^\phi\setminus \bigcup^\infty_{\ell=1}B^{\phi:\ell}\tag{2}\label{two} \end{align} where $\phi:\ell$ is the string of length $n+1$ obtained by concatenation, that is, if $\phi=(k_1,\ldots,k_n)$, then $\phi:\ell=(k_1,\ldots,k_n,\ell)$. It is clear that $$E^\phi= \bigcup^\infty_{\ell=1}E^{\phi:\ell}\subset\bigcup^\infty_{\ell=1}B^{\phi:\ell}\in\mathcal{B}. $$ Hence, $C^\phi$ is of first category. Since $\mathbb{N}^{<\omega}$ is countable, the set \begin{align} C:=\bigcup_{\phi\in\mathbb{N}^{<\omega}}C^{\phi}\tag{3}\label{three} \end{align} is of first category.
Claim: $B^\emptyset\setminus C\subset E^\emptyset$. Notice first that from \eqref{two} and \eqref{three}, $$B^\emptyset\setminus C\subset B^\emptyset\setminus C^\emptyset\subset \bigcup^\infty_{\ell=1}B^{(\ell)}.$$ Thus, if $x\in B^\emptyset\setminus C$, then there is $f_1\in\mathbb{N}$ such that $x\in B^{(f_1)}$. By induction, suppose we have found $\sigma=(f_1,\ldots,f_n)$ such that $x\in B^{(f_1,\ldots f_n)}$. Again, by \eqref{two} and \eqref{three}, $$ B^{(f_1,\ldots f_n)}\setminus C\subset B^{(f_1,\ldots f_n)}\setminus C^{(f_1,\ldots f_n)}\subset \bigcup^\infty_{\ell=1}B^{(f_1,\ldots f_n,\ell)}, $$ thus there is $f_{n+1}\in\mathbb{N}$ such that $x\in B^{(f_1,\ldots,f_n,f_{n+1})}$. Continuing this way, we obtain a function $f\in\mathcal{N}$ such that $$ x\in \bigcap^\infty_{\ell}B^{f|_\ell}\subset\bigcap^\infty_{\ell=1}E_{f|_\ell}\subset S(E)=E^\emptyset. $$ This concludes the proof of the claim.
To finish the proof of Theorem S, observe that $B^\emptyset\setminus E^\emptyset$ is of first category since $B^\emptyset\setminus E^\emptyset\subset C$. Consequently, $ B^\emptyset\setminus E^\emptyset\in\mathcal{B}$ and so, $B^\emptyset\setminus(B^\emptyset\setminus E^\emptyset)=B^\emptyset\cap E^\emptyset= E^\emptyset\in \mathcal{B}$.
Proof of Proposition B: Let $\{B_n:n\in\mathbb{N}\}$ be a countable basis for $X$. Let $V=\bigcup \{B_n: B_n\cap A \,\text{is of first category}\}$, and define $A_1=X\setminus V$. Clearly $A_1$ is closed in $X$ and $A\setminus A_1=A\cap V$ is of first category, for it is the countable union of sets of first category. The set $B=A\cup A_1=A_1\cup(A\setminus A_1)$ has the Baire property.
Suppose $B'$ is another set that contains $A$ and that has the Baire property. As the collection of sets that have the Baire property form a $\sigma$-algebra, this is part (2) of the problem in the OP, $C=B\setminus B'$ has the Baire property.
Claim: $C$ is of first category. Otherwise, $C$ is of second category category and by part (1) and (2) of the problem in the OP
$$A\cup (X\setminus A_1)\subset X\setminus C=F\triangle P.$$ Let $B_n$ be an element in the basis such that $B_n\cap F=\emptyset$. Then $B_n\setminus C$ is of first category, which in turn implies that $B_n\cap A$ is of first category. Since $X$ is a Polish space and $B_n\setminus C$ is of first category, $\operatorname{int}(B_n\setminus C)=\emptyset$; hence, $B_n\cap C\neq\emptyset$. As $C\subset A_1$, $B_n\cap A_1\neq\emptyset$ and so $B_n\cap A$ is of second category, this is a contradiction!
The set $B$ defined above satisfies the desired properties.
Comment: As suggested by Pelota, part (4) follows by considering an analytic $A$ set whose complement is not analytic.