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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

  1. $A$ has the Baire property if and only if $A=F\triangle P$ where $F$ is closed and $P$ is of first category.
  2. The collection of all sets that have the Baire property is a $\sigma$-algbera.
  3. 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.
  4. 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!

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    $\begingroup$ You are very close to solving point 4: just think about this: by point 2, the complement of a "Baire set" is Baire; is the complement of an analytic set analytic? $\endgroup$
    – Pelota
    Apr 28, 2023 at 13:43
  • $\begingroup$ @OliverDíaz Any set which is $\mathbf{\Pi}^1_1$ but not $\mathbf{\Sigma}^1_1$ will work for $4$, that's what Pelota is suggesting $\endgroup$ Apr 29, 2023 at 5:27
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    $\begingroup$ Question 3 is considerably harder. There are different proofs of this result but it's hard to guess which one the book expects you to figure out. The most elementary one is probably showing that analytic sets are closed under the Souslin operation, see for example the descriptive set theory notes by Anush Tserunyan $\endgroup$ Apr 29, 2023 at 5:48

1 Answer 1

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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:

Theorem M: Suppose $X$ is a Polish space, and let $\mathcal{F}$ be the collection of closed subsets of $X$. A set $A\subset X$ is analytic iff $A=S(E)$ for some Souslin scheme $E$ on $\mathcal{F}$.

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.

Proposition B: For any $A\subset X$, there exists a set $B$ containing $A$ that has the Baire property and such that if $B'$ has the Baire property and $A\subset B'$, then $B\setminus B'$ is of first category.

A proof of this result is given at the end.

We now prove the following result from which part (3) of the problem follows:

Theorem S: Suppose $X$ is a Polish space and $\mathcal{B}$ is the collection of sets that have the Baire property. If $E$ is a Souslin scheme on $\mathcal{B}$, then $S(E)\in\mathcal{B}$.

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.

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