The purpose of this post is to try to get my head wrapped around a couple of “analytic” concepts: the Baire property, meagreness and so forth. I’ll attempt to give full definitions and proofs, since this is mostly for my reference.

## 1. Meagreness

Let $$X$$ be a topological space. A subset $$N \subset X$$ is nowhere dense when its closure $$\overline{N}$$ has empty interior (the symbol for interior is typically something like $$N^o$$. Unaccountably, I dislike this, so I will just write “interior” when I mean it.) More generally, we’ll say that $$M \subset X$$ is meagre if there exist nowhere dense sets $$N_1,N_2,\ldots$$ such that $$M \subset \bigcup_{n\in\mathbb{N}} N_n$$. Meagreness (as the name implies) is a notion of smallness. A set is non-meagre when it is … not meagre, shocking.

On the other side, say that a set $$R \subset X$$ is residual if there are open, dense sets $$U_1,U_2,\ldots$$ such that $$R \supset \bigcap_{n \in \mathbb{N}} U_n$$. A little set-theoretic algebra should convince you that the residual sets are precisely the comeagre sets, i.e. the sets whose complement is meagre.

If we consider the real line $$\mathbb{R}$$ with the usual topology, we can see directly that the set of rational numbers $$\mathbb{Q}$$, being a countable set of points, is meagre, so the irrational numbers are comeagre, hence residual. It seems intuitively obvious that it would be difficult to write a subset of $$\mathbb{R}$$ containing a nonempty open interval as a countable union of nowhere dense sets, but proving this seems challenging.

Fortunately, we have a tool that does this for us!

### 1.1. The Baire category theorem

Subsets of $$X$$ which are meagre are often said to be “of the first category”. I’m at this moment not 100% clear on whether “of the second category” means “non-meagre” or “comeagre”—I don’t particularly need to know either, but it goes to show you that terminology can be bad! We’ll avoid “first/second category” here.

Anyway:

Theorem (Baire category theorem). A nonempty space $$X$$ is said to be a Baire space (not to be confused with the Baire space—yep, mathematicians are awful at names, sorry about it) if either of the following equivalent conditions are satisfied.

1. Nonempty open sets are non-meagre.
2. Residual sets are dense.

Suppose that $$X$$ is nonempty and completely metrizable. Then $$X$$ is Baire.

First let’s prove the equivalence of the two conditions. Suppose that residual sets are dense, but that $$U$$ is open and meagre, say contained in $$\bigcup_{n\in\mathbb{N}} N_n$$, where each $$N_n$$ is nowhere dense. Now, saying that $$N_n$$ is nowhere dense precisely says that $$V_n = X - N_n$$ contains an open and dense set. So the intersection $$\bigcap_{n\in\mathbb{N}} V_n$$ is residual, but this intersection is precisely $$X - \bigcup_{n \in \mathbb{N}} N_n$$. Therefore if residual sets are dense, $$U$$ must be empty.

Conversely, if $$R$$ is residual but not dense, since residual sets are comeagre, we have that the complement of $$R$$ is meagre but contains a nonempty open set in its interior.

Now suppose that $$X$$ is completely metrizable, in fact, choose a complete metric $$d$$ on $$X$$ generating the topology, and suppose that $$U_1,U_2,\ldots$$ are dense and open. To prove that any residual set is dense, it suffices to show that sets of the form $$\bigcap_{n\in\mathbb{N}} U_n$$ are dense. Supposing such a set is not dense, there exists a point $$x$$ with an open neighborhood $$V$$ entirely contained in the complement. Notice that because each $$U_n$$ is dense, $$x$$ is a limit point of $$U_n$$, so in particular $$V$$ contains a point $$x_n$$ of each $$U_n$$.

In fact, we can be slightly more clever: since $$V \cap U_1$$ is open and nonempty by assumption and $$X$$ is a metric space, it contains a ball $$B_1$$ of some radius $$\epsilon_1 < 1$$ centered at $$x_1$$. But then $$B_1 \cap U_2$$ is open and nonempty by the denseness of $$U_2$$ applied to $$x_1$$, so it contains a ball of some radius $$\epsilon_2 < \frac{1}{2}$$ centered at some point $$x_2$$, and we can continue so that for each $$n$$, we have that $$B_n \cap U_{n+1}$$ contains a ball of some radius $$\epsilon_n < \frac{1}{2^n}$$ centered at some point $$x_n$$. But the sequence $$\{x_n\}$$ is clearly Cauchy by construction, so by the completeness of $$d$$ it converges to some point $$x'$$. This point $$x'$$ is clearly contained in $$V$$ but also in $$\bigcap_{n \in \mathbb{N}} U_n$$, contradicting our assumption that our residual set was not dense.

## 2. The property of Baire

Neato! Okay so it gets worse: I mentioned offhand that there was, in addition to the concept of a Baire space, the Baire space, $$\mathbb{N}^\mathbb{N}$$. We’re not done with Baire, sadly. Say that a subset $$A$$ of $$X$$ has the Baire property if we may write $$A$$ as the symmetric difference $$A = U \triangle M$$, where $$U$$ is open and $$M$$ is meagre. (The symmetric difference is surprisingly easy to describe in words and annoying to describe in symbols: if $$A$$ and $$B$$ are subsets of $$X$$, their symmetric difference is just the set of points in $$A$$ or $$B$$ but not both.) Observe that $$A \triangle (A \triangle B)$$ is the set of points in $$A$$—or those points that are in $$A$$ or $$B$$ but not both—but not both. Said less confusingly, if a point is in $$A$$ and $$B$$, then it is in $$A$$ but not $$A \triangle B$$ so it belongs to $$A \triangle (A \triangle B)$$. Similarly if a point is in $$B$$ but not $$A$$, then it belongs to $$A \triangle (A \triangle B)$$. Thus we may affectionately refer to $$A \triangle (A \triangle B)$$ by its more common name, $$B$$!

Okay, dizzying set-theoretic algebra aside, another way of phrasing the property of Baire is to say that $$A$$ has the property of Baire if there exists an open set $$U$$ such that $$A \triangle U$$ is meagre. Thus open sets clearly have the property of Baire. Actually, so do closed sets: the point is that the boundary (closure minus interior) of any open set is nowhere dense; the proof is more set-theoretic trickery, but it’s not too hard so I’ll leave it to you. Therefore a closed set is either nowhere dense (hence meagre) or differs from its interior by a meagre set. Actually, this shows that the complement of a set with the property of Baire has it too: if $$A = U \triangle M$$, it turns out that $$X - A$$ is nothing but $$(X - U) \triangle M$$.

Actually, countable unions of sets with the property of Baire have it too: say $$B_i = U_i \triangle M_i$$. Write $$B = \bigcup_{i\in\mathbb{N}} B_i$$, $$U = \bigcup_{i\in\mathbb{N}} U_i$$ and $$M = \bigcup_{i \in \mathbb{N}} M_i$$. It is clear that $$U$$ is open and $$M$$ is meagre. So consider $$M' = B \triangle U$$. If $$x$$ is a point of $$M'$$ contained in $$B$$, it must not be contained in $$U$$, hence not in any $$U_i$$, hence it is contained in some $$M_i$$, hence in $$M$$. Conversely, if $$x$$ is a point of $$U$$ not contained in $$B$$, it is necessarily contained in some $$U_i$$ but not in any $$B_i$$, so in particular it must be contained in $$M_i$$, hence in $$M$$. Therefore $$B \triangle U$$ is meagre, being a subset of a meagre set.

Cool! If you know the words, you can sing along with me: sets with the property of Baire include all Borel sets and in fact form a $σ$-algebra.

### 2.1. Analytic sets

Now suppose that $$X$$ is Polish and that $$A$$ is the image of a Polish space $$Y$$ under some continuous map $$f\colon Y \to X$$. We say that $$A$$ is analytic. We claim that $$A$$ has the property of Baire. For this we need a lemma.

Lemma. Suppose $$X$$ is Polish and $$S$$ is a subset. There exists a subset $$A \supset S$$ having the property of Baire such that whenever $$Z \subset A - S$$ has the Baire property, then $$Z$$ is meagre.

Indeed, recall that Polish spaces are second countable (consider open balls of rational radius centered at a countable dense subset), so let $$\mathscr{U} = \{ U_1, U_2, \ldots \}$$ be a countable basis. Given $$S$$, write

$D(S) = \{x \in X : \text{ if } x \in U_i,\ U_i \cap S \text{ is non-meagre. }\}$

The complement of $$D(S)$$ is the set of those points $$x \in X$$ for which there exists $$U_i \in \mathscr{U}$$ containing $$x$$ such that $$U_i \cap S$$ is meagre. That is, if $$X - D(S)$$ contains $$x$$, it actually contains some $$U_i$$ containing it, so $$D(S)$$ is closed. If we can show that $$S - D(S)$$ is meagre, we will have shown that $$A = (S - D(S)) \cup D(S)$$ has the Baire property. But $$S - D(S)$$ is precisely the set of those $$S \cap U_i$$ which are meagre, so by countability of $$\mathscr{U}$$, $$S - D(S)$$ is meagre.

Suppose that $$Z \subset A - S$$ has the property of Baire. Then $$Z = U \triangle M$$ for some open set $$U$$ and meagre set $$M$$. The set $$U$$ contains $$U_i$$ because $$\mathscr{U}$$ is a basis. The set $$U_i - Z$$ is contained in $$M$$ by the kind of silly dance we did in the previous section, so is meagre. It follows that $$U_i \cap S$$ is meagre, being contained in $$U_i - Z$$. But $$U_i \cap Z \ne \varnothing$$ and $$Z \subset D(S)$$ by construction, we conclude that $$U_i \cap S$$ cannot be meagre by the definition of $$D(S)$$. The lemma follows.

Phew!

Now let’s prove that if $$A$$ is the image of a Polish space $$Y$$ in $$X$$, it has the Baire property! I’m following an exercise/proof Thomas Jech leaves for the reader in chapter 11 of Set Theory. Fix once and for all the following countable basis $$\mathscr{U}$$ for $$Y$$. We index $$\mathscr{U}$$ by the (countable) collection of finite sequences of natural numbers. Fix a complete metric on $$Y$$ inducing the topology. Begin with the empty sequence as all of $$Y$$. Then inductively, to extend a sequence $$\alpha$$ of length $$n$$, choose a countable collection of points in $$U_\alpha$$ and balls of radius at most $$\frac{1}{2^{n+1}}$$ centered at these points so that $$U_\alpha$$ is contained in the union of these balls. (This is possible because $$Y$$ has a countable dense subset.)

Given $$U_\alpha \in \mathscr{U}$$, write $$A_\alpha$$ for $$f(U_\alpha)$$. By the lemma applied to $$S = A_\alpha$$, there is a set $$B_\alpha$$ having the property of Baire containing $$A_\alpha$$ such that whenever $$Z$$ is contained in $$B_\alpha - A_\alpha$$ and has the Baire property, the set $$Z$$ is meagre. Indeed, since $$\overline{A_\alpha}$$ has the Baire property, being closed, we may (must, actually) fit $$B_\alpha$$ between $$A_\alpha$$ and $$\overline{A_\alpha}$$.

Write $$B = B_\varnothing$$. then $$B$$ has the Baire property, and it suffices to show that $$B - A$$ is meagre, since sets with the property of Baire form a $σ$-algebra containing the meagre sets.

Now, notice that in $$Y$$, each point $$y \in Y$$ is the unique point of intersection of $\{y\} = \bigcap_{n=0}^\infty U_{\omega |_n}$ where $$\omega \in \mathbb{N}^{\mathbb{N}}$$ is some infinite sequence and $$\omega|_n$$ is the sequence restricted to the first $$n$$ elements. (This uses completeness!) Therefore we have the trivially true fact that $Y = \bigcup_{\omega \in \mathbb{N}^{\mathbb{N}}} \bigcap_{n=0}^\infty U_{\omega |_n}.$ This silly dodge is called the Suslin operation, it’s often written $$\mathscr{A}$$ when considered over sets indexed by finite sequences of naturals and perhaps surprisingly it turns out to be useful! Indeed, it is similarly obvious that $$A = \mathscr{A}\{ A_\alpha \}$$ and $$A = \mathscr{A}\{ \overline{A_\alpha} \}$$, so also true that $$A = \mathscr{A}\{ B_\alpha \}$$.

Therefore we have that $B - A = B - \bigcup_{\omega\in\mathbb{N}^{\mathbb{N}}}\bigcap_{n=0}^\infty B_{\omega|_n}.$ We claim that this is a subset of $\bigcup_{\alpha}\left(B_\alpha - \bigcup_{n\in\mathbb{N}} B_{\alpha,n}\right).$ Here by $$B_{\alpha,n}$$ I mean the set obtained by appending $$n$$ to the sequence $$\alpha$$. Indeed, suppose $$x \in B$$ is not an element of this latter set. This means that whenever $$x$$ belongs to some $$B_\alpha$$, there actually exists some $$n$$ such that $$x$$ belongs to $$B_{\alpha,n}$$. But by induction, we conclude that $$x$$ belongs to some $$\bigcap_{n = 0}^\infty B_{\omega|_n}$$, so actually $$x$$ doesn’t belong to the former set either, proving the claim.

To complete the proof, it suffices to show that each set in this latter union is meagre. But this is clear, because each such set is sandwiched between $$A_\alpha$$ and $$B_\alpha$$, so is meagre by the lemma! (Indeed, each of these sets has the property of Baire, being part of the $σ$-algebra generated by the $$B_\alpha$$.)

Therefore we conclude that $$A$$ has the property of Baire.

## 3. Pettis’ Lemma

I want to conclude by proving a theorem of Pettis on topological groups.

Theorem (Pettis’ Lemma). Suppose that $$G$$ is a (Hausdorff) topological group which is a Baire space and that $$A$$ is a non-meagre subset having the property of Baire. Then $$A^{-1} A$$ is an identity neighborhood.

Indeed, Pettis states and proves it for any set merely containing a non-meagre subset with the property of Baire. His proof is … well, it’s old. It took me all of yesterday to understand what he was saying, which is why I’m writing this post. But! It’s reasonably clear now. Let’s dive in.

Since $$A$$ has the property of Baire, we may write $$A = W \triangle M$$, where $$W$$ is open and $$M$$ is meagre. Observe that $$W$$ is nonempty because $$A$$ was non-meagre. By continuity of the map $$(g,h) \mapsto gh^{-1}$$, we see that every identity neighborhood $$V$$ contains an open identity neighborhood $$U$$ such that $$UU^{-1} \subset V$$. Since general open sets are left-translates of open identity neighborhoods, we have that $$W$$ contains $$gUU^{-1}$$ for some $$g \in G$$. We claim that $$U \subset A^{-1} A$$. Indeed, take $$h \in U$$ and consider the residual set $$E_h = (G - M) \cap [(G - M)h^{-1}]$$. Since $$G$$ is a Baire space, so is the open set $$gUh^{-1}$$, which therefore intersects $$E_h$$, say in some element $$x$$. We claim that both $$x$$ and $$xh$$ are in $$W \triangle M = A$$, so that $$h$$ is in $$A^{-1} A$$. Indeed, since $$x \in E_h$$, we see that neither $$x$$ nor $$xh$$ are in $$M$$. Since $$h \in U$$, the open set $$gUh^{-1}$$ is in $$g U U^{-1} \subset W$$, so we see that $$x$$ and $$xh$$ are in $$W$$, hence in $$A$$.

Corollary. If $$A$$ is an analytic and non-meagre subset of a Polish group $$G$$, then $$A^{-1}A$$ is an identity neighborhood.