# Polish spaces and Baire spaces

Jordan Bell
June 27, 2014

## 1 Introduction

These notes consist of me working through those parts of the first chapter of Alexander S. Kechris, Classical Descriptive Set Theory, that I think are important in analysis. Denote by $\mathbb{N}$ the set of positive integers. I do not talk about universal spaces like the Cantor space $2^{\mathbb{N}}$, the Baire space $\mathbb{N}^{\mathbb{N}}$, and the Hilbert cube $[0,1]^{\mathbb{N}}$, or “localization”, or about Polish groups.

If $(X,\tau)$ is a topological space, the Borel $\sigma$-algebra of $X$, denoted by $\mathscr{B}_{X}$, is the smallest $\sigma$-algebra of subsets of $X$ that contains $\tau$. $\mathscr{B}_{X}$ contains $\tau$, and is closed under complements and countable unions, and rather than talking merely about Borel sets (elements of the Borel $\sigma$-algebra), we can be more specific by talking about open sets, closed sets, and sets that are obtained by taking countable unions and complements.

###### Definition 1.

An $F_{\sigma}$ set is a countable union of closed sets.

A $G_{\delta}$ set is a complement of an $F_{\sigma}$ set. Equivalently, it is a countable intersection of open sets.

If $(X,d)$ is a metric space, the topology induced by the metric $d$ is the topology generated by the collection of open balls. If $(X,\tau)$ is a topological space, a metric $d$ on the set $X$ is said to be compatible with $\tau$ if $\tau$ is the topology induced by $d$. A metrizable space is a topological space whose topology is induced by some metric, and a completely metrizable space is a topological space whose topology is induced by some complete metric. One proves that being metrizable and being completely metrizable are topological properties, i.e., are preserved by homeomorphisms.

If $X$ is a topological space, a subspace of $X$ is a subset of $X$ which is a topogical space with the subspace topology inherited from $X$. Because any topological space is a closed subset of itself, when we say that a subspace is closed we mean that it is a closed subset of its parent space, and similarly for open, $F_{\sigma}$, $G_{\delta}$. A subspace of a compact Hausdorff space is compact if and only if it is closed; a subspace of a metrizable space is metrizable; and a subspace of a completely metrizable space is completely metrizable if and only if it is closed.

A topological space is said to be separable if it has a countable dense subset, and second-countable if it has a countable basis for its topology. It is straightforward to check that being second-countable implies being separable, but a separable topological space need not be second-countable. However, one checks that a separable metrizable space is second-countable. A subspace of a second-countable topological space is second-countable, and because a subspace of a metrizable space is metrizable, it follows that a subspace of a separable metrizable space is separable.

A Polish space is a separable completely metrizable space. My own interest in Polish spaces is because one can prove many things about Borel probability measures on a Polish space that one cannot prove for other types of topological spaces. Using the fact (the Heine-Borel theorem) that a compact metric space is complete and totally bounded, one proves that a compact metrizable space is Polish, but for many purposes we do not need a metrizable space to be compact, only Polish, and using compact spaces rather than Polish spaces excludes, for example, $\mathbb{R}$.

## 2 Separable Banach spaces

Let $K$ denote either $\mathbb{R}$ or $\mathbb{C}$. If $X$ and $Y$ are Banach spaces over $K$, we denote by $\mathscr{B}(X,Y)$ the set of bounded linear operators $X\to Y$. With the operator norm, this is a Banach space. We shall be interested in the strong operator topology, which is the initial topology on $\mathscr{B}(X,Y)$ induced by the family $\{T\mapsto Tx:x\in X\}$. One proves that the strong operator topology on $\mathscr{B}(X,Y)$ is induced by the family of seminorms $\{T\mapsto\left\|Tx\right\|:x\in X\}$, and because this is a separating family of seminorms, $\mathscr{B}(X,Y)$ with the strong operator topology is a locally convex space. A basis of convex sets for the strong operator topology consists of those sets of the form

 $\{S\in\mathscr{B}(X,Y):\left\|Sx_{1}-T_{1}x_{1}\right\|<\epsilon,\ldots,\left% \|Sx_{n}-T_{n}x_{n}\right\|<\epsilon\},$

for $x_{1},\ldots,x_{n}\in X$, $\epsilon>0$, $T_{1},\ldots,T_{n}\in\mathscr{B}(X,Y)$.

We prove conditions under which the closed unit ball in $\mathscr{B}(X,Y)$ with the strong operator topology is Polish.11 1 Alexander S. Kechris, Classical Descriptive Set Theory, p. 14.

###### Theorem 2.

Suppose that $X$ and $Y$ are separable Banach spaces. Then the closed unit ball

 $B_{1}=\{T\in\mathscr{B}(X,Y):\left\|T\right\|\leq 1\}$

with the subspace topology inherited from $\mathscr{B}(X,Y)$ with the strong operator topology is Polish.

###### Proof.

Let $E$ be $\mathbb{Q}$ or $\{a+ib:a,b\in\mathbb{Q}\}$, depending on whether $K$ is $\mathbb{R}$ or $\mathbb{C}$, let $D_{0}$ be a countable dense subset of $X$, and let $D$ be the span of $D_{0}$ over $K$. $D$ is countable and $Y$ is Polish, so the product $Y^{D}$ is Polish. Define $\Phi:B_{1}\to Y^{D}$ by $\Phi(T)=T\circ\iota$, where $\iota:D\to X$ is the inclusion map. If $\Phi(S)=\Phi(T)$, then because $D$ is dense in $X$ and $S,T:X\to Y$ are continuous, $X=Y$, showing that $\Phi$ is one-to-one. We check that $\Phi(B_{1})$ consists of those $f\in Y^{D}$ such that both (i) if $x,y\in D$ and $a,b\in E$ then $f(ax+by)=af(x)+bf(y)$, and (ii) if $x\in D$ then $\left\|f(x)\right\|\leq\left\|x\right\|$. One proves that $\Phi(B_{1})$ is a closed subset of $Y^{D}$, and because $Y^{D}$ is Polish this implies that $\Phi(B_{1})$ with the subspace topology inherited from $Y^{D}$ is Polish. Then one proves that $\Phi:B_{1}\to\Phi(B_{1})$ is a homeomorphism, where $B_{1}$ has the subspace topology inherited from $\mathscr{B}(X,Y)$ with the strong operator topology, which tells us that $B_{1}$ is Polish. ∎

If $X$ is a Banach space over $K$, where $K$ is $\mathbb{R}$ or $\mathbb{C}$, we write $X^{*}=\mathscr{B}(X,K)$. The strong operator topology on $\mathscr{B}(X,K)$ is called the weak-* topology on $X^{*}$. Keller’s theorem22 2 Alexander S. Kechris, Classical Descriptive Set Theory, p. 64, Theorem 9.19. states that if $X$ is a separable infinite-dimensional Banach space, then the closed unit ball in $X^{*}$ with the subspace topology inherited from $X^{*}$ with the weak-* topology is homeomorphic to the Hilbert cube $[0,1]^{\mathbb{N}}$.

## 3 G-delta sets

If $(X,d)$ is a metric space and $A$ is a subset of $X$, we define

 $\mathrm{diam}(A)=\sup\{d(x,y):x,y\in A\},$

with $\mathrm{diam}(\emptyset)=0$, and if $x\in X$ we define

 $d(x,A)=\inf\{d(x,y):y\in A\},$

with $d(x,\emptyset)=\infty$. We also define

 $B_{d}(A,\epsilon)=\{x\in X:d(x,A)<\epsilon\}.$

If $X$ and $Y$ are topological spaces and $f:X\to Y$ is a function, the set of continuity of $f$ is the set of all points in $X$ at which $f$ is continuous. To say that $f$ is continuous is equivalent to saying that its set of continuity is $X$.

If $X$ is a topological space, $(Y,d)$ is a metric space, $A\subset X$, and $f:A\to Y$ is a function, for $x\in X$ we define the oscillation of $f$ at $x$ as

 $\mathrm{osc}_{f}(x)=\inf\{\mathrm{diam}(f(U\cap A)):\textrm{U is an open % neighborhood of x}\}.$

To say that $f:A\to Y$ is continuous at $x\in A$ means that for every $\epsilon>0$ there is some open neighborhood $U$ of $x$ such that $y\in U\cap A$ implies that $d(f(y),f(x))<\epsilon$, and this implies that $\mathrm{diam}(f(U\cap A))\leq 2\epsilon$. Hence if $f$ is continuous at $x$ then $\mathrm{osc}_{f}(x)=0$. On the other hand, suppose that $\mathrm{osc}_{f}(x)=0$ and let $\epsilon>0$. There is then some open neighborhood $U$ of $x$ such that $\mathrm{diam}(f(U\cap A))<\epsilon$, and this implies that $d(f(y),f(x))<\epsilon$ for every $y\in U\cap A$, showing that $f$ is continuous at $x$. Therefore, the set of continuity of $f:A\to Y$ is

 $\{x\in A:\mathrm{osc}_{f}(x)=0\}.$

As well, if $x\in X\setminus\overline{A}=\overline{A}^{c}$, then $\overline{A}^{c}$ is an open neighborhood of $x$ and $f(\overline{A}^{c}\cap A)=f(\emptyset)=\emptyset$ and $\mathrm{diam}(\emptyset)=0$, so in this case $\mathrm{osc}_{f}(x)=0$.

The following theorem shows that the set of points where a function taking values in a metrizable space has zero oscillation is a $G_{\delta}$ set.33 3 Alexander S. Kechris, Classical Descriptive Set Theory, p. 15, Proposition 3.6.

###### Theorem 3.

Suppose that $X$ is a topological space, $Y$ is a metrizable space, $A\subset X$, and $f:A\to Y$ is a function. Then $\{x\in X:\mathrm{osc}_{f}(x)=0\}$ is a $G_{\delta}$ set.

###### Proof.

Let $d$ be a metric on $Y$ that induces its topology and let $A_{\epsilon}=\{x\in X:\mathrm{osc}_{f}(x)<\epsilon\}$. For $x\in A_{\epsilon}$, there is an open neighborhood $U$ of $x$ such that $\mathrm{osc}_{f}(x)\leq\mathrm{diam}(f(U\cap A))<\epsilon$. But if $y\in U$ then $U$ is an open neighborhood of $y$ and $\mathrm{diam}(f(U\cap A))<\epsilon$, so $\mathrm{osc}_{f}(y)<\epsilon$ and hence $y\in A_{\epsilon}$, showing that $A_{\epsilon}$ is open. Finally,

 $\{x\in X:\mathrm{osc}_{f}(x)=0\}=\bigcap_{n\in\mathbb{N}}A_{1/n},$

which is a $G_{\delta}$ set, completing the proof. ∎

In a metrizable space, the only closed sets that are open are $\emptyset$ and the space itself, but we can show that any closed set is a countable intersection of open sets.44 4 Alexander S. Kechris, Classical Descriptive Set Theory, p. 15, Proposition 3.7.

###### Theorem 4.

If $X$ is a metrizable space, then any closed subset of $X$ is a $G_{\delta}$ set.

###### Proof.

Let $d$ be a metric on $X$ that induces its topology. Suppose that $A$ is a nonempty subset of $X$ and that $x,y\in X$. We have $d(x,A)\leq d(x,y)+d(y,A)$ and $d(y,A)\leq d(y,x)+d(x,A)$, so

 $|d(x,A)-d(y,A)|\leq d(x,y).$

It follows that $B_{d}(A,\epsilon)$ is open. But if $F$ is a closed subset of $X$ then check that

 $F=\bigcap_{n\in\mathbb{N}}B_{d}(F,1/n),$

which is an $F_{\sigma}$ set, completing the proof. (If we did not know that $F$ was closed then $F$ would be contained in this intersection, but need not be equal to it.) ∎

Kechris attributes the following theorem55 5 Alexander S. Kechris, Classical Descriptive Set Theory, p. 16, Theorem 3.8. to Kuratowski. It and the following theorem are about extending continuous functions from a set to a $G_{\delta}$ set that contains it, and we will use the following theorem in the proof of Theorem 7.

###### Theorem 5.

Suppose that $X$ is metrizable, $Y$ is completely metrizable, $A$ is a subspace of $X$, and $f:A\to Y$ is continuous. Then there is a $G_{\delta}$ set $G$ in $X$ such that $A\subset G\subset\overline{A}$ and a continuous function $g:G\to Y$ whose restriction to $A$ is equal to $f$.

###### Proof.

Let $G=\overline{A}\cap\{x\in X:\mathrm{osc}_{f}(x)=0\}$. Theorem 4 tells us that the first set is $G_{\delta}$ and Theorem 3 tells us that the second set is $G_{\delta}$, so $G$ is $G_{\delta}$. Because $f:A\to Y$ is continuous, $A\subset\{x\in X:\mathrm{osc}_{f}(x)=0\}$, and hence $A\subset G$.

Let $x\in G\subset\overline{A}$, and let $x_{n},t_{n}\in A$ with $x_{n}\to x$ and $t_{n}\to x$. Because $\mathrm{osc}_{f}(x)=0$, for every $\epsilon>0$ there is some open neighborhood $U$ of $x$ such that $\mathrm{diam}(f(U\cap A))<\epsilon$. But then there is some $n$ such that $k\geq n$ implies that $x_{k},t_{k}\in U$, and thus $\mathrm{diam}(f(\{x_{k},t_{k}:k\geq n\}))<\epsilon$. Hence $\mathrm{diam}(f(\{x_{k},t_{k}:k\geq n\}))\to 0$ as $n\to\infty$, and this is equivalent to the sequence $f(x_{1}),f(t_{1}),f(x_{2}),f(t_{2}),\ldots$ being Cauchy. Because $Y$ is completely metrizable this sequence converges to some $y\in Y$ and therefore the subsequence $f(x_{n})$ and the subsequence $f(t_{n})$ both converge to $y$. Thus it makes sense to define $g:G\to Y$ by

 $g(x)=\lim_{n\to\infty}f(x_{n}),$

and the restriction of $g$ to $A$ is equal to $f$. It remains to prove that $g$ is continuous.

If $U$ is an open subset of $X$, then $g(U\cap G)\subset\overline{f(U\cap A)}$, hence

 $\mathrm{diam}(g(U\cap G))\leq\mathrm{diam}(\overline{f(U\cap A)})=\mathrm{diam% }(f(U\cap A)).$

For any $x\in G$ this and $\mathrm{osc}_{f}(x)=0$ yield

 $\mathrm{osc}_{g}(x)\leq\mathrm{osc}_{f}(x)=0,$

showing that the set of continuity of $g$ is $G$, i.e. that $g$ is continuous. ∎

The following shows that a homeomorphism between subsets of metrizable spaces can be extended to a homeomorphism of $G_{\delta}$ sets.66 6 Alexander S. Kechris, Classical Descriptive Set Theory, p. 16, Theorem 3.9.

###### Theorem 6 (Lavrentiev’s theorem).

Suppose that $X$ and $Y$ are completely metrizable spaces, that $A$ is a subspace of $X$, and that $B$ is a subspace of $Y$. If $f:A\to B$ is a homeomorphism, then there are $G_{\delta}$ sets $G\supset A$ and $H\supset B$ and a homeomorphism $G\to H$ whose restriction to $A$ is equal to $f$.

###### Proof.

Theorem 5 tells us that there is a $G_{\delta}$ set $G_{1}\supset A$ and a continuous function $g_{1}:G_{1}\to Y$ whose restriction to $A$ is equal to $f$, and there is a $G_{\delta}$ set $H_{1}\supset B$ and a continuous function $h_{1}:H_{1}\to X$ whose restriction to $B$ is equal to $f^{-1}$. Let

 $R=\{(x,y)\in G_{1}\times Y:y=g_{1}(x)\},\qquad S=\{(x,y)\in X\times H_{1}:x=h_% {1}(y)\}.$

Because $g_{1}:G_{1}\to Y$ is continuous, $R$ is a closed subset of $X\times Y$, and because $h_{1}:H_{1}\to X$ is continuous, $S$ is a closed subset of $X\times Y$. Let

 $G=\pi_{X}(R\cap S),\qquad H=\pi_{Y}(R\cap S),$

where $\pi_{X}:X\times Y\to X$ and $\pi_{Y}:X\times Y\to Y$ are the projection maps. If $x\in A$ then $h_{1}(g_{1}(x))=f^{-1}(f(x))=x$, and hence $x\in G$, and if $y\in B$ then $g_{1}(h_{1}(y))=f(f^{-1}(y))=y$, and hence $y\in H$, so we have

 $A\subset G\subset G_{1},\qquad B\subset H\subset H_{1}.$

The map $E_{1}:G_{1}\to X\times Y$ defined by $E_{1}(x)=(x,g_{1}(x))$ is continuous because $g_{1}:G_{1}\to Y$ is continuous, and hence

 $E_{1}^{-1}(S)=\{x\in G_{1}:x=h_{1}(g_{1}(x))\}=G$

is a closed subset of $G_{1}$, and thus by Theorem 4 is a $G_{\delta}$ set in $G_{1}$. But $G_{1}$ is a $G_{\delta}$ subset of $X$, so $G$ is a $G_{\delta}$ set in $X$ also. Define $E_{2}:H_{1}\to X\times Y$ by $E_{2}(y)=(h_{1}(y),y)$, which is continuous because $h_{1}$ is continuous. Then

 $E_{2}^{-1}(R)=\{y\in H_{1}:y=g_{1}(h_{1}(y))\}=H$

is a closed subset of $H_{1}$, and hence is $G_{\delta}$ in $H_{1}$. But $H_{1}$ is a $G_{\delta}$ subset of $Y$, so $H_{1}$ is a $G_{\delta}$ set in $Y$ also.

Check that the restriction of $g_{1}$ to $G_{1}$ is a homeomorphism $G_{1}\to H_{1}$ whose restriction to $A$ is equal to $f$, completing the proof. ∎

If a topological space has some property and $Y$ is a subset of $X$, one wants to know conditions under which $Y$ with the subspace topology inherited from $X$ has the same property. For example, a subspace of a compact Hausdorff space is compact if and only if it is closed, and a subspace of a completely metrizable space is completely metrizable if and only if it is closed. The following theorem shows in particular that a subspace of a Polish space is Polish if and only if it is $G_{\delta}$.77 7 Alexander S. Kechris, Classical Descriptive Set Theory, p. 17, Theorem 3.11. (The statement of the theorem is about completely metrizable spaces and we obtain the conclusion about Polish spaces because any subspace of a separable metrizable space is itself separable.)

###### Theorem 7.

Suppose that $X$ is a metrizable space and $Y$ is a subset of $X$ with the subspace topology. If $Y$ is completely metrizable then $Y$ is a $G_{\delta}$ set in $X$. If $X$ is completely metrizable and $Y$ is a $G_{\delta}$ set in $X$ then $Y$ is completely metrizable.

###### Proof.

Suppose that $Y$ is completely metrizable. The map $\mathrm{id}_{Y}:Y\to Y$ is continuous, so Theorem 5 tells us that there is a $G_{\delta}$ set $Y\subset G\subset\overline{Y}$ and a continuous function $g:G\to Y$ whose restriction to $Y$ is equal to $\mathrm{id}_{Y}$. For $x\in G\subset\overline{Y}$, there are $y_{n}\in Y$ with $y_{n}\to x$, and because $g$ is continuous we get $\mathrm{id}_{Y}(y_{n})=g(y_{n})\to g(x)$, i.e. $y_{n}\to g(x)$, hence $g(x)=x$. But $g:G\to Y$ so $x\in Y$, showing that $G=Y$ and hence that $Y$ is a $G_{\delta}$ set.

Suppose that $X$ is completely metrizable and that $Y$ is a $G_{\delta}$ subset of $X$, and let $d$ be a complete metric on $X$ that is compatible with the topology of $X$; if we restrict this metric to $Y$ then it is a metric on $Y$ that is compatible with the subspace topology on $Y$ inherited from $X$, but it need not be a complete metric. Let $U_{n}$ be open sets in $X$ with $Y=\bigcap_{n\in\mathbb{N}}U_{n}$, let $F_{n}=X\setminus U_{n}$, and for $x,y\in Y$ define

 $d_{1}(x,y)=d(x,y)+\sum_{n\in\mathbb{N}}\min\left\{2^{-n},\left|\frac{1}{d(x,F_% {n})}-\frac{1}{d(y,F_{n})}\right|\right\}.$

One proves that $d_{1}$ is a metric on $Y$ and that it is compatible with the subspace topology on $Y$. Suppose that $y_{n}\in Y$ is Cauchy in $(Y,d_{1})$. Because $d\leq d_{1}$, this is also a Cauchy sequence in $(X,d)$, and because $(X,d)$ is complete, there is some $y\in X$ such that $y_{n}\to y$ in $(X,d)$. Then one proves that $y_{n}\to y$ in $(Y,d_{1})$, from which we have that $(Y,d_{1})$ is a complete metric space. ∎

## 4 Continuous functions on a compact space

If $X$ and $Y$ are topological spaces, we denote by $C(X,Y)$ the set of continuous functions $X\to Y$. If $X$ is a compact topological space and $(Y,\rho)$ is a metric space, we define

 $d_{\rho}(f,g)=\sup_{x\in X}\rho(f(x),g(x)),\qquad f,g\in C(X,Y),$

which is a metric on $C(X,Y)$, which we call the $\rho$-supremum metric. One proves that $d_{\rho}$ is a complete metric on $C(X,Y)$ if and only if $\rho$ is a complete metric on $Y$.88 8 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 124, Lemma 3.97. It follows that if $Y$ is a Banach space then so is $C(X,Y)$ with the supremum norm $\left\|f\right\|_{\infty}=\sup_{x\in X}\left\|f(x)\right\|_{Y}$.

Suppose that $X$ is a compact topological space and that $Y$ is a metrizable space. If $\rho_{1},\rho_{2}$ are metrics on $Y$ that induce its topology, then $d_{\rho_{1}},d_{\rho_{2}}$ are metrics on $C(X,Y)$, and it can be proved that they induce the same topology,99 9 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 124, Lemma 3.98. which we call the topology of uniform convergence.

Finally, if $X$ is a compact metrizable space and $Y$ is a separable metrizable space, it can be proved that $C(X,Y)$ is separable.1010 10 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 125, Lemma 3.99.

Thus, using what we have stated above, suppose that $X$ is a compact metrizable space and that $Y$ is a Polish space. Because $X$ is a compact metrizable space and $Y$ is a separable metrizable space, $C(X,Y)$ is separable. Because $X$ is a compact topological space and $Y$ is a completely metrizable space, $C(X,Y)$ is completely metrizable, and hence Polish.

## 5 C([0,1])

$C^{1}(\mathbb{R})$ consists of those functions $F:\mathbb{R}\to\mathbb{R}$ such that for each $x_{0}\in\mathbb{R}$, there is some $F^{\prime}(x_{0})\in\mathbb{R}$ such that

 $F^{\prime}(x_{0})=\lim_{x\to x_{0}}\frac{F(x)-F(x_{0})}{x-x_{0}},$

and such that this function $F^{\prime}$ belongs to $C(\mathbb{R})$. We define $C^{1}([0,1])$ to be those functions $[0,1]\to\mathbb{R}$ that are the restriction to $[0,1]$ of some element of $C^{1}(\mathbb{R})$. We shall prove that $C^{1}([0,1])$ is an $F_{\sigma\delta}$ set in $C([0,1])$.1111 11 Alexander S. Kechris, Classical Descriptive Set Theory, p. 70.

Suppose that $f\in C^{1}([0,1])$. For each $x\in[0,1]$,

## 6 Meager sets and Baire spaces

Let $X$ be a topological space. A subet $A$ of $X$ is called nowhere dense if the interior of $\overline{A}$ is $\emptyset$. A subset $A$ of $X$ is called meager if it is a countable union of nowhere dense sets. A meager set is also said to be of first category, and a nonmeager is said to be of second category. Meager is a good name for at least two reasons: it is descriptive and the word is not already used to name anything else. First category and second category are bad names for at least four reasons: the words describe nothing, they are phrases rather than single words, they suggests an ordering, and they conflict with reserving the word “category” for category theory. A complement of a meager is said to be comeager.

If $X$ is a set, an ideal on $X$ is a collection of subsets of $X$ that includes $\emptyset$ and is closed under subsets and finite unions. A $\sigma$-ideal on $X$ is an ideal that is closed under countable unions.

###### Lemma 8.

The collection of meager subsets of a topological space is a $\sigma$-ideal.

If $X$ is a topological space and $x\in X$, we say that $x$ is isolated if $\{x\}$ is open. We say $X$ is perfect if it has no isolated points, and a $T_{1}$ space if $\{x\}$ is closed for each $x\in X$. Suppose that $X$ is a perfect $T_{1}$ space and let $A$ be a countable subset of $X$. For each $x\in A$, because $X$ is $T_{1}$, the closure of $\{x\}$ is $\{x\}$, and because $X$ is perfect, the interior of $\{x\}$ is $\emptyset$, and hence $\{x\}$ is nowhere dense. $A=\bigcup_{x\in A}\{x\}$ is a countable union of nowhere dense sets, hence is meager. Thus we have proved that any countable subset of a perfect $T_{1}$ space is meager.

Suppose that $X$ is a topological space. If every comeager set in $X$ is dense, we say that $X$ is a Baire space.

###### Lemma 9.

A topological space is a Baire space if and only if the intersection of any countable family of dense open sets is dense.

We prove that open subsets of Baire spaces are Baire spaces.1212 12 Alexander S. Kechris, Classical Descriptive Set Theory, p. 41, Proposition 8.3.

###### Theorem 10.

If $X$ is a Baire space and $U$ is an open subspace of $X$, then $U$ is a Baire space.

###### Proof.

Because $U$ is open, an open subset of $U$ is an open subset of $X$ that is contained in $U$. Suppose that $U_{n}$, $n\in\mathbb{N}$, are dense open subsets of $U$. So they are each open subsets of $X$, and $U_{n}\cup(X\setminus\overline{U})$ is a dense open subset of $X$ for each $n\in\mathbb{N}$. Then because $X$ is a Baire space,

 $\bigcap_{n\in\mathbb{N}}(U_{n}\cup(X\setminus\overline{U}))=\left(\bigcap_{n% \in\mathbb{N}}U_{n}\right)\cup(X\setminus\overline{U})$

is dense in $X$. It follows that $\bigcap_{n\in\mathbb{N}}U_{n}$ is dense in $U$, showing that $U$ is a Baire space. ∎

The following is the Baire category theorem.1313 13 Alexander S. Kechris, Classical Descriptive Set Theory, p. 41, Theorem 8.4.

###### Theorem 11 (Baire category theorem).

Every completely metrizable space is a Baire space. Every locally compact Hausdorff space is a Baire space.

###### Proof.

Let $X$ be a completely metrizable space and let $d$ be a complete metric on $X$ compatible with the topology. Suppose that $U_{n}$ are dense open subsets of $X$. To show that $\bigcap_{n\in\mathbb{N}}U_{n}$ is dense it suffices to show that for any nonempty open subset $U$ of $X$,

 $\bigcap_{n\in\mathbb{N}}(U_{n}\cap U)=U\cap\bigcap_{n\in\mathbb{N}}U_{n}\neq\emptyset.$

Because $U$ is a nonempty open set it contains an open ball $B_{1}$ of radius $<1$ with $\overline{B_{1}}\subset U$. Since $U_{1}$ is dense and $B_{1}$ is open, $B_{1}\cap U_{1}\neq\emptyset$ and is open because both $B_{1}$ and $U_{1}$ are open. As $B_{1}\cap U_{1}$ is a nonempty open set it contains an open ball $B_{2}$ of radius $<\frac{1}{2}$ with $\overline{B_{2}}\subset B_{1}\cap U_{1}$. Suppose that $n>1$ and that $B_{n}$ is an open ball of radius $<\frac{1}{n}$ with $\overline{B_{n}}\subset B_{n-1}\cap U_{n-1}$. Since $U_{n}$ is dense and $B_{n}$ is open, $B_{n}\cap U_{n}\neq\emptyset$ and is open because both $B_{n}$ and $U_{n}$ are open. As $B_{n}\cap U_{n}$ is a nonempty open set it contains an open ball $B_{n+1}$ of radius $<\frac{1}{n+1}$ with $\overline{B_{n+1}}\subset B_{n}\cap U_{n}$. Then, we have $B_{n+1}\subset B_{n}$ for each $n\in\mathbb{N}$. Letting $x_{i}$ be the center of $B_{i}$, we have $d(x_{j},x_{i})<\frac{1}{i}$ for $j>i$, and hence $x_{i}$ is a Cauchy sequence. Since $(X,d)$ is a complete metric space, there is some $x\in X$ such that $x_{i}\to x$. For any $m$ there is some $i_{0}$ such that $i\geq i_{0}$ implies that $d(x_{i},x)<\frac{1}{m}$, and hence $x\in B_{m}=\bigcap_{n=1}^{m}B_{n}$. Therefore

 $x\in\bigcap_{n\in\mathbb{N}}B_{n}\subset\bigcap_{n\in\mathbb{N}}(U_{n}\cap U),$

which shows that $\bigcap_{n\in\mathbb{N}}U_{n}$ is dense and hence that $X$ is a Baire space.

Let $X$ be a locally compact Hausdorff space. Suppose that $U_{n}$ are dense open subsets of $X$ and that $U$ is a nonempty open set. Let $x_{1}\in U$, and because $X$ is a locally compact Hausdorff space there is an open neighborhood $V_{1}$ of $x_{1}$ with $\overline{V_{1}}$ compact and $\overline{V_{1}}\subset U$. Since $U_{1}$ is dense and $V_{1}$ is open, there is some $x_{2}\in V_{1}\cap U_{1}$. As $V_{1}\cap U_{1}$ is open, there is an open neighborhood $V_{2}$ of $x_{2}$ with $\overline{V_{2}}$ compact and $\overline{V_{2}}\subset V_{1}\cap U_{1}$. Thus, $\overline{V_{n}}$ are compact and satisfy $\overline{V_{n+1}}\subset\overline{V_{n}}$ for each $n$, and hence

 $\bigcap_{n\in\mathbb{N}}\overline{V_{n}}\neq\emptyset.$

This intersection is contained in $\bigcap_{n\in\mathbb{N}}(U_{n}\cap U)$ which is therefore nonempty, showing that $\bigcap_{n\in\mathbb{N}}U_{n}$ is dense and hence that $X$ is a Baire space. ∎

## 7 Nowhere differentiable functions

From what we said in §4, because $[0,1]$ is a compact metrizable space and $\mathbb{R}$ is a Polish space, $C([0,1])=C([0,1],\mathbb{R})$ with the topology of uniform convergence is Polish. This topology is induced by the norm $\left\|f\right\|_{\infty}=\sup_{x\in[0,1]}|f(x)|$, with which $C([0,1])$ is thus a separable Banach space.

For a function $F:\mathbb{R}\to\mathbb{R}$ to be differentiable at a point $x_{0}$ means that there is some $F^{\prime}(x_{0})\in\mathbb{R}$ such that

 $\lim_{x\to x_{0}}\frac{F(x)-F(x_{0})}{x-x_{0}}=F^{\prime}(x_{0}).$

If $f:[0,1]\to\mathbb{R}$ is a function and $x_{0}\in[0,1]$, we say that $f$ is differentiable at $x_{0}$ if there is some function $F:\mathbb{R}\to\mathbb{R}$ that is differentiable at $x_{0}$ and whose restriction to $[0,1]$ is equal to $f$, and we write $f^{\prime}(x_{0})=F^{\prime}(x_{0})$. The purpose of speaking in this way is to be precise about what we mean by $f$ being differentiable at the endpoints of the interval $[0,1]$.

If $f:[0,1]\to\mathbb{R}$ is differentiable at $x_{0}\in[0,1]$, then there is some $\delta>0$ such that if $0<|x-x_{0}|<\delta$ and $x\in[0,1]$, then

 $\left|\frac{f(x)-f(x_{0})}{x-x_{0}}-f^{\prime}(x_{0})\right|<1,$

and hence

 $|f(x)-f(x_{0})|<(1+|f^{\prime}(x_{0})|)|x-x_{0}|.$

On the other hand, if $f\in C([0,1])$ then $\{x\in[0,1]:|x-x_{0}|\geq\delta\}$ is a compact set on which $x\mapsto\frac{f(x)-f(x_{0})}{x-x_{0}}$ is continuous, and hence the absolute value of this function is bounded by some $M$. Thus, if $|x-x_{0}|\geq\delta$ and $x\in[0,1]$, then

 $\left|\frac{f(x)-f(x_{0})}{x-x_{0}}\right|\leq M,$

hence

 $|f(x)-f(x_{0})|\leq M|x-x_{0}|.$

Therefore, if $f\in C([0,1])$ is differentiable at $x_{0}\in[0,1]$ then there is some positive integer $N$ such that

 $|f(x)-f(x_{0})|\leq N|x-x_{0}|,\qquad x\in[0,1].$

For $N\in\mathbb{N}$, let $E_{N}$ be those $f\in C([0,1])$ for which there is some $x_{0}\in[0,1]$ such that

 $|f(x)-f(x_{0})|\leq N|x-x_{0}|,\qquad x\in[0,1].$

We have established that if $f\in C([0,1])$ and there is some $x_{0}\in[0,1]$ such that $f$ is differentiable at $x_{0}$, then there is some $N\in\mathbb{N}$ such that $f\in E_{N}$. Therefore, the set of those $f\in C([0,1])$ that are differentiable at some point in $[0,1]$ is contained in

 $\bigcup_{N\in\mathbb{N}}E_{N},$

and hence to prove that the set of $f\in C([0,1])$ that are nowhere differentiable is comeager in $C([0,1])$, it suffices to prove that each $E_{N}$ is nowhere dense. To show this we shall follow the proof in Stein and Shakarchi.1414 14 Elias M. Stein and Rami Shakarchi, Functional Analysis, p. 163, Theorem 1.5.

###### Lemma 12.

For each $N\in\mathbb{N}$, $E_{N}$ is a closed subset of the Banach space $C([0,1])$.

###### Proof.

$C([0,1])$ is a metric space, so to show that $E_{N}$ is closed it suffices to prove that if $f_{n}\in E_{N}$ is a sequence tending to $f\in C([0,1])$, then $f\in E_{N}$. For each $n$, let $x_{n}\in[0,1]$ be such that

 $|f_{n}(x)-f_{n}(x_{n})|\leq N|x-x_{n}|,\qquad x\in[0,1].$

Because $x_{n}$ is a sequence in the compact set $[0,1]$, it has subsequence $x_{a(n)}$ that converges to some $x_{0}\in[0,1]$. For all $x\in[0,1]$ we have

 $\displaystyle|f(x)-f(x_{0})|$ $\displaystyle\leq$ $\displaystyle|f(x)-f_{a(n)}(x)|+|f_{a(n)}(x)-f_{a(n)}(x_{0})|$ $\displaystyle+|f_{a(n)}(x_{0})-f(x_{0})|.$

Let $\epsilon>0$. Because $\left\|f_{n}-f\right\|_{\infty}\to 0$, there is some $n_{0}$ such that when $n\geq n_{0}$, the first and third terms on the right-hand side are each $<\epsilon$. For the second term on the right-hand side, we use

 $|f_{a(n)}(x)-f_{a(n)}(x_{0})|\leq|f_{a(n)}(x)-f_{a(n)}(x_{a(n)})|+|f_{a(n)}(x_% {a(n)})-f_{a(n)}(x_{0})|.$

But $f_{a(n)}\in E_{N}$, so this is $\leq$

 $N|x-x_{a(n)}|+N|x_{a(n)}-x_{0}|.$

Putting everything together, for $n\geq n_{0}$ we have

 $|f(x)-f(x_{0})|<2\epsilon+N|x-x_{a(n)}|+N|x_{a(n)}-x_{0}|.$

Because $x_{a(n)}\to x_{0}$, we get

 $|f(x)-f(x_{0})|\leq 2\epsilon+N|x-x_{0}|.$

But this is true for any $\epsilon>0$, so

 $|f(x)-f(x_{0})|\leq N|x-x_{0}|,$

showing that $f\in E_{N}$. ∎

For $M\in\mathbb{N}$ let $P_{M}$ be the set of those $f\in C([0,1])$ that are piecewise linear and whose line segments have slopes with absolute value $\geq M$. If $M,N\in\mathbb{N}$, $M>N$, and $f\in P_{M}$, then for any $x_{0}\in[0,1]$, this $x_{0}$ is the abscissa of a point on at least one line segment whose slope has absolute value $\geq M$ (the point will be on two line segments when it is their common endpoint), and then there is another point on this line segment, with abscissa $x$, such that $|f(x)-f(x_{0})|\geq M|x-x_{0}|>N|x-x_{0}|$, and the fact that for every $x_{0}\in[0,1]$ there is such $x\in[0,1]$ means that $f\not\in E_{N}$. Therefore, if $M>N$ then $P_{M}\cap E_{N}=\emptyset$.

###### Lemma 13.

For each $M\in\mathbb{N}$, $P_{M}$ is dense in $C([0,1])$.

###### Proof.

Let $f\in C([0,1])$ and $\epsilon>0$. Because $f$ is continuous on the compact set $[0,1]$ it is uniformly continuous, so there is some positive integer $n$ such that $|x-y|\leq\frac{1}{n}$ implies that $|f(x)-f(y)|\leq\epsilon$. We define $g:[0,1]\to\mathbb{R}$ to be linear on the intervals $[\frac{k}{n},\frac{k+1}{n}]$, $k=0,\ldots,n-1$ and to satisfy

 $g\left(\frac{k}{n}\right)=f\left(\frac{k}{n}\right),\qquad k=0,\ldots,n.$

This nails down $g$, and for any $x\in[0,1]$ there is some $k=0,\ldots,n-1$ such that $x$ lies in the interval $[\frac{k}{n},\frac{k+1}{n}]$. But since $g$ is linear on this interval and we know its values at the endpoints, for any $y$ in this interval we have

 $\displaystyle g(y)$ $\displaystyle=$ $\displaystyle\frac{f\left(\frac{k+1}{n}\right)-f\left(\frac{k}{n}\right)}{% \frac{k+1}{n}-\frac{k}{n}}y+f\left(\frac{k}{n}\right)-\frac{f\left(\frac{k+1}{% n}\right)-f\left(\frac{k}{n}\right)}{\frac{k+1}{n}-\frac{k}{n}}\cdot\frac{k}{n}$ $\displaystyle=$ $\displaystyle n\left(f\left(\frac{k+1}{n}\right)-f\left(\frac{k}{n}\right)% \right)y+f\left(\frac{k}{n}\right)-k\left(f\left(\frac{k+1}{n}\right)-f\left(% \frac{k}{n}\right)\right),$

so

 $\displaystyle|g(x)-f(x)|$ $\displaystyle\leq$ $\displaystyle|g(x)-g(k/n)|+|g(k/n)-f(k/n)|+|f(k/n)-f(x)|$ $\displaystyle=$ $\displaystyle|g(x)-f(k/n)|+|f(k/n)-f(x)|$ $\displaystyle=$ $\displaystyle n\left|\left(f\left(\frac{k+1}{n}\right)-f\left(\frac{k}{n}% \right)\right)\left(x-\frac{k}{n}\right)\right|+|f(k/n)-f(x)|$ $\displaystyle\leq$ $\displaystyle\left|f\left(\frac{k+1}{n}\right)-f\left(\frac{k}{n}\right)\right% |+|f(k/n)-f(x)|$ $\displaystyle\leq$ $\displaystyle 2\epsilon.$

This is true for all $x\in[0,1]$, so

 $\left\|g-f\right\|_{\infty}\leq 2\epsilon.$

Now that we know that we can approximate any $f\in C([0,1])$ with continuous piecewise linear functions, we shall show that we can approximate any continuous piecewise linear function with elements of $P_{M}$, from which it will follow that $P_{M}$ is dense in $C([0,1])$. Let $g$ be a continuous piecewise linear function. We can write $g$ in the following way: there is some positive integer $n$ and $a_{0},\ldots,a_{n-1},b_{0},\ldots,b_{n-1}\in\mathbb{R}$ such that $g$ is linear on the intervals $[\frac{k}{n},\frac{k+1}{n}]$, $k=0,\ldots,n-1$, and satisfies $g(x)=a_{k}x+b_{k}$ for $x\in[\frac{k}{n},\frac{k+1}{n}]$; this can be satisfied precisely when $a_{k}\frac{k+1}{n}+b_{k}=a_{k+1}\frac{k+1}{n}+b_{k+1}$ for each $k=0,\ldots,n-1$. For $\epsilon>0$, let

 $\phi_{\epsilon}(x)=g(x)+\epsilon,\qquad\psi_{\epsilon}(x)=g(x)-\epsilon,\qquad x% \in[0,1].$

We shall define a function $h:[0,1]\to\mathbb{R}$ by describing its graph. We start at $(0,g(0))$, and then the graph of $h$ is a line segment of slope $M$ until it intersects the graph of $\phi_{\epsilon}$, at which point the graph of $h$ is a line segment of slope $-M$ until it intersects the graph of $\psi_{\epsilon}$. We repeat this until we hit the point $(\frac{1}{n},h(\frac{1}{n}))$; we remark that it need not be the case that $h(\frac{1}{n})=g(\frac{1}{n})$. If $(\frac{1}{n},h(\frac{1}{n}))$ lies on the graph of $\phi_{\epsilon}$ then we start a line segment of slope $-M$, and if it lies on the graph of $\psi_{\epsilon}$ then we start a line segment of slope $M$, and otherwise we continue the existing line segment until it intersects $\phi_{\epsilon}$ or $\psi_{\epsilon}$ and we repeat this until the point $(\frac{2}{n},h(\frac{2}{n}))$, and then repeat this procedure. This constructs a function $h\in P_{M}$ such that $\left\|h-g\right\|_{\infty}\leq\epsilon$. But for any $f\in C([0,1])$ and $\epsilon>0$, we have shown that there is some continuous piecewise linear $g$ such that $\left\|g-f\right\|_{\infty}<\epsilon$, and now we know that there is some $h\in P_{M}$ such that $\left\|h-g\right\|_{\infty}<\epsilon$, so $\left\|h-f\right\|_{\infty}<2\epsilon$, showing that $P_{M}$ is dense in $C([0,1])$. ∎

Let $N\in\mathbb{N}$, suppose that $f\in E_{N}$, and let $\epsilon>0$. Let $M>N$, and because $P_{M}$ is dense in $C([0,1])$, there is some $h\in P_{M}$ such that $\left\|f-h\right\|_{\infty}<\epsilon$. But $P_{M}\cap E_{N}=\emptyset$ because $M>N$, so $h\not\in E_{N}$, showing that there is no open ball with center $f$ that is contained in $E_{N}$, which shows that $E_{N}$ has empty interior. But we have shown that $E_{N}$ is closed, so the interior of the closure of $E_{N}$ is empty, namely, $E_{N}$ is nowhere dense, which completes the proof.

## 8 The Baire property

Suppose that $X$ is a topological space and that $\mathscr{I}$ is the $\sigma$-ideal of meager sets in $X$. For $A,B\subset X$, write

 $A\bigtriangleup B=(A\setminus B)\cup(B\setminus A).$

We write $A=^{*}B$ if $A\bigtriangleup B\in\mathscr{I}$. One proves that if $A=^{*}B$ then $X\setminus A=^{*}X\setminus B$, and that if $A_{n}=^{*}B_{n}$ then $\bigcap_{n\in\mathbb{N}}A_{n}=^{*}\bigcap_{n\in\mathbb{N}}B_{n}$ and $\bigcup_{n\in\mathbb{N}}A_{n}=^{*}\bigcup_{n\in\mathbb{N}}B_{n}$. A subset $A$ of $X$ is said to have the Baire property if there is an open set $U$ such that $A=^{*}U$. (It is a common practice to talk about things that are equal to a thing that is somehow easy to work with modulo things that are considered small.) The following theorem characterizes the collection of subsets with the Baire property of a topological space.1515 15 Alexander S. Kechris, Classical Descriptive Set Theory, p. 47, Proposition 8.22.

###### Theorem 14.

Let $X$ be a topological space and let $\mathscr{B}$ be the collection of subsets of $X$ with the Baire property. Then $\mathscr{B}$ is a $\sigma$-algebra on $X$, and is the algebra generated by all open sets and all meager sets.

###### Proof.

If $F$ is closed, then $F\setminus\mathrm{Int}(F)$ is closed and has empty interior, so is nowhere dense and therefore meager. Thus, if $F$ is closed then $F=^{*}\mathrm{Int}(F)$.

$\emptyset=^{*}\emptyset$ and $\emptyset$ is open so $\emptyset$ has the Baire property, and so belongs to $\mathscr{B}$. Suppose that $B\in\mathscr{B}$. This means that there is some open set $U$ such that $B=^{*}U$, which implies that $X\setminus B=^{*}X\setminus U$. But $X\setminus U$ is closed, hence $X\setminus U=^{*}\mathrm{Int}(X\setminus U)$, so $X\setminus B=^{*}\mathrm{Int}(X\setminus U)$. As $\mathrm{Int}(X\setminus U)$ is open, this shows that $X\setminus B$ has the Baire property, that is, $X\setminus B\in\mathscr{B}$.

Suppose that $B_{n}\in\mathscr{B}$. So there are open sets $U_{n}$ such that $B_{n}=^{*}U_{n}$, and it follows that $\bigcup_{n\in\mathbb{N}}B_{n}=^{*}\bigcup_{n\in\mathbb{N}}U_{n}$. The union on the right-hand side is open, so $\bigcup_{n\in\mathbb{N}}$ has the Baire property and thus belongs to $\mathscr{B}$. This shows that $\mathscr{B}$ is a $\sigma$-algebra.

Suppose that $\mathscr{A}$ is an algebra containing all open sets and all meager sets, and let $B\in\mathscr{B}$. Because $B$ has the Baire property there is some open set $U$ such that $B=^{*}U$, which means that $M=B\bigtriangleup U=(B\setminus U)\cup(U\setminus B)$ is meager. But $B=M\bigtriangleup U=(M\setminus U)\cup(U\setminus M)$, and because $\mathscr{A}$ is an algebra and $U,M\in\mathscr{A}$ we get $B\in\mathscr{A}$, showing that $\mathscr{B}\subset\mathscr{A}$. ∎

If $X_{n}$ is a sequence of sets, we call $A\subset\prod_{n\in\mathbb{N}}X_{n}$ a tail set if for all $(x_{n})\in A$ and $(y_{n})\in\prod_{n\in\mathbb{N}}X_{n}$, $\{n\in\mathbb{N}:y_{n}\neq x_{n}\}$ being finite implies that $(y_{n})\in A$. The following theorem states is a topological zero-one law,1616 16 Alexander S. Kechris, Classical Descriptive Set Theory, p. 55, Theorem 8.47. whose proof uses the Kutatowski-Ulam theorem,1717 17 Alexander S. Kechris, Classical Descriptive Set Theory, p. 53, Theorem 8.41. which is about meager sets in a product of two second-countable topological spaces. Since, from the Baire category theorem, any completely metrizable space is a Baire space and a separable metrizable space is second-countable, we can in particular use the following theorem when the $X_{n}$ are Polish spaces.

###### Theorem 15.

Suppose that $X_{n}$ is a sequence of second-countable Baire spaces. If $A\subset\prod_{n\in\mathbb{N}}X_{n}$ has the Baire property and is a tail set, then $A$ is either meager or comeager.