Baire Category

Definition 4.1   Let $X$ be a metric space. A set $E \subset X$ is nowhere dense if $\bar E$ (the closure of $E$ ) has empty interior.

Notice that if $E$ is nowhere dense, then $E^c$ is a dense set.

Example: $\mathbb{Q}$ is not nowhere dense, but the (regular) Cantor set is nowhere dense.

Definition 4.2   $E \subset X$ is of first category if $E$ is a countable union of nowhere dense sets.
A set $E \subset X$ is of second category if it is not first category.

Synonyms: A meager set is a set of first category. A non-meager set is of second category.
Examples: $\mathbb{Q}$ is first category, as is the Cantor set (hence we can have uncountable sets which are first category).

Some key theorems:

Theorem 18   Baire's Theorem:
Let $X$ be a complete metric space. Suppose for each $k \in \mathbb{N}$ , $G_k$ is a dense open set. Then, $\bigcap_{k=1}^\infty G_k$ is dense.

Proof. Fix $x \in X$ and $\epsilon > 0$ . $G_1$ is dense so there exists a $y_1 \in G_1$ and $0 < r_1 < \frac12 \epsilon$ such that $\overline{B(y_1,r_1)} \subset B_\epsilon (x) \cap G_1$ . Now, $B(r_1,y_1) \cap G_2$ is open and nonempty so there exists a $y_2$ and $0 < r_2 < \frac12 r_1$ such that

$$\overline{ B(r_2,y_2) }\subset B(r_1,y_1) \cap G_2.$$

Continue this process inductively to obtain the sequence $\{ y_i \}_{i=1}^\infty$ in $X$ such that $\overline{B(r_k,y_k)} \subset B(r_{k-1},y_{k-1}) \cap G_k$ and $0 < r_k < \frac12 r_{k-1}$ .
Claim: $y_k$ is Cauchy. For $n,m \in \mathbb{N}$ , $n, m > N$ , $y_n, y_m \in B(r_N,y_N)$ so

$$d(y_n,y_m) \leq d(y_n,y_N) + d(y_N,y_m)$$

$$< 2 r_N < 2 \left( \frac12 \right)^N \epsilon = \left( \frac12 \right)^{N-1} \epsilon .$$

Hence, there exists $y \in X$ such that $y = \displaystyle \lim_{k \rightarrow \infty} y_k$ .
For each $N \in \mathbb{N}$ ,

$$y \in \overline{ B(r_N, y_N) } \subset B(r_{N-1}, y_{N-1}) \cap G_n$$

Therefore, $y \in G_n$ for all $N$ . Further,

$$\overline{B(r_N,y_N)} \subset B(r_1,y_1) \subset B_\epsilon (x)$$

so $$\bigcap_{N=1}^\infty G_N$$ is dense in $X$ . $\qedsymbol$

Observation: Suppose that, for all $n$ , $H_n$ is a dense $G_\delta$ -set in the complete metric space $X$ . Then, $$\bigcap_{n=1}^\infty H_n$$ is dense in $X$ .

Proof. Let

$$H_n = \bigcap_{m=1}^\infty G_{n,m}$$

where $G_{n,m}$ is open. Since $H_n$ is dense, so is $G_{n,m}$ for all $n,m$ . Thus,

$$\bigcap_{n=1}^\infty H_n = \bigcap_{n,m} G_{n,m}$$

is dense by Baire's Theorem.
$\qedsymbol$

Theorem 19   Baire Category Theorem:
Let $X$ be a complete metric space. Then every open subset of $X$ is second category in $X$

Proof. Let $G$ be a non-empty open set. Let $E_n$ be a countable collection of nowhere dense sets. Then $X \setminus \bar E_n$ is dense and open, so $\bigcap_{i=1}^\infty (X \setminus \bar E_n)$ is dense. Then,

$$G \cap \left( \bigcap_{i=1}^\infty X \setminus \bar E_n \right) \neq \emptyset$$

Hence $G$ is not a subset of $\bigcup_{n=1}^\infty \bar E_n$ , so in particular $G \neq \bigcup_{n=1}^\infty E_n$ . Thus, $G$ is of second category. $\qedsymbol$

Examples:
Let $C$ be the set of real valued continuous functions on $[0,1]$ . Make $C$ a metric space by defining

$$d(f,g) = \displaystyle \sup_{x \in [0,1]} \vert f(x) - g(x)\vert.$$

Then if $\epsilon > 0$ and $f \in C$ , then the sequence $d(f_n,f) \rightarrow 0$ if and only if $f_n \rightarrow f$ uniformly on $[0,1]$ .

Recall: The derivates of $f$ :

$$(D^+ f)(x_0) = \limsup_{x \rightarrow x_0^+} \frac{f(x) - f(x_0)}{x - x_0}$$

$$(D^- f)(x_0) = \limsup_{x \rightarrow x_0^-} \frac{f(x) - f(x_0)}{x - x_0}$$

$$(D_\pm f)(x_0) = \liminf_{x \rightarrow x_0^\pm} \frac{f(x) - f(x_0)}{x-x_0}$$

Not that $f$ is differentiable at $x_0 \in (0,1)$ if and only if the derivates all agree.

Define:

$$\mathcal{F}= \{ f \in C : (D^+f)(x_0)$$    is finite for some $$x_0 \in [0,1) \}$$

Proposition 4.1   $\mathcal{F}$ is $1^{st}$ category in $C$ .

Proof. Let

$$\mathcal{F}_n = \{ f \in C : \exists x_0 \in [0,1 - \frac1n]$$    s.t. $$\vert f(x) - f(x_0)\vert \leq n(x - x_0) \forall x \in [x_0,1) \}$$

Observe that $\mathcal{F}_n \subset \mathcal{F}_{n+1} \subset \mathcal{F}$ . Actually,

$$\bigcup_{n=1}^\infty F_n = F.$$

To see this, let $f \in \mathcal{F}$ . Then $\vert D^+ f(x_0)\vert < \infty$ for some $x_0 \in [0,1)$ . Hence,

$$\vert D^+f(x_0) \vert < N_0 \in \mathbb{N}$$

Also,

$$x \in [0,1 - \frac{1}{N_1}]$$

for some $N_1$ . Hence, $f \in F_n$ if $n \geq \max\{N_0, N_1\}$ .

Claim: $\mathcal{F}_n$ is nowhere dense in $C$ . Once we show that $F_n$ is closed, given any $f \in C$ , we can add a bunch of ``zig-zags" that all have slope greater than $n$ so we remain within $\epsilon$ of $f$ . Hence, there is no nearby points of $\mathcal{F}_n$ . i.e.,

If $F_n$ is not nowhere dense, then $\bar F_n = F_n$ contains an open ball, say $B(\epsilon , f) \subset F_n$ .

We now show that $\mathcal{F}_n$ is closed. Suppose $f_k \in \mathcal{F}_n$ and $f_k \rightarrow f \in C$ . Then there exists $x_k \in [0, 1 - \frac1n]$ such that

$$\vert f_k(x) - f_k (x_k) \vert \leq n(x - x_k), \forall \in [x_k 1 )$$

By passing to a subsequence if necessary, we may assume that $x_k$ converges; put $x_0 - \lim_{k \rightarrow \infty} x_k$ . Then, $x_0 \in [0,1 - \frac1n]$ . For any $x \in [x_0, 1)$ ,

$$\vert f(x) - f(x_0)\vert \leq \vert f(x) - f_k(x)\vert + \vert f_k(x) - f_k(x_k)\vert + \vert f_k(x_k) - f(x_k)\vert + \vert f(x_k) - f(x_0)\vert$$

Given $x \in [x_0, 1)$ , if $x = x_0$ then $\vert f(x) - f(x_0)\vert \leq n (x - x_0)$ is trivial. Assume $x_0 < x < 1$ . Let $\epsilon > 0$ . Since $x_k \rightarrow x_0$ and $f$ is continuous, find $N \in \mathbb{N}$ such that if $k \geq N$ ,

$$\vert f(x_k) - f(x_0) \vert < \frac\epsilon 4$$

and

$$\vert x_k - x_0\vert < \frac\epsilon {4n}$$

and $x_k < x$ . Choose $k > N$ such that $\vert f_k(t) - f(t) \vert < \frac\epsilon 4$ for all $t \in [0,1]$ . We can do this as convergence is uniform in $C$ . Apply our inequality to get

$$\vert f(x) - f(x_0)\vert < \frac\epsilon 4 + n(x - x_k) + \frac\e... ...4 + \frac\epsilon 4 \leq \frac{3\epsilon }4 + n(x - x_0) + n\vert x_k - x\vert$$

$$\leq \epsilon + n(x - x_0).$$

$\qedsymbol$

We again would like to use the Baire Category theorem to show the existence of a differentiable function $h: \mathbb{R}\rightarrow \mathbb{R}$ such that for all $(a,b) \subset \mathbb{R}$ , there exists $x,y \in (a,b)$ such that $h'(x) > 0 > h'(y)$ .

Let

$$D = \{ f : \mathbb{R}\rightarrow \mathbb{R}: f$$    is bounded and $$\exists F : \mathbb{R}\rightarrow \mathbb{R}$$ with $$F' = f \}$$

For $f, g \in D$ , define $d(f,g) = \displaystyle \sup_{x \in \mathbb{R}} \vert f(x) - g(x) \vert$ .

Claim: $D$ is a complete metric space. If $f_n \in D$ is Cauchy, then $\{ f_n(0) \}$ converges. Let $F_n : \mathbb{R}\rightarrow \mathbb{R}$ satisfy $F_n(0) = 0$ and $F_n' = f_n$ . By a theorem from 826, $F_n$ converges uniformly to a differentiable function $F$ and $F'$ is the uniform limit of $f_n'$ . (Rudin, 7.17).

Let

$$D_0 = \{ f \in D : f^{-1}(\{ 0 \})$$    is dense in $$\mathbb{R}\}$$

Given $f \in D_0$ , let $Z_f = f^{-1}(\{ 0 \})$ .

Claim: $Z_f$ is a dense $G_\delta$ set. Proof is an exercise.

Thus, if $f_n \in D_0$ and $f_n \rightarrow f \in D$ then

$$\bigcup_{n=1}^\infty Z_{f_n} \subset Z_f$$

hence by Baire, $Z_f$ is dense and so is $f \in D_0$ . Therefore, $D_0$ is a complete metric space. If $f, g \in D_0$ then $f + g \in D_0$ . Thus,

$$Z_{f + g} \supset Z_f \cap Z_g$$

and $f + g \in D_0$ . It is obvious that if $f \in D_0$ , then $\lambda f \in D_0$ for all $\lambda \in \mathbb{R}$ . Hence, $D_0$ is a vector space.

Claim: $D_0 \neq \{ 0 \}$ .
Let $\{ d_n \}_{n = 1}^\infty$ be a countable dense subset of $(-1,1)$ . For $x \in [-1,1]$ , let

$$F(x) = \sum_{n=1}^\infty \frac{(x - d_n)^{\frac13}}{2^n}.$$

Notice the series converges uniformly, and so the limit is a continuous function. The function is increasing, so we know that it has a derivative almost everywhere. Then,

$$\frac{F(x) - F(t) }{x - t} = \sum_{n=1}^\infty = \frac{1}{[(t - d_n)^{\frac23} + (t - d_n)^\frac13(x - d_n)^\frac13 + (x - d_n)^\frac23] 2^n }$$

For $a,b \in \mathbb{R}$ ,

$$\frac12(a^2 + b^2) \leq a^2 + ab + b^2 \leq \frac32 (a^2 + b^2)$$

Let

$$K = \{ x \in (-1,1) : \sum_{n=1}^\infty \frac{1}{3(x - d_n)^\frac23 2^n}$$    converges $$\}$$

Using the DCT, one can show that if $x \in K$ , then $F'(x)$ exists and

$$F'(x) = \sum_{n=1}^\infty \frac{1}{3 (x - d_n)^{\frac23} 2^n}$$

Also if $x \in K$ ,

$$\lim_{t \rightarrow x} \frac{F(x) - F(t)}{x - t} = \infty$$

Finally, for all $x \in (-1,1)$

$$\lim_{t \rightarrow x} \frac{F(t) - F(x) }{t - x} > \frac{\sqrt{2}}3.$$

Let $G :(F(-1), F(1) ) \rightarrow (-1,1)$ be the inverse of $F$ . Then, $G$ is differentiable on $(F(-1),F(1))$ and $G'(F(x)) = \frac{1}{F'(x)}$ . Notice that $G'$ is bounded and $G'(x) = 0$ if and only if $x \in F(K)$ . So $G' = 0$ on a dense subset of $(F(-1),F(1))$ .

Monday, 2-21-2005

Let

$$g(x) = \begin{cases}(x + 1)^2 (x - 1)^2, & x \in (F(-1),F(1)) \ 0 & \text{ otherwise } \end{cases}$$

Then, $g'$ exists for all $x \in \mathbb{R}$ ; consider

$$H(x) = \begin{cases}g(G(x)), & x \in (F(-1),F(1)) \ 0, & \text{ otherwise } \end{cases}$$

Then, $H'(x) = g'(G(x))G'(x)$ for $x \in (F(-1),F(1))$ and 0 otherwise. Thus $H$ is a nonzero function in $D_0$ .
Now, let

$$E = \{ f \in D_0 : \exists$$    an interval $(a,b) \subset \mathbb{R}$ with $f(a,b) \subset [0,+\infty)$ or $f(a,b) \subset (-\infty, 0]$  $$\}$$

Claim: $E$ is first category in $D_0$ . Once we show this, there must be a function $D_0$ whose antiderivative is monotone on no subinterval on $\mathbb{R}$ . By Baire then, $D_0 \setminus E \neq \emptyset$ .
To show this, let $\{ I_n \}_{n=1}^\infty$ be an enumeration of the open intervals with rational endpoints, and put

$$A_n = \{ f \in E : f(I_n) \subset [0, + \infty) \};$$

$$B_n = \{ f \in E : f(I_n) \subset (-\infty, 0] \}.$$

Notice that

$$E = \bigcup_{n=1}^\infty A_n \cup \bigcup_{n=1}^\infty B_n,$$

so we just need to show that the $A_n$ and $B_n$ are nowhere dense.

Notice that both $A_n$ and $B_n$ are closed, so we really just need to show that the interior is empty. Let $\epsilon > 0$ and $f \in A_n$ .

Claim: $B(\epsilon , f)$ is not a subset of $A_n$ .
Since $f \in A_n$ and $Z_f$ is dense, there exists $x_0 \in I_n$ with $f(x_0) = 0$ . By translating a nonzero element of $H$ of $D_0$ , we find an element $g \in D_0$ such that $g(x_0) < 0$ (multiply by $-1$ if necessary). Let

$$M > \sup{ x \in \mathbb{R}} \vert g(x) \vert;$$

then,

$$d(f,\frac{\epsilon g}{m} + f) ) < \epsilon .$$

As both $g$ and $f$ are in $D_0$ ,

$$\frac{\epsilon g}{m} + f \in D_0.$$

But,

$$\left( \frac{\epsilon g}{M} + f\right)(x0) < 0$$

is not in $A_n$ . Finally, we get that $A_n$ is nowhere dense. Similarly, $B_n$ is nowhere dense, so $E$ is first category.

Putting everything together, we finish the proof.

Example: A complete metric space with no isolated points is uncountable ($t \in X$ is isolated if $\{ t \}$ is open).
If $X = \{ x_n \}_{n=1}^\infty$ is countable, then since each $\{ x_n \}$ is nowhere dense, we'd have $X$ first category.

Theorem 20   Uniform Boundedness:
Let $\mathcal{F}$ be a family of real valued continuous functions on a complete metric space $X$ . Suppose $\forall x \in X$ ,

$$\sup \{ \vert f(x)\vert : f \in \mathcal{F}\} =: M_x < \infty$$

Then, there exists an open set $G \subset X$ and a constant $C > 0$ such that

$$\vert f(x) \vert < C, \forall x \in G$$

and for all $f \in sF$ .

Proof. Let

$$E_{m,f} = \{ x \in X : \vert f(x)\vert \leq m \}.$$

Then $E_{m,f}$ is closed and

$$\bigcap_{f \in \mathcal{F}} E_{m,f} =: E_m$$

is also closed. By hypothesis,

$$\bigcup_{f \in \mathcal{F}} E_m = X.$$

By Baire, at least one of the $E_m$ 's has non-empty interior. Then if $E_{m^0} =: G \neq \emptyset$ , the theorem holds for this open set. $\qedsymbol$

Suppose $X,Y$ are metric spaces, $\mathcal{F}$ is a family of functions from $X$ into $Y$ . We say that $\mathcal{F}$ is equicontinuous if for all $\epsilon > 0$ , there exists a $\delta > 0$ such that $d_Y(f(x),f(y)) < \epsilon$ whenever $f \in \mathcal{F}$ and $y \in X$ with $d_X(x,y) < \delta$ .
The family is equicontinuous on $X$ if it is equicontinuous at each $x \in X$ .

Theorem 21   Ascoli-Arzela
Let $X$ be a separable metric space and $Y$ be a complete metric space. Let $\mathcal{F}$ be an equicontinuous family of functions from $X$ into $Y$ . Suppose that $(f_n)_{n=1}^\infty$ is a sequence in $\mathcal{F}$ such that $\{ f_n(x) : 1 \leq n \leq \infty \}$ is compact. Then there exists a subsequence $\{ f_{n_k} \}$ of $\{ f_n \}$ such that $f_{n_k}$ converges pointwise to a continuous function $f: X \rightarrow Y$ and if $K \subset X$ is a compact set, $f_{n_k} \rightarrow f$ uniformly on $K$ .

Remark: Notice that if $X$ is itself compact and $Y = \mathbb{R}$ . Take $\mathcal{C}\subset \mathcal{F}\subset \mathcal{C}(X)$ such that $\mathcal{F}$ is equicontinuous and pointwise bounded. Then, $\bar \mathcal{C}$ is a compact set.

Proof. Let $(f_n)$ be a sequence in $\mathcal{C}$ . By the Ascoli-Arzela theorem, there exists a subsequence which converges uniformly on $X$ . So as then $\bar \mathcal{C}$ is a metric space such that every sequence has a convergent subsequence, we conclude that $\bar \mathcal{C}$ is a compact subset of $\mathcal{C}(X)$ . $\qedsymbol$

Proof of Ascoli-Arzela is same as it has always been.