Chapter 08 Additional Topics

8.2 Metric Spaces and the Baire Category Theorem

Definition 8.2.1.

Given a set $X$, a function $d : X × X → R$ is a metric on $X$ if for all $x,y ∈ X$:

(i) $d(x,y) ≥ 0$ with $d(x,y) = 0$ if and only if $x = y$,

(ii) $d(x,y) = d(y,x)$, and

(iii) for all $z ∈ X$, $d(x,y) ≤ d(x,z)+d(z,y)$.

A metric space is a set $X$ together with a metric $d$.

Exercise 8.2.1.

Decide which of the following are metrics on $X = R^2$. For each, we let $x = (x_1,x_2)$ and $y = (y_1,y_2)$ be points in the plane.

(a) $d(x,y) = \sqrt[]{(x_1−y_1)^2 +(x_2−y_2)^2}$.

Proof: Yes.

(i)

$$\sqrt[]{(x_1−y_1)^2 +(x_2−y_2)^2} \geq 0$$

It's equal to $0$, only when $x_1 = y_1, x_2 = y_2$.

(ii) yes.

(iii) Yes. It's easier to see with a plot. Triangle 2 edges are longer than 1 edge.

$\square$

(b) $d(x,y) = \max \{|x_1−y_1|,|x_2−y_2|\}.$

(i) yes.

(ii) yes.

(iii)

$$d(x, z) + d(z,y) = \max \{|x_1 - z_1|,|x_2 - z_2|\} + \max \{|y_1 - z_1|,|y_2 - z_2|\} \\ \geq \max \{ |x_1 - z_1| + |y_1 - z_1|, |x_2 - z_2| + |y_2 - z_2| \} \\ \geq \max \{ |x_1 - y_1|, |x_2 - y_2| \} = d(x,y)$$

$\square$

(c) $d(x,y) = |x_1 x_2 +y_1 y_2|$.

(i) No, Consider $x = (1, 1), y = (1, -1)$. $x_1 x_2 +y_1 y_2 = 0$.

$\square$

Exercise 8.2.2.

Let $C[0,1]$ be the collection of continuous functions on the closed interval $[0,1]$. Decide which of the following are metrics on $C[0,1]$.

(a) $d(f,g) = \sup \{|f(x)−g(x)| : x ∈ [0,1]\}$.

Proof:

(i) $d(f,g) \geq 0$. If $d(f,g) = 0$, then $\sup \{|f(x)−g(x)| : x ∈ [0,1]\} = 0$, that means $|f(x) - g(x)| = 0$ for all $x \in [0,1]$.

(ii) $d(f,g) = d(g,f)$

(iii)

$$d(f,h) + d(h, g) = \sup \{|f(x)−h(x)| : x ∈ [0,1]\} + \sup \{|h(x)−g(x)| : x ∈ [0,1]\} \\ \geq \sup \{ |f(x)−h(x)| + |h(x)−g(x)| : x ∈ [0,1] \} \\ \geq \sup \{ |f(x)−g(x)| : x ∈ [0,1] \} \\ = d(f,g)$$

So yes.

$\square$

(b) $d(f,g) = |f(1)−g(1)|$.

(i) it does not hold. So it's not a metric.

$\square$

(c) $d(f,g) = \int_{0}^{1} |f-g|$

(i) $\int_{0}^{1} |f-g| \geq 0$. If $\int_{0}^{1} |f-g| = 0$, since $f, g$ are continuous functions, then $f-g = 0$, i.e. $f = g$.

(ii) $\int_{0}^{1} |f-g| = \int_{0}^{1} |g-f|$

(iii)

$$\int_{0}^{1} |f - g| \leq \int_{0}^{1} |f-h| + |h-g| = \int_{0}^{1} |f-h| + \int_{0}^{1} |h - g| = d(f,h) + d(h, g)$$

So yes.

The following distance function is called the discrete metric and can be defined on any set $X$. For any $x,y ∈ X$, let

$$ρ(x,y) = \begin{cases} 1 &\text{if } x \neq y\\ 0 &\text{if } x = y \\ \end{cases}$$

Exercise 8.2.3.

Verify that the discrete metric is actually a metric.

Proof:

(i) $d(x,y) \geq 0$. Also if $d(x,y) = 0$, then $x = y$.

(ii) if $x = y$, then $d(x,y) = d(x, x) = 0 = d(y, y) = d(y,x)$.

If $x \neq y$, then $d(x,y) = 1 = d(y,x)$.

(iii) $d(x,y) \leq 1$, $d(x,z) + d(z,y) \geq 1$, $d(x,y) \leq d(x,z) + d(z,y)$.

$\square$

Definition 8.2.2.

Let $(X,d)$ be a metric space. A sequence $(x_n) ⊆ X$ converges to an element $x ∈ X$ if for all $ϵ > 0$ there exists an $N ∈ \mathbf{N}$ such that $d(x_n,x) < ϵ$ whenever $n ≥ N$.

Definition 8.2.3.

A sequence $(x_n)$ in a metric space $(X,d)$ is a Cauchy sequence if for all $ϵ > 0$ there exists an $N ∈ \mathbf{N}$ such that $d(x_m,x_n) < ϵ$ whenever $m,n ≥ N$.

Exercise 8.2.4.

Show that a convergent sequence is Cauchy.

Proof:

$$d(x_m,x_n) \leq d(x_m,x) + (x,x_n) \leq \epsilon /2 + \epsilon /2 = \epsilon$$

$\square$

Definition 8.2.4.

A metric space $(X,d)$ is complete if every Cauchy sequence in $X$ converges to an element of $X$.

Exercise 8.2.5.

(a) Consider $\mathbf{R}^2$ with the discrete metric $ρ(x,y)$ examined in Exercise 8.2.3. What do Cauchy sequences look like in this space? Is $\mathbf{R}^2$ complete with respect to this metric?

Proof: Cauchy sequences will eventually become a constant. Yes, it's complete.

(b) Show that $C[0,1]$ is complete with respect to the metric in Exercise 8.2.2 (a).

Proof:

Given a cauchy function sequence $(f_n(x))$, and given $\epsilon$, we can find $N$, and $m,n > N$, $d(f_m, f_n) < \epsilon$.

Since $d(f_m, f_n) = \sup \{|f_m(x) - f_n(x)| : x \in [0,1]\}$, then $|f_m(x) - f_n(x)| < \epsilon$ for $x \in [0,1]$. Then we can apply Theorem 6.2.5 (Cauchy Criterion for Uniform Convergence), to get $f(x)$ is also continuous. Then $f(x) \in C[0,1]$.

$\square$

(c) Define $C^1[0,1]$ to be the collection of differentiable functions on $[0,1]$ whose derivatives are also continuous. Is $C^1[0,1]$ complete with respect to the metric defined in Exercise 8.2.2 (a)?

Proof:

Let's consider the following function sequence we encounter in the end of chapter 5.4

$$f_n(x) = \sum_{k = 0}^{n} (\frac{1}{2})^k \cos (2^k x)$$

where $x \in [0,1]$.

This is a Cauchy sequence, because

$$\left| f_m(x) - f_n(x) \right| \leq \left| (\frac{1}{2})^{m+1} + \cdots + (\frac{1}{2})^n \right| < \frac{1}{2^m}$$

So it converges uniformly to a function $f$ which is continuous.

However, from the discussion at the end of chapter 5.4, the results from G.H. Hardy show since $ab = (\frac{1}{2}) \cdot 2 = 1$, then $f(x)$ is a nowhere-differentiable function. That is $f \not\in C^1[0,1]$. So it's not complete w.r.t. to the metric defined in Exercise 8.2.2 (a).

Another example I asked Google AI mode is similar to exercise 8.5.4 (b), which uses $\sin$ and $\cos$ function to approximate $f(x) = |x|$. Again, the convergence is uniform, (see my solution on that one), so it's a Cauchy sequence, but $|x|$ is not differentiable at 0.

$\square$

Because completeness is a prerequisite for doing anything significant in the way of analysis, the metric in Exercise 8.2.2 (a) is the most natural metric to consider when working with $C[0,1]$. The notation

$$\left\| f-g \right\|_{\infty} = d(f,g) = \sup \{|f(x)−g(x)| : x ∈ [0,1]\}$$

is standard, and setting $g = 0$ gives the so-called “sup norm”

$$\left\| f \right\|_{\infty} = d(f,0) = \sup \{|f(x)| : x ∈ [0,1]\}$$

In all upcoming discussions, it is assumed that the space $C[0,1]$ is endowed with this metric unless otherwise specified.

Definition 8.2.5 (Continuous function).

Let $(X,d_1)$ and $(Y,d_2)$ be metric spaces. A function $f : X → Y$ is continuous at $x ∈ X$ if for all $ϵ > 0$ there exists a $δ > 0$ such that $d_2(f(x),f(y)) < ϵ$ whenever $d_1(x,y) < δ$.

Exercise 8.2.6.

Which of these functions from $C[0,1]$ to $\mathbf{R}$ (with the usual metric) are continuous?

(a) $g(f) = \int_{0}^{1}fk$ where $k$ is some fixed function in $C[0,1]$.

Proof:

$$g(f_1) - f(f_2) = \int_{0}^{1} f_1 k - \int_{0}^{1} f_2 k \\ = \int_{0}^{1} (f_1 - f_2) k$$

Since $k$ is a fixed function in $C[0,1]$, then it's continuous. We can assume $|k| \leq M$. Then if $d_1(f_1, f_2) < \epsilon / M$,

$$|g(f_1) - f(f_2)| = \left| \int_{0}^{1} (f_1 - f_2) k \right| \leq \int_{0}^{1} | (f_1 - f_2) k | < \int_{0}^{1} \epsilon / M \cdot M = \epsilon$$

So $g$ is continuous.

$\square$

(b) $g(f) = f(1/2)$.

Proof: Let $d_1(f_1, f_2) < \epsilon$, then

$$|g(f_1) - g(f_2)| = |f_1(1/2) - f_2(1/2)| \leq d_1(f_1, f_2) < \epsilon$$

$\square$

(c) $g(f) = f(1/2)$, but this time with respect to the metric on $C[0,1]$ from Exercise 8.2.2 (c).

Proof: Let

$$f_n(x) = \begin{cases} nx - n(1/2 - 1/n) &\text{if } 1/2 - 1/n \leq x \leq 1/2 \\ -n x + n (1/2 + 1/n) &\text{if } 1/2 \leq x \leq 1/2 + 1/n \\ 0 & \text{otherwise } \end{cases}$$

Note the $f_n$'s shape is a triangle centered at $1/2$ with a height of $n$.

Let $f = 0$,

$$d_1(f_n, f) = \int_{0}^{1} | f_n - f | = \int_{0}^{1} f_n = \frac{1}{n} \rightarrow 0$$

But $g(f_n) - g(f) = f_n(1/2) - f(1/2) = 1 - 0 = 1$.

$\square$

Topology on Metric Spaces

Definition 8.2.6. Given $ϵ > 0$ and an element $x$ in the metric space $(X,d)$, the $ϵ$-neighborhood of $x$ is the set $V_ϵ(x) = \{y ∈ X : d(x,y) < ϵ\}$.

Exercise 8.2.7.

Describe the $ϵ$-neighborhoods in $\mathbf{R}^2$ for each of the different metrics described in Exercise 8.2.1. How about for the discrete metric?

Solution:

(a) $d(x,y) = \sqrt[]{(x_1−y_1)^2 +(x_2−y_2)^2}$.

The $ϵ$-neighborhood of $x$ is a cicle of radius $\epsilon$ centered at $x$.

(b) $d(x,y) = \max \{|x_1−y_1|,|x_2−y_2|\}.$

The $ϵ$-neighborhood of $x$ is a square with edge length $2 \epsilon$ centered at $x$.

(c) the discrete metric.

The $ϵ$-neighborhood of $x$ is only itself if $ϵ < 1$. It's $X$ if $ϵ \geq 1$.

$\square$

open sets, limit points, and closed sets

With the definition of an $ϵ$-neighborhood, we can now define open sets, limit points, and closed sets exactly as we did before.

• A set $O ⊆ X$ is open if for every $x ∈ O$ we can find a neighborhood $V_ϵ(x) ⊆ O$.
• A point $x$ is a limit point of a set $A$ if every $V_ϵ(x)$ intersects $A$ in some point other than $x$.
• A set $C$ is closed if it contains its limit points.

Exercise 8.2.8.

Let $(X,d)$ be a metric space.

(a) Verify that a typical $ϵ$-neighborhood $V_ϵ(x)$ is an open set. Is the set

$$C_{\epsilon}(x) = \{y \in X: d(x,y) \leq \epsilon \}$$

a closed set

Proof: Let $y \in V_{\epsilon}(x)$, and $\delta = \epsilon - d(x,y) > 0$. Consider $z \in V_{\delta}(y)$.

$$d(z, x) \leq d(z, y) + d(y, x) \leq \delta + d(y,x) = \epsilon - d(y,x) + d(y,x) = \epsilon$$

So $z \in V_{\epsilon}(x)$, so $V_{\delta}(y) \subseteq V_{\epsilon}(x)$, then $V_{\epsilon}(x)$ is an open set.

Assume $d(x,y) > \epsilon$, we want to show $y$ is not a limit point of $C_{\epsilon}(x)$.

Let $\delta = \frac{d(x,y)-\epsilon }{2}$, consider $V_{\delta}(y)$, and let $z \in V_{\delta}(y)$.

if $z \in C_{\epsilon}(x)$, then

$$d(x, y) \leq d(x,z) + d(z,y) \leq \epsilon + \delta < \epsilon + 2 \delta = d(x,y)$$

We got a contradiction. So $z \not\in C_{\epsilon}(x)$. Then $V_{\delta}(y) \cap C_{\epsilon}(x) = \emptyset$, i.e. $y$ is not a limit point of $C_{\epsilon}(x)$. So $C_{\epsilon}(x)$ is a closed set.

$\square$

(b) Show that a set $E ⊆ X$ is open if and only if its complement is closed.

Proof:

$\Rightarrow$ Assume $E$ is open, and $x \in E$. Then we can find $V_{\epsilon}(x) \subseteq E$. So $V_{\epsilon}(x) \cap E^c = \emptyset$, so $x$ is not a limit point of $E^c$. That means $E$ does not contain any limit point of $E^c$. So $E^c$ contains all its limit point. So $E^c$ is closed.

$\Leftarrow$ Assume $E^c$ is closed. And $x \in E$. Since it's not a limit point of $E^c$, we can find $V_{\epsilon}(x) \cap E^c = \emptyset$. So $V_{\epsilon}(x) \subseteq E$. Then $E$ is open.

Exercise 8.2.9.

(a) Show that the set $Y= \{f ∈ C[0,1] : |f|_∞ ≤ 1\}$ is closed in $C[0,1]$.

Proof: we can see that $Y = C_{1}(0)$, so it's closed.

(b) Is the set $T= \{f ∈ C[0,1] : f(0) = 0\}$ open, closed, or neither in $C[0,1]$?

Proof:

It's not open. Given $f \in T$, and $\epsilon > 0$. Consider

$$g = f + \frac{\epsilon}{2}(x-1)$$

Then $d(f,g) = \sup \{ |\frac{\epsilon}{2}(x-1)| : x \in [0, 1] \} = \frac{\epsilon }{2}$.

So $g \in V_{\epsilon}(f)$, but $g \not\in T$.

On the other hand, it $g$ is a limit point of $T$, then given $\epsilon$, $|g(0)| \leq \sup \{ |f(x) - g(x)| : x \in [0,1]\} < \epsilon$. So $g(0) = 0$, then $g \in T$.

$\square$

Definition 8.2.7.

A subset $K$ of a metric space $(X,d)$ is compact if every sequence in $K$ has a convergent subsequence that converges to a limit in $K$.

Exercise 8.2.10.

(a) Supply a definition for bounded subsets of a metric space $(X,d)$.

Solution: If a non empty subset $S$ of $(X,d)$ is bounded, we can find $x \in S$ and a real number $d > 0$, such that $S \subseteq V_{d}(x)$.

(b) Show that if $K$ is a compact subset of the metric space $(X,d)$, then $K$ is closed and bounded.

Proof:

Assume $K$ is not bounded, and let $x \in K$. Let $x_{n} \in K$, such that $d(x_n, x) > n$.

We can find a subsequence of $(x_n)$ such that $(x_{n_k})$ converges to $y \in K$. Let $d = d(x, y)$, we can find $n_k$, such that $n_k > d + 1$ and $d(y, x_{n_k}) < 1$.

Then

$$d(x_{n_k}, x) \leq d(x_{n_k}, y) + d(y, x) < 1 + d < n_k$$

On the other hand, $d(x_{n_k}, x) > n_k$. We have a contradiction.

So $K$ is bounded.

Assume $x \in K$ is a limit point of $K$. Let $x_n \in K \cap V_{1/n}(x)$. So $(x_n) \rightarrow x$. On the other hand, a subsequence $(x_{n_k}) \rightarrow c \in K$, so $c = x$. $x \in K$.

So K is closed.

(c) Show that $Y ⊆ C[0,1]$ from Exercise 8.2.9 (a) is closed and bounded but not compact.

Solution: Consider

$$f_n = x^n$$

$(f_n) \rightarrow f$ is not continuous. So $Y$ is not compact.

$\square$

Definition 8.2.8.

Given a subset $E$ of a metric space $(X,d)$, the closure $E$ is the union of $E$ together with its limit points. The interior of $E$ is denoted by $E^◦$ and is defined as

$$E^◦ = \{x ∈ E : \text{ there exists }V_ϵ(x) ⊆ E\}.$$

Exercise 8.2.11.

(a) Show that $E$ is closed if and only if $E = \overline{E}$. Show that $E$ is open if and only if $E^o = E$.

Proof:

$\Rightarrow$ $E$ is closed. Apparantly, $\overline{E} \supseteq E$. Since $E$ has all its limit points, so $\overline{E} \subseteq E$. So $\overline{E} = E$.

$\Leftarrow$ if $x$ is a limit point of $E$, then $x \in \overline{E} = E$. so $E$ is closed.

The second part.

$\Rightarrow$ $E$ is open. Apparantly, $E^o \subseteq E$. If $x \in E$, then we can find $V_ϵ(x) ⊆ E$. Then $x \in E^o$. So $E^o \supseteq E$. Then $E^o = E$.

$\Leftarrow$ If $x \in E = E^o$, we can find $V_ϵ(x) ⊆ E$. So $E$ is open.

$\square$

(b) Show that $\overline{E}^c = (E^c)^o$, and similarly that $(E^o)^c = \overline{E^c}$.

Proof:

First part.

• $\overline{E}^c \subseteq (E^c)^o$.

$x \in \overline{E}^c$, then $x \in E^c$ and is not a limit point of $E$. So we can find $V_ϵ(x) \cap E = \emptyset$, i.e. $V_ϵ(x) ⊆ E^c$. So $x \in (E^c)^o$.

• $\overline{E}^c \supseteq (E^c)^o$.

$x \in (E^c)^o$. Then we can find $V_ϵ(x) ⊆ E^c$. So $x$ is not a limit point of $E$. Also $(E^c)^o ⊆ E^c$, so $x \not\in E$. then $x \not\in \overline{E}$. Then $x \in (\overline{E})^c$.

Second part.

• $(E^o)^c \subseteq \overline{E^c}$.

$x \in (E^o)^c$, then $x \not\in E^o$. For any $V_ϵ(x)$, $V_ϵ(x) \not \subseteq E$, i.e. $V_ϵ(x) \cap E^c \neq \emptyset$, then either $x \in E^c$, or $x$ is a limit point of $E^c$. Either way, $x \in \overline{E^c}$. Then $(E^o)^c \subseteq \overline{E^c}$.

• $(E^o)^c \supseteq \overline{E^c}$.

$x$ in $\overline{E^c}$. If $x \in E^c$, then $x \not\in E$, so $x \not\in E^o$. So $x \in (E^o)^c$.

If $x$ is a limit point of $E^c$. Then for all $V_ϵ(x)$, $V_ϵ(x) \cap E^c \neq \emptyset$. Then $V_ϵ(x) \not \subseteq E$. So $x \not\in E^o$. Then $x \in (E^o)^c$. Then $(E^o)^c \supseteq \overline{E^c}$.

$\square$

Exercise 8.2.12.

(a) Show

$$\overline{V_ϵ(x)} \subseteq \{ y \in X : d(x,y) \leq \epsilon \},$$

in an arbitrary metric space $(X,d)$.

Proof:

Let $A = \{y \in X : d(x,y) \leq \epsilon\}$. If $x \in V_ϵ(x)$, then $d(x,y) < \epsilon \subseteq A$.

If $x$ is a limit point of $V_ϵ(x)$. Then it's a limit point of $A$. We have proved in exercise 8.2.8 (a), that $A$ is a closed set. So $x \in A$. That means $\overline{V_ϵ(x)} \subseteq A$.

(b) To keep things from sounding too familiar, find an example of a specific metric space where

$$\overline{V_ϵ(x)} \neq \{ y \in X : d(x,y) \leq \epsilon \}$$

Proof:

Consider the discrete metric. $V_1(x) = \{x\}$. But $\{y \in X : d(x,y) \leq \epsilon\} = X$.

$\square$

Definition 8.2.9.

A set $A ⊆ X$ is dense in the metric space $(X,d)$ if $\overline{A} = X$. A subset $E$ of a metric space $(X,d)$ is nowhere-dense in $X$ if $E^o$ is empty.

Exercise 8.2.13.

If $E$ is a subset of a metric space $(X,d)$, show that $E$ is nowhere-dense in $X$ if and only if $\overline{E}^c$ is dense in $X$.

Proof:

$\Rightarrow$ Assume $E$ is nowhere-dense in $X$.

So $\overline{E}^o = \emptyset$, then $(\overline{E}^o)^c = X$.

From exercise 8.2.11, $(\overline{E}^o)^c = \overline{(\overline{E}^c)}$.

So we have $\overline{E}^c$ is dense.

$\Leftarrow$ Assume $\overline{E}^c$ is dense in X.

Then $\overline{(\overline{E}^c)} = X$, so $(\overline{E}^o)^c = X$, then $\overline{E}^o$ is empty. So $E$ is nowhere-dense in $X$.

$\square$

The Baire Category Theorem

Theorem 8.2.10.

Let $(X,d)$ be a complete metric space, and let ${O_n}$ be a countable collection of dense, open subsets of $X$. Then, $\cap_{n=1}^{\infty} O_n$ is not empty.

Proof:

Pick $x_1 \in O_1$. Because $O_1$ is open, there exists an $ϵ_1 > 0$ such that $V_{ϵ_1} (x_1) ⊆ O_1$.

Exercise 8.2.14.

(a) Give the details for why we know there exists a point $x_2 \in V_{ϵ_1} (x_1) \cap O_2$ and an $ϵ_2 > 0$ satisfying $ϵ_2 < ϵ_1/2$ with $V_{ϵ_2} (x_2)$ contained in $O_2$ and

$$\overline{V_{ϵ_2} (x_2)} \subseteq V_{ϵ_1} (x_1).$$

Proof:

Since $O_2$ is dense, then $\overline{O_2} = X$. Assume $V_{ϵ_1} (x_1) \cap O_2 = \emptyset$, and let $x_2$ be an arbitrary point in $V_{ϵ_1} (x_1)$. Then $x_2$ is a limit point of $O_2$, for every $V_{ϵ_2} (x_2)$, $V_{ϵ_2} (x_2) \cap O_2 \neq \emptyset$.

But since $x_2 \in V_{ϵ_1} (x_1)$ and $V_{ϵ_1} (x_1) \cap O_2 = \emptyset$, so we can find a $V_{ϵ_2} (x_2)$, $V_{ϵ_2} (x_2) \cap O_2 = \emptyset$.

We have a contradiction.

Since both $O_2, V_{ϵ_1} (x_1)$ are open, we have $O_2 \cap V_{ϵ_1} (x_1)$ is open. So we can find $V_{ϵ_2'} (x_2) \subseteq O_2 \cap V_{ϵ_1} (x_1)$.

Let $ϵ_2 = ϵ_2'/2$. From exercise 8.2.12 (a)

$$\overline{V_{ϵ_2} (x_2)} \subseteq C_{ϵ_2} (x_2) \subseteq V_{ϵ_2'} (x_2) \subseteq O_2 \cap V_{ϵ_1} (x_1)$$

$\square$

(b) Proceed along this line and use the completeness of $(X,d)$ to produce a single point $x ∈ O_n$ for every $n ∈ N$.

Proof: We produced a sequence of closed set

$$\overline{V_{ϵ_1} (x_1)} \supseteq \overline{V_{ϵ_2} (x_2)} \supseteq \overline{V_{ϵ_3} (x_3)} \supseteq \cdots$$

Consider the sequence $(x_n)$. It's a Cauchy sequence, since $ϵ_{n+1} < ϵ_n / 2$. Then $(x_n) \rightarrow x$, so $x$ is a limit point of all $V_{ϵ_n} (x_n)$, then $x \in \overline{V_{ϵ_n} (x_n)} \subseteq O_n$ for every $n ∈ N$.

$\square$

Theorem 8.2.11 (Baire Category Theorem).

A complete metric space is not the union of a countable collection of nowhere-dense sets.

Exercise 8.2.15.

Complete the proof of the theorem.

Proof: Consider $\{E_n\}$ is a countable collection of nowhere-dense sets.

We consider

$$O_n = (E_n^c)^o$$

From exercise 8.2.11, $(E_n^c)^o = \overline{E_n}^c$. From exercise 8.2.13, since $E_n$ is nowhere dense in $X$, then $\overline{E_n}^c$ is dense in $X$. That is $O_n$ is dense. Since it's also open, we can apply exercise 8.2.14., there exists $x \in O_n$ for all $n \in \mathbf{N}$. Then $x \not\in \cup E_n$.

$\square$

• This result is called the Baire Category Theorem because it creates two categories of size for subsets in a metric space. A set of “first category” is one that can be written as a countable union of nowhere-dense sets. These are the small, intuitively thin subsets of a metric space. We now see that if our metric space is complete, then it is necessarily of “second category,” meaning it cannot be written as a countable union of nowhere-dense sets.

Theorem 8.2.12.

The set

$$D = \{ f \in C[0,1]: f'(x) \text{ exists for some } x \in [0,1] \}$$

is a set of first category in $C[0,1]$.

Proof: For each pair of natural numbers $m,n$, define

$$A_{m,n} = \left\{ f \in C[0,1] : \text{ there exists } x \in [0,1] \text{ where } \left| \frac{f(x) - f(t)}{x-t} \leq n \right| \text{ whenever } 0 < |x-t| < \frac{1}{m} \right\}$$

The set $A_{m,n}$ contains any function in $C[0,1]$ for which it is possible to find at least one point $x$ where the slopes through $(x,f(x))$ and points on the function nearby—within $1/m$ to be precise—are bounded by $n$.

Exercise 8.2.16.

Show that if $f ∈ C[0,1]$ is differentiable at a point $x ∈ [0,1]$, then $f ∈ A_{m,n}$ for some pair $m,n ∈ \mathbf{N}$.

Proof:

If $f$ is differentiable at $x$, then let $f'(x) = c$. pick $n > |c|$ and $d = n - |c|$, we can find $\delta$ such that if $|t-x| < \delta$

$$\left| \frac{f(t)-f(x)}{t-x} - c \right| < d$$

So

$$\left| \frac{f(t)-f(x)}{t-x} \right| < d+|c| = n$$

Also let $1/m < \delta$.

$\square$

• Fix $m$ and $n$. The first order of business is to prove that $A_{m,n}$ is a closed set. To this end, let $(f_k)$ be a sequence in $A_{m,n}$ and assume $(f_k) \rightarrow f \in C[0,1]$. We need to show $f ∈ A_{m,n}$.
• Because $f_k \in A_{m,n}$ then for each $k ∈ N$ there exists a point $x_k ∈ [0,1]$ where

$$\left| \frac{ f_k(x_k) - f_k(t) }{x_k - t} \right| \leq n \text{ for all } 0 < |x_k - t| < 1/m$$

Exercise 8.2.17.

(a) The sequence $(x_k)$ does not necessarily converge, but explain why there exists a subsequence $(x_{k_l})$ that is convergent. Let $x = \lim(x_{k_l}).$

Proof: This is because $(x_k)$ is bounded, so it must have convergent subsequence.

$\square$

(b) Prove that $f_{k_l}(x_{k_l}) \rightarrow f(x)$.

Proof: Since $(f_k) \rightarrow f$, we can find $N_1$, if $k_l > N_1$, $d(f_{k_l}, f) < \epsilon / 2$. Since $(x_{k_l}) \rightarrow x$, and $f$ is continuous, we can find $N_2$, if $k_l > N_2$, then $|f(x_{k_l}) - f(x)| < \epsilon / 2$. Then let $N = \max \{N_1, N_2\}$. If $k_l > N$

$$|f_{k_l}(x_{k_l}) - f(x)| \leq |f_{k_l}(x_{k_l}) - f(x_{k_l})| + |f(x_{k_l}) - f(x)| < \epsilon / 2 + \epsilon / 2 = \epsilon$$ $\square$

(c) Now finish the proof that $A_{m,n}$ is closed.

Proof:

Fix $t$, such that $0 < |x-t| < 1/m$. We can find $N$, such that $k_l > N$, then $0 < |x_{k_l} - t | < 1/m$.

$$\left| \frac{f(x_{k_l}) - f(t)}{x_{k_l}-t}\right| \leq n$$

Then

$$\lim_{k_l \to \infty} \left| \frac{f(x_{k_l}) - f(t)}{x_{k_l}-t} \right| = \left| \frac{f(x) - f(t)}{x-t} \right| \leq n$$

$\square$

Exercise 8.2.18.

A continuous function is called polygonal if its graph consists of a finite number of line segments.

(a) Show that there exists a polygonal function $p \in C[0,1]$ satisfying $\left\| f-p \right\|_{\infty} < \epsilon / 2$.

Proof: We have proved a similar result in exercise 6.7.2. So we just copy the proof here and replace $[a,b]$ with $[0,1]$.

Since $f$ is continuous on $[0, 1]$, then $f$ is uniformly continuous on $[0, 1]$, so we can find $\delta$, as long as $\left| x - y \right| < \delta$, $\left| f(x) - f(y) \right| < \epsilon/4$.

Then we can find

$$0 = x_0 < x_1 < x_2 < \cdots < x_{n} = 1$$

such that $x_{i+1} - x_i < \delta$

Assume $c \in [x_{i}, x_{i+1}]$

$$\left| f(c) - p (c) \right| \leq |f(c) - f(x_i)| + |f(x_i) - p (x_i)| + |p (x_i) - p (c)| \leq \\ |f(c) - f(x_i)| + |f(x_i) - p (x_i)| + |p (x_i) - p (x_{i+1})| \\ = |f(c) - f(x_i)| + 0 + |f(x_i) - f(x_{i+1})| \\ < \epsilon /2 + 0 + \epsilon / 2 = \epsilon$$

$\square$

(b) Show that if $h$ is any function in $C[0,1]$ that is bounded by $1$, then the function

$$g(x) = p(x) + \frac{\epsilon }{2} h(x)$$

satisfies $g ∈ V_ϵ(f)$.

Proof:

$$\left| g(x) - f(x) \right| = \left| p(x) + \frac{\epsilon }{2} h(x) -f(x) \right| \leq \left| f(x) - p(x) \right| + \left| \frac{\epsilon }{2} h(x) \right| \\ < \epsilon / 2 + \epsilon / 2 = \epsilon$$

$\square$

(c) Construct a polygonal function $h(x)$ in $C[0,1]$ that is bounded by $1$ and leads to the conclusion $g \not\in A_{m,n}$, where $g$ is defined as in (b). Explain how this completes the argument for Theorem 8.2.12.

Proof:

Since $p$ is polygonal, then assume $p$ can be divided to $n$ segments,

$$0 = x_0 < x_1 < x_2 < \cdots < x_{n} = 1$$

In each segment, $p(x) = a_i x + b_i$. Choose an even integer $r \in \mathbf{N}$, such that $\frac{\epsilon}{2} \cdot r - |a_i| > n$. Then construct a periodic triangle function with period of $2/r$. This is the first period.

$$h(x) = \begin{cases} rx &\text{if } 0 \leq x < 1/r \\ -rx + 2 &\text{if } 1/r \leq x < 2/r \\ \end{cases}$$

Then in each segment of $[x_{i-1}, x_i]$, the slop is either $\frac{\epsilon}{2} \cdot r + a_i$ or $-\frac{\epsilon}{2} \cdot r - a_i$. In any case, the slop is bigger than $n$.

So $g \not\in A_{m,n}$.

If $f \in A_{m,n}$, then $f \not\in A_{m,n}^o$, because for any $V_ϵ(f)$, we can find $g \in V_ϵ(f)$, such that $g \not\in A_{m,n}$. So $A_{m,n}^o$ is an empty set. Also since $A_{m,n} = \overline{A_{m,n}}$, then $\overline{A_{m,n}}^o$ is empty. So $A_{m,n}$ is nowhere dense. Together with Exercise 8.2.16., $D$ is the union a countable union of nowhere-dense sets.

$\square$