Baire’s Theorem

A metric space is a pair ${(X,d)}$ where ${X}$ is a set and ${d : X \times X \rightarrow {\mathbb R}}$ is a nonnegative function such that for each ${x,y,z \in X}$ we have

(i) ${d(x,y) = 0}$ if and only if ${x = y}$ ;
(ii) ${d(x,y) = d(y,x)}$ ;
(iii) ${d(x,z) \le d(x,y) + d(y,x)}$ .

For example, the set of real numbers ${{\mathbb R}}$ comes equipped with a standard distance function ${d}$ defined by ${d(x,y) = |x-y|}$ where ${|\cdot|}$ denotes the absolute value. More generally, one can define a distance function on ${{\mathbb R}^n}$ by

$\displaystyle d(x,y) = \sqrt{(x_1 - y_1)^2 + \cdots + (x_n - y_n)^2}$

if ${x = (x_1, \ldots, x_n)}$ and ${y = (y_1, \ldots, y_n)}$ .

For another example, we equip the space ${m}$ of all bounded infinite sequences of real numbers with a distance function ${d}$ given by

$\displaystyle d(x,y) = \sup_{k} |x_k - y_k|$

if ${x = (x_1, x_2, \ldots)}$ and ${y = (y_1, y_2, \ldots)}$ .

Let ${x}$ be a point of a metric space ${X}$ . The open ball around ${x}$ of radius ${\epsilon}$ is the set ${B_\epsilon(x)}$ of points in ${X}$ whose distance is less than ${\epsilon}$ from ${x}$ :

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

Let ${M}$ be a subset of ${X}$ . A point ${x}$ in ${X}$ is called a contact point of ${M}$ if for each ${\epsilon > 0}$ the intersection ${B_\epsilon(x) \cap M}$ is nonempty. We denote by ${\overline{M}}$ the set of all contact points of ${M}$ , and we call ${\overline{M}}$ the closure of ${M}$ .

A subset ${M}$ of a metric space ${X}$ is called closed if ${\overline{M} = M}$ . In particular, for ${x \in X}$ and ${\epsilon > 0}$ , the reader can verify that the set of points ${\{y \in X : |x - y | \le \epsilon\}}$ coincides with the closure ${\overline{B_\epsilon(x)}}$ and is hence a closed set. We call the set ${\overline{B_\epsilon(x)}}$ the closed ball of radius ${\epsilon}$ about ${x}$ .

Let ${A}$ and ${B}$ be two subsets of a metric space ${X}$ . We say that ${A}$ is dense in ${B}$ if the inclusion ${B \subset \overline{A}}$ is satisfied. In particular, ${A}$ is said to be everywhere dense in ${X}$ if ${X \subset \overline{A}}$ . The set ${A}$ is said to be nowhere dense in ${X}$ if ${A}$ is dense in no open ball at all.

A metric space is called separable if it has a countable everywhere dense subset.

For example, the space ${({\mathbb R}, d)}$ equipped with the usual distance on ${{\mathbb R}}$ has a countable everywhere dense subset given by the rational numbers ${{\mathbb Q}}$ . More generally, one can show that the space ${({\mathbb R}^n, d)}$ equipped with the usual Euclidean distance is separable.

As a non-example, consider the space ${m}$ of all bounded infinite sequences of real numbers, defined above. We claim that ${m}$ is not separable. To show this, we construct an uncountable subset ${E}$ of ${m}$ that is spaced far enough apart. Indeed, let ${E}$ denote the subset of ${m}$ consisting only of those sequences of ${0}$ ‘s and ${1}$ ‘s. Then distinct elements of ${E}$ are exactly distance ${1}$ apart. Consider the subset of ${m}$ given by

$\displaystyle U = \bigcup_{e \in E} B_{1/2}(e).$

If ${A}$ were an everywhere dense subset of ${m}$ , then in particular we would have ${U \subset \overline{A}}$ . It follows that ${A}$ would contain at least one point in each ball of the form ${B_{1/2}(e)}$ for ${e \in E}$ . This implies that ${A}$ must be uncountable.

Let ${\{x_n\}_{n=1}^\infty}$ be a sequence of points in a metric space ${(X,d)}$ .

The sequence is said to converge to a point ${x \in X}$ if for each ${\epsilon > 0}$ , there is a positive integer ${N > 0}$ such that ${d(x,x_n) < \epsilon}$ for each ${n > N}$ .
The sequence is said to be a Cauchy sequence if for each ${\epsilon > 0}$ there is a positive integer ${N}$ such that ${d(x_n,x_m) < \epsilon}$ for each ${m,n > N}$ .

Clearly, any convergent sequence is a Cauchy sequence. The converse is in general not true, and we have a special name for a metric space all of whose Cauchy sequences converge.

A metric space ${X}$ is called complete if any Cauchy sequence of points in ${X}$ converges to a point in ${X}$ . Otherwise, ${X}$ is said to be incomplete.

For example, it is well-known that the space of real numbers equipped with the standard distance function is complete.

Proposition 1 The space ${m}$ of bounded infinite sequences of real numbers is complete.

Proof: Let ${\{x^{(n)}\}}$ be a Cauchy sequence of elements of ${m}$ . If ${\epsilon >0}$ there is an ${N_\epsilon > 0}$ such that

$\displaystyle |x^{(n)}_k - x^{(n')}_k| < \epsilon$

for each ${n,n' > N_\epsilon}$ and each ${k}$ . It follows that for each ${k = 1,2, \ldots,}$ the sequence of real numbers ${\{x^{(n)}_k\}_{n=1}^\infty}$ is Cauchy. Since ${{\mathbb R}}$ is complete, this sequence converges to some real number ${x_k \in {\mathbb R}}$ . That is, if ${\epsilon > 0}$ , there is an ${N_k > 0}$ such that ${|x^{(n)}_k - x_k| < \epsilon}$ for each ${n > N_k}$ .

We now claim that the sequence ${\{x^{(n)}\}}$ converges to ${x = (x_1, x_2, \ldots)}$ . Let ${\epsilon > 0}$ be arbitrary. Since the sequence ${\{x^{(n)}\}}$ is Cauchy, there is an ${N > 0}$ such that

$\displaystyle |x^{(n)}_k - x^{(n')}_k| < \epsilon/2$

for each ${n,n' > N}$ and each ${k}$ . Taking the limit as ${n' \rightarrow \infty}$ , we see that

$\displaystyle |x_k^{(n)} - x_k| \le \epsilon/2 < \epsilon$

for each ${n > N}$ and each ${k}$ . This shows that ${\{x^{(n)}\}}$ converges to ${x}$ .

We also claim that ${x = (x_1, x_2, \ldots)}$ is a bounded sequence of real numbers. Since the sequence ${\{x^{(n)}\}}$ converges to ${x}$ , there is an ${N > 0}$ such that

$\displaystyle | x_k^{(n)} - x_k| < 1$

for each ${n > N}$ and each ${k}$ . Since ${x^{(N+1)}}$ is bounded, there is an ${M > 0}$ such that

$\displaystyle \sup_{k}|x^{(N+1)}_k| < M.$

By the triangle inequality, it follows that

$\displaystyle \sup_{k} |x_k| \le \sup_{k}|x_k^{(N+1)}| + \sup_{k}|x_k - x_k^{(N+1)}| < M+1,$

as claimed. $\Box$

We now find another way to characterize complete spaces. First we need some notation.

A sequence ${\{A_n\}_{n=1}^\infty}$ of subsets of a set ${X}$ is said to be nested if

$\displaystyle A_1 \supset A_2 \supset \cdots$

Theorem 2 (Nested Sphere Theorem) Let ${(X,d)}$ be a metric space. A necessary and sufficient condition for ${X}$ to be complete is that every nested sequence of closed balls ${\overline{B_{\epsilon_n}(x_n)}}$ such that ${\epsilon_n \rightarrow 0}$ as ${n \rightarrow \infty}$ has nonempty intersection.

Proof: Suppose that ${X}$ is complete. Let ${\{\overline{B_{\epsilon_n}(x_n)}\}}$ be a nested sequence of closed balls such that ${\epsilon_n \rightarrow 0}$ as ${n \rightarrow \infty}$ . We claim that the sequence of centers ${\{x_n\}}$ is Cauchy. Indeed, let ${\epsilon > 0}$ be arbitrary. Choose ${N}$ so large that ${\epsilon_n < \epsilon/2}$ for each ${n \ge N}$ . Then for each ${m,n > N}$ , the centers ${x_n, x_m}$ belongs to ${B_{\epsilon_N}(x_N)}$ and hence are no more that ${2\epsilon_N < \epsilon}$ apart.

Now suppose that the latter condition holds. We will show that ${X}$ is complete. Let ${\{x_n\}}$ be a Cauchy sequence of points in ${X}$ . Let ${n_1}$ be such that

$\displaystyle d(x_n, x_m) < \frac{1}{2}$

for each ${m,n > n_1}$ . Let ${B_1}$ denote the closed ball about ${x_{n_1}}$ of radius ${1}$ . Once ${n_j}$ has been constructed, let ${n_{j+1}}$ be such that

$\displaystyle d(x_n,x_m) < \frac{1}{2^{j+1}}$

for each ${m,n > n_{j+1}}$ , and let ${B_{j+1}}$ denote the closed ball of radius ${2^{j}}$ about ${x_{n_{j+1}}}$ . Then the sequence of balls ${\{B_j\}}$ is a nested sequence with radii converging to zero. It follows that their intersection is nonempty. Let ${x}$ be a point of this intersection. It is clear that the subsequence ${\{x_{n_j}\}}$ converges to ${x}$ . Moreover, we claim that the original Cauchy sequence converges to ${x}$ . Indeed, let ${\epsilon > 0}$ be arbitrary. Choose ${N_1 > 0}$ so that ${d(x_m,x_n) < \epsilon/2}$ for each ${m,n \ge N_1}$ . Choose ${N_2 > 0}$ so large that ${n_{N_2} \ge N_1}$ and that ${d(x_{n_j},x) < \epsilon/2}$ for each ${j \ge N_2}$ . Then if ${n \ge \max\{N_1, n_{N_2}\}}$ , we see that

$\displaystyle d(x_n,x) \le d(x_n, x_{n_{N_2}}) + d(x_{n_{N_2}},x) < \epsilon/2 + \epsilon/2 = \epsilon,$

completing the proof. $\Box$

Lemma 3 Let ${X}$ be a metric space and let ${Y}$ be a subset of ${X}$ . Suppose that ${A}$ is nowhere dense in ${Y}$ . Then there is a ball ${B_\epsilon(x) \subset Y}$ which intersects ${A}$ trivially.

Proof: Suppose to the contrary that we may find no such ball. Let ${y}$ be an arbitrary point of ${Y}$ . By supposition, for each ${\epsilon > 0}$ , the ball ${B_\epsilon(y)}$ intersects ${A}$ . It follows that ${y}$ is a contact point of ${A}$ and hence ${Y \subset \overline{A}}$ . This contradicts the assumption that ${A}$ is nowhere dense in ${Y}$ . $\Box$

We are now in the position to prove an important tool in topology and analysis.

Theorem 4 (Baire’s Theorem) A complete metric space ${X}$ cannot be the union of a countable number of nowhere dense sets.

Proof: Suppose that we have found a countable number of nowhere dense sets ${\{A_n\}}$ such that

$\displaystyle X = \bigcup_n A_n.$

Let ${B_0}$ be a closed ball of radius ${1}$ . Since ${A_1}$ is nowhere dense in ${B_0}$ , by the Lemma there is a closed ball ${B_1}$ of radius less than ${1/2}$ such that ${B_1 \subset B_0}$ and ${B_1}$ is disjoint from ${A_1}$ . Since ${A_2}$ is nowhere dense in ${S_1}$ , there is a closed ball ${B_2}$ of radius less than ${1/3}$ such that ${B_2 \subset B_1}$ and ${B_2}$ is disjoint from ${A_2}$ . In this way, we obtain a nested sequence of closed balls

$\displaystyle B_0 \supset B_1 \supset B_2 \supset \cdots$

with radii tending to zero such that ${B_n}$ is disjoint from ${A_n}$ . The nested sphere theorem gives a point ${x}$ in the intersection of these balls. Since ${x}$ belongs to each ball, it does not belong to any of the ${A_n}$ and thus does not belong to their union, which was assumed to be the whole space ${X}$ , which is absurd. $\Box$

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Baire’s Theorem

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