User:TakuyaMurata/Metric spaces

2 Lemma Proof: Suppose a sequence $$y_n \in E$$ converges to $$y \in \overline{E}$$. Since $$\operatorname{dist}(x, E) \le \operatorname{dist}(x, y_n)$$, we have:
 * (i) $$\operatorname{dist}(\cdot, E) = \operatorname{dist}(\cdot, \overline{E})$$.
 * (ii) $$\operatorname{dist}(\cdot, E)$$ is a 1-Lipschitz continuous.
 * $$\operatorname{dist}(x, E) \le \operatorname{dist}(x, y)$$.

Taking inf over $$y$$ gives (i). For the second, since $$\operatorname{dist}(x, E) \le \operatorname{dist}(x, y) + \operatorname{dist}(y, z)$$, taking inf over z,
 * $$\operatorname{dist}(x, E) \le \operatorname{dist}(x, y) + \operatorname{dist}(y, E)$$.

By symmetry in the formula, we obtain (ii).$$\square$$

2 Theorem A metric space is completely regular Hausdorff.

Proof: Let $$x \ne y$$, and $$r = d(x, y)$$. It follows that $$B(x, r/2)$$ and $$B(y, r/2)$$ are disjoint to each others. In fact, if they intersect at, say, $$z$$, then
 * $$r = d(x, y) \le d(x, z) + d(z, y) < r / 2 + r / 2 = r$$,

which is absurd. Finally, a metric space is completely regular because $$\operatorname{dist}$$, which is continuous by the above lemma, separates a closed set and a point.$$\square$$

2 Theorem ''Let $$f:X \to Y$$ be a function. Suppose $$X$$ is metric. Then $$f$$ is continuous if and only if $$x_n \to x$$ implies $$f(x_n) \to f(x)$$.''

Proof: Both the statements are equivalent to:
 * $$f(\overline{E}) \subset \overline{f(E)}$$

for every subset $$E \subset X$$. $$\square$$

2 Corollary ''$$f$$ is continuous if and only if $$f$$ is continuous on every compact subset of $$X$$. (Recall that in Chapter 1 we proved the same result for locally compact spaces.)''

Proof: ($$\Rightarrow$$) is obvious. For ($$\Leftarrow$$), suppose $$x_n \to x$$. Let $$K = \{ x, x_1, x_2, ... \}$$. Then $$K$$ is compact; in fact, if $$U_j$$ is an open cover of $$K$$, then $$x \in U_N$$ for some $$N$$ and, by convergence, $$U_N$$ contains all but finitely many terms of the sequence $$x_n$$. Now, since $$f$$ is continuous on $$K$$, $$f(x_n) \to f(x)$$. $$\square$$

2 Theorem ''Let $$K$$ be a metric space. Then the following are equivalent:'' Proof: To show (i) $$\Rightarrow$$ (ii), suppose (ii) is invalid. That is to say, there exists a sequence $$x_n \in K$$ with no convergent subsequence. For each $$n$$, we can then find $$\delta_n > 0$$ such that $$B(x_n, \delta_n)$$ contains only $$x_n$$ and no other point in the sequence. Since the sequence $$x_n$$ consists of isolated points, it is closed, and thus it is a compact subset whose open cover doesn't admit a finite subcover, falsifying (i). (ii) $$\Rightarrow$$ (iii). By (ii), every Cauchy sequence contains a convergent subsequence, and so, the sequence itself converges. $$K$$ is thus complete. To show the totally boundedness, suppose $$K$$ is not totally bounded. We can then find a sequence consisting of isolated points, which obviously doesn't have a convergent subsequence and so contradicts (ii). We conclude (ii) $$\Rightarrow$$ (iii). For (iii) $$\Rightarrow$$ (i), since $$K$$ is complete, it suffices to show that every ultrafilter is Cauchy, but this is immediate from the definitions of the totally boundedness and a Cauchy filter. $$\square$$
 * (i) $$K$$ is compact.
 * (ii) Every sequence of $$K$$ admits a convergent subsequence. (i.e., $$K$$ is sequentially compact)
 * (iii) $$K$$ is complete and totally bounded.

2 Corollary Every compact metric space $$K$$ is separable.

Proof: Since $$K$$ is totally bounded, for each $$n > 0$$, there is a finite subset $$F_n \subset K$$ such that
 * $$K = \bigcup_{x \in F_n} B(x, n^{-1})$$

Let $$E$$ be the union of $$F_n$$ over all $$n$$. Evidently, $$E$$ is dense. $$\square$$

2 Theorem ''Let $$f_k \in C(X)$$ be a sequence. Then $$f_k$$ converges pointwise if and only if $$f_k(x_k) \to f(x)$$ for any sequence $$x_k \to x$$.

Proof: ($$\Rightarrow$$) Let $$K$$ be a compact set containing $$x_k$$. Then

2 Theorem (Baire) ''Let $$X$$ be a complete metric space. If $$X = \bigcup_n E_n$$, then some $$E_n$$ contains an interior point.''

Proof: Let $$V_n$$ be the interior of the closure of $$E_n$$. Then $$V_n$$ is dense by hypothesis. The theorem is then equivalent to say that the intersection of $$V_n$$ over all $$n$$ is dense. To prove that, let $$W \subset X$$ be an open subset. By denseness, there is $$x_1$$ and $$r_1 > 0$$ such that:
 * $$\overline{B}(x_1, r_1) \subset W \cap V_1$$.

By iteration, we can find $$x_n$$ and $$r_n > 0$$ such that:
 * $$\overline{B}(x_n, r_n) \subset B(x_{n-1}, r_{n-1}) \cap V_n$$ as well as $$r_n < n^{-1}$$

Since $$x_n \in B(x_m, r_m)$$ when $$n > m$$, we have that $$x_n$$ is Cauchy, and $$x_n$$ converges to some limit $$x$$. For any $$n$$, by closedness,
 * $$x \in \overline{B}(x_{n-1}, r_{n-1}) \subset B(x_n, r_n)$$.

Hence, $$x \in W$$ and $$x \in V_n$$ for all $$n$$. $$\square$$

2 Theorem (Banach fixed point theorem) ''Let $$X$$ be a complete metric space, and let $$f:X \to X$$ be a function. If there is a constant $$0 < c < 1$$ such that:
 * $$d(f(x), f(y)) \le c d(x, y)$$ for all $$x, y \in X$$

then $$f$$ admits a unique fixed point; i.e., $$f(x) = x$$ for some $$x$$. Moreover, $$\lim_{n \to \infty} f^n(y) = f(x)$$ for every $$y \in X$$.

Proof: See Banach fixed point theorem

Define $$\operatorname{dist}(x, E) = \inf \{ d(x, y) | y \in E \}$$ for a nonempty subset $$E \subset X$$.

A topological space homeomorphic to a separable complete metric space is called a Polish space.

2 Theorem ''Let $$X$$ be a topological space. Then $$X$$ is Polish if and only if it is homeomorphic to a $$G_\delta$$ subset of the Hilbert cube.

2 Theorem Every Polish space is isomorphic to one of (i) R, (ii) Z or (iii) a finite space.

Let $$\mathcal{C}_c(X) = \Gamma_c (X, \mathcal{C})$$, and $$\mathcal{C}_0(X)$$ be the space of all $$f \in \mathcal{C}(X)$$ such that $$\{ x \in X | f(x) \ge c \}$$ is compact. Clearly, $$\mathcal{C}_c(X) \subset \mathcal{C}_0(X)$$. When $$X$$ is an open subset of $$\mathbf{R}^n$$, $$\mathcal{C}(X)$$ consists of those $$f$$ such that $$\lim_{|x| \to \infty} f(x) = 0$$.

2 Theorem Let X'' be a locally compact space. Then $$\mathcal{C}_c(X)$$ is dense in $$\mathcal{C}_0(X)$$.''

Proof: Let $$f \in \mathcal{C}_0(X)$$, and $$K_n = \{ x \in X | f(x) \ge n^{-1} \}$$. Using Lemma 2.something, choose $$\psi_n \in \mathcal{C}_c(X)$$ such that $$\psi_n = 1$$ on $$K_n$$. Then $$\psi_n f \to f$$. $$\square$$

Reference

 * Descriptive Set Theory