KindiakMath

Several Metrization Lemmas

Problem 0. Prove that the function f : [0, \infty) \to \mathbb R defined by f(x) = x/(1+x) is subadditive i.e. for any r, s \in [0, \infty), f(r+s) \leq f(r) + f(s) with equality if and only if rs = 0.

(Click for Solution)

Solution. Since r , s \geq 0,

\begin{aligned} f(r+s) &= \frac{r+s}{1+(r+s)} \\ &= \frac r{1+(r+s)} + \frac s{1+(r+s)} \\ &\leq \frac r{1+r} + \frac s{1+s} = f(r) + f(s). \end{aligned}

The equality condition holds since f'(x) = 1/(1+x)^2 > 0 so that f is strictly increasing.

Problem 1. Let K be metrizable with a metric d. Prove that there exists a metric \rho bounded by 1 on K that generates the same metric topology on K.

(Click for Solution)

Solution. Define \rho := f \circ d. Since f([0, 1)) \subseteq [0, 1), f \circ d is bounded by 1. To check that the topologies generated are the same, let \mathcal B_d and \mathcal B_\rho denote the metric topologies induced by the metrics d, \rho respectively.

To prove that \mathcal T_d \subseteq \mathcal T_\rho, fix B_d(x, \epsilon) \in \mathcal B_d. We can assume \epsilon < 1 without loss of generality. Since f is injective, B_\rho(x, f(\epsilon)) \subseteq B_d(x, \epsilon), establishing (\subseteq ). To prove that \mathcal T_d \supseteq \mathcal T_\rho, fix B_\rho(x, \epsilon) \in \mathcal B_\rho, assuming \epsilon < 1 once again. Then B_d(x, f^{-1}(\epsilon)) \subseteq B_\rho(x, \epsilon), establishes (\supseteq ).

Problem 2. For each n, let K_n be metrizable with a bounded metric \rho_n as made possible by Problem 1. Define K := \prod_{n=1}^\infty K_n equipped with the product topology. Prove that the map d : K \times K \to \mathbb R defined by

\displaystyle d(\mathbf x, \mathbf y) := \sum_{n=1}^\infty \rho_n(x_n, y_n) \cdot 2^{-n}

is a well-defined metric on K and generates the product topology on K.

(Click for Solution)

Solution. Since \rho_n is bounded by 1, for any \mathbf x, \mathbf y \in K, d(\mathbf x, \mathbf y) converges absolutely and is bounded by 1 too. Hence, d is well-defined.

To check the product topology claim, fix any \mathbf x \in K and basis element B of K that contains \mathbf x. Assume B = B_1(x_1,\epsilon_1) \times \cdots \times B_n(x_n,\epsilon_n) \times \prod_{i=n+1}^\infty K_i and \epsilon_i < 1 without loss of generality. Define \epsilon = \min\{\epsilon_1 \cdot 2^{-1},\dots,\epsilon_n \cdot 2^{-n}\} , so that B_d(\mathbf x, \epsilon) \subseteq B.

On the other hand, fix \mathbf x \in K and B_d(\mathbf x, \epsilon), and assume \epsilon < 1 without loss of generality. Find N such that 1/2^N < \epsilon/2. Define

\displaystyle B = B_1(x_1,\epsilon/2) \times \cdots \times B_N(x_n,\epsilon/2) \times \prod_{i=N+1}^\infty K_i.

Then B \subseteq B_d(\mathbf x, \epsilon) since for any \mathbf y \in B,

\displaystyle \begin{aligned} d(\mathbf x, \mathbf y) &:= \sum_{n=1}^\infty \rho_n(x_n, y_n) \cdot 2^{-n} \\ &\leq \frac{\epsilon}2 \cdot \sum_{n=1}^N 2^{-n} + \sum_{n=N+1}^\infty 2^{-n} \\ &< \frac{\epsilon}2 \cdot \frac {2^{-1}}{1 - 2^{-1}} + \frac{2^{-(N+1)}}{1 - 2^{-1}} \\ &= \frac {\epsilon}2 \cdot 1 + 2^{-N} < \frac{\epsilon}2 + \frac{\epsilon}2 = \epsilon.\end{aligned}

Problem 3. Briefly explain why \mathbb R^\omega is metrizable.

(Click for Solution)

Solution. Since each \mathbb R is metrizable, by Problem 2, \mathbb R^\omega := \prod_{n=1}^\infty \mathbb R equipped with the product topology is metrizable.

Problem 4. Let K, L be topological spaces and f : K \to L be a continuous injection that is a homeomorphism to f(K). Prove that if L is metrizable, then so is K.

(Click for Solution)

Solution. Given the metric d on L, define the metric \rho on K by \rho(x, y) := d(f(x),f(y)) and verify that d generates the relevant topology.

Problem 5. Conclude that for any topological space K, if a map f : K \to \mathbb R^\omega described in Problem 4 exists, then K is metrizable.

(Click for Solution)

Solution. By Problem 3, \mathbb R^\omega is metrizable. By Problem 4, K is metrizable.

—Joel Kindiak, 25 Apr 25, 1325H

,

Published by

joelkindiak

Leave a comment