Exercise 2.3

Question

Let $X$ be a metric space, with metric $\rho$. For any nonempty $E \subset
    X$, define
    \begin{equation*}
      \rho_E(x) = \inf\{\rho(x,y) : y \in E\}.
    \end{equation*}
    Show that $\rho_E$ is a uniformly continuous function on $X$. If $A$ and $B$
    are disjoint nonempty closed subsets of $X$, examine the relevance of the
    function
    \begin{equation*}
      f(x) = \frac{\rho_A(x)}{\rho_A(x) + \rho_B(x)}
    \end{equation*}
    to Urysohn's lemma.

Amit Awasthi · Accepted Answer

\begin{proof} Let $\epsilon > 0$ be given. Then, if $ ho(x,z) < \epsilon$, by the triangle inequality we have \begin{align*} &\lvert ho_E(x) - ho_E(z) vert\ &\quad= \left\lvert \inf_{y \in E} ho(x,y) - \inf_{y \in E} ho(z,y) ight vert\ &\quad\le \left\lvert ho(x,z) + \inf_{y \in E} ho(z,y) - \inf_{y \in E} ho(z,y) ight vert\ &\quad= ho(x,z) < \epsilon. \end{align*} \par Now let $K \subset X$ be compact, and $V$ be an open set containing $K$. Let $\delta = \inf_{x \in K} ho_{X \setminus V}(x) > 0$, which is positive since $\delta = ho_{X \setminus V}(x_0)$ for some $x_0 \in K$ by compactness of $K$, and $ ho_{X \setminus V}(x_0) = 0$ implies $x_0 \in X \setminus V$ by openness of $V$. We then claim that setting $B = K$ and \begin{equation*} A = \{x \in X \mid ho_K(x) \ge \delta/3\}, \end{equation*} in the definition of $f(x)$, we have $K \prec f \prec V$. Note the first part of the problem implies $A$ is closed, and that $f(x)$ is continuous. By definition, clearly $0 \le f(x) \le 1$. Now if $x \in K$, then $f(x) = 1$ trivially. Moreover, \begin{align*} \{x \in X \mid f(x) e 0\} &= \{x \in X \mid ho_K(x) < \delta/3\}\ &\subset \{x \in X \mid ho_K(x) \le 2\delta/3\} \subset V \end{align*} and $\{x \in X \mid ho_K(x) \le 2\delta/3\}$ is closed, hence the support of $f$ is contained in $V$. \end{proof}

Exercise 2.3

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Exercise 2.3

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