Exercise 4.20

Question

Exercise 20:
    If $E$ is a nonempty subset of a metric space $X$, define the distance from $x\in X$ to $E$ by $$\rho_E(x)=\inf_{x\in E}d(x,z).$$
    \begin{itemize}
        \item[(a)] Prove that $\rho_E(x)=0$ if and only if $x\in\bar E$.
        \item[(b)] Prove that $\rho_E$ is a uniformly continuous function on $X$, by showing that $$\big|\rho_E(x)-\rho_E(y)\big|\le d(x,y)$$ for all $x\in X$, $y\in X$.
    \end{itemize}

analambanomenos · Accepted Answer

(a): The condition $\rho_E(x)=0$ is equivalent to having every neighborhood of $x$ of radius $\delta>0$ having some element $z$ of $E$, that is, $x\in\bar E$.

(b): Following the hint, if $z\in E$, then $\rho_E(x)\le d(x,y)+d(y,z)$ by the triangle inequality, so that $\rho_E(x)\le d(x,y)+\rho_E(y)$. Symmetrically, $\rho_E(y)\le
    d(x,y)+\rho_E(x)$, so that $\big|\rho_E(x)-\rho_E(y)\big|\le d(x,y)$ for all $x\in X$, $y\in X$.

Amit Awasthi · Answer

\begin{proof}[Proof of $(a)$] First prove that $ ho_E(x) = 0$ if $x \in \overline{E}$. Since $E \subset X$, this means $x \in \overline{E} \subset X$. Then, since $x \in \overline{E}$, given $\epsilon > 0$ we can find a $z \in E$ such that $d(x,z) < \epsilon$. So, $\mathrm{inf}_{z \in E} d(x,z) = ho_E(x) = 0$. \par Now prove the converse, that if $ ho_E(x) = 0$, $x \in \overline{E}$. This means that $\mathrm{inf}_{z \in E} d(x,z) = 0$ which is only possible when $z = x$. However, for $z = x$, $x \in E$ must be true. Since $E \subset \overline{E}$, this also means $x \in \overline{E}$. \end{proof} \begin{proof}[Proof of $(b)$] From the definition of $ ho_E$, it follows that \begin{equation*} ho_E(x) \le d(x,z) \le d(x,y) + d(y,z) \end{equation*} for any $z \in E$. Then, \begin{align*} ho_E(x) &= \mathrm{inf}_{z \in E} d(x,z)\ &\le \mathrm{inf}_{z \in E} (d(x,y)+d(y,z)) = d(x,y) + ho_E(y), \end{align*} so \begin{equation}\label{4.20a} ho_E(x) - ho_E(y) \le d(x,y). \end{equation} Also, \begin{equation*} ho_E(y) \le d(y,z) \le d(y,x) + d(x,z) \end{equation*} for any $z \in E$. Then, \begin{align*} ho_E(y) &= \mathrm{inf}_{z \in E}(d(y,z))\ &\le \mathrm{inf}_{z \in E}(d(x,y)+d(x,z)) = d(x,y) + ho_E(x), \end{align*} so \begin{equation}\label{4.20b} ho_E(y) - ho_E(x) \le d(x,y). \end{equation} Since both \eqref{4.20a} and \eqref{4.20b} have to be true, it follows that \begin{multline*} \mathrm{max}\left( ho_E(x) - ho_E(y), ho_E(y) - ho_E(x) ight)\ = | ho_E(x) - ho_E(y)| \le d(x,y) \end{multline*} \par Now prove that $ ho_E$ is a uniformly continuous on $X$. For every $\epsilon > 0$, let $\delta = \epsilon$, so $d(x,y) < \delta$ implies $d(x,y) < \epsilon$, so $| ho_E(x) - ho_E(y)| < d(x,y) < \epsilon$ for all $x,y \in X$. Thus, $ ho_E$ is uniformly continuous on $X$. \end{proof}

Exercise 4.20

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Exercise 4.20

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