Proof of The Distance to a Subset of a Metric Space: Bounds, the Lipschitz Estimate, Continuity, and Vanishing Exactly on the Closure

Throughout, for $x\in X$ we write $T_{x}=\{d(x,a):a\in A\}$, so that $\operatorname{dist}(x,A)=\inf T_{x}$ by Distance from a Point to a Nonempty Subset of a Metric Space (i), the infimum being the greatest lower bound in the sense of Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (i) and Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv).

We begin with the bounds (i). Since $\operatorname{dist}(x,A)$ is a lower bound of $T_{x}$, we have $\operatorname{dist}(x,A)\le d(x,a)$ for every $a\in A$. Distances are nonnegative (Metric Space (ii)), so $0$ is a lower bound of $T_{x}$, and therefore $0\le\operatorname{dist}(x,A)$. Now let $r>\operatorname{dist}(x,A)$. If $r$ were a lower bound of $T_{x}$, it would satisfy $r\le\operatorname{dist}(x,A)$, the infimum being the greatest lower bound. Hence $r$ is not a lower bound of $T_{x}$, which means that $d(x,a)<r$ for some $a\in A$.

For the Lipschitz estimate (ii), let $x,y\in X$. For every $a\in A$, (i) and Metric Space (v) give

$$\operatorname{dist}(x,A)\le d(x,a)\le d(x,y)+d(y,a).$$

Thus $\operatorname{dist}(x,A)-d(x,y)$ is a lower bound of $T_{y}$, and so

$$\operatorname{dist}(x,A)-d(x,y)\le\operatorname{dist}(y,A).$$

Exchanging the roles of $x$ and $y$ gives $\operatorname{dist}(y,A)-d(x,y)\le\operatorname{dist}(x,A)$, and the two inequalities together are the claim.

Continuity (iii) follows at once. Let $K\subseteq X$, let $x\in K$ and $\varepsilon>0$, and take $\delta=\varepsilon$. For every $y\in K$ with $d(x,y)<\delta$, (ii) gives

$$d_{\mathbb{R}}\bigl(\operatorname{dist}(y,A),\operatorname{dist}(x,A)\bigr)=\bigl|\operatorname{dist}(y,A)-\operatorname{dist}(x,A)\bigr|\le d(x,y)<\varepsilon,$$

where $d_{\mathbb{R}}(s,t)=|s-t|$ is the metric of the real line (Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (iv)). This is continuity at $x$ relative to $K$ in the sense of Continuous Map Between Metric Spaces (i); as $x\in K$ was arbitrary, the function is continuous on $K$ (Continuous Map Between Metric Spaces (ii)).

For (iv), recall from Closure of a Subset of a Metric Space (i) and Open Ball in a Metric Space (ii) that $x$ is a closure point of $A$ if and only if

$$\text{for every real }r>0\text{ there is }a\in A\text{ with }d(x,a)<r.$$

If $\operatorname{dist}(x,A)=0$, this condition is the last part of (i). Conversely, if it holds, then (i) gives $\operatorname{dist}(x,A)\le d(x,a)<r$ for every $r>0$, so $\operatorname{dist}(x,A)$ cannot be positive (take $r=\operatorname{dist}(x,A)$); together with $\operatorname{dist}(x,A)\ge0$ this gives $\operatorname{dist}(x,A)=0$.

Finally, for (v), suppose that $A$ is closed and $x\in X\setminus A$. By Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (iv), $A$ equals its closure $\operatorname{cl}A$ in the sense of Closure of a Subset of a Metric Space (ii), so $x$ is not a closure point of $A$, and $\operatorname{dist}(x,A)\neq0$ by (iv). Since $\operatorname{dist}(x,A)\ge0$ by (i), we conclude that $\operatorname{dist}(x,A)>0$.