Uniform Continuity of the Dist. Between Pts. and Subs. in a Met. Sp.

# Uniform Continuity of the Distance Between Points and Subsets in a Metric Space

Recall from The Distance Between Points and Subsets in a Metric Space page that if $(S, d)$ is a metric space, $A \subseteq S$ is nonempty, and $x \in S$ then we can define a function $f_A : S \to \mathbb{R}$ for all $x \in S$ by:

(1)\begin{align} \quad f_A(x) = \inf \{ d(x, y) : y \in A \} \end{align}

We defined the distance between the point $x$ and the subset $A$ as the number $f_A(x)$.

We will now look at a very nice theorem which will tell us that these functions $f_A$ are actually uniformly continuous.

Theorem 1: Let $(S, d)$ be a metric space and let $A \subseteq S$ be nonempty. Then the function $f_A : S \to \mathbb{R}$ defined for all $x \in S$ by $f_A(x) = \inf \{ d(x, y) : y \in A \}$ is uniformly continuous. |

**Proof:**Let $\epsilon > 0$ be given and let $x, y \in S$. We want to show that there exists a $\delta > 0$ such that whenever $d(x, y) < \delta$ then:

\begin{align} \quad \mid f_A(x) - f_A(y) \mid < \epsilon \end{align}

- Now for any $z \in A$ we have that:

\begin{align} \quad d(x, z) \leq d(x, y) + d(y, z) \end{align}

- Taking the infimum of both sides as $z$ ranges through $A$ and:

\begin{align} \quad \inf \{ d(x, z) : z \in A \} \leq d(x, y) + \inf \{ d(y, z) : z \in A \} \\ \quad f_A(x) \leq d(x, y) + f_A(y) \\ \quad f_A(x) - f_A(y) \leq d(x, y) \quad (*) \end{align}

- Furthermore we have that:

\begin{align} \quad d(y, z) \leq d(x, y) + d(x, z) \end{align}

- Taking the infimum of both sides as $z$ ranges through $A$ and;

\begin{align} \quad \inf \{ d(y, z) : z \in A \} \leq d(x, y) + \inf \{ d(x, z) : z \in A \} \\ \quad f_A(y) \leq d(x, y) + f_A(x) \\ \quad f_A(y) - f_A(x) \leq d(x, y) \\ \quad -[f_A(x) - f_A(y)] \leq d(x, y) \quad (**) \end{align}

- Combining $(*)$ and $(**)$ shows us that:

\begin{align} \quad \mid f_A(x) - f_A(y) \mid \leq d(x, y) \end{align}

- So set $\delta = \epsilon$. Then if $d(x, y) < \delta = \epsilon$ we have that:

\begin{align} \quad \mid f_A(x) - f_A(y) \mid \leq d(x, y) < \delta = \epsilon \end{align}

- Therefore $f_A$ is uniformly continuous on $S$ which completes our proof. $\blacksquare$