If every real valued continuous function on $A\subseteq \mathbb R^n$ is uniformly continuous , then $A$ is bounded?

Question

Let $A \subseteq \mathbb R^n$ be such that every real valued continuous function on $A$ is uniformly continuous , then $A$ closed and bounded . If $a \in \bar A \setminus A$ , then using the function $f: A \to \mathbb R$ , $f(x)=||x-a||^2$ I can reach a contradiction , showing $A$ is closed . To show bounded ness I was trying to use the function $f(x)=||x||^2$ and show if $A$ is not bounded , then $f$ is not uniformly continuous , but I'm unable to do so , please help . Also , if every real valued continuous function on a metric space $X$ is uniformly continuous , then for some fixed $b \in X $ , by using the function $f(x)=(d(x,b))^2$ , can we show that $X$ is bounded ?

Answer 1

Hint: let $\mathbb Z^n$ denote the integer lattice in $\mathbb R^n$. Which functions $f : \mathbb Z^n \to \mathbb R$ are uniformly continuous?