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**Drexel28** This is a lot harder than I initially thought. The idea behind the following is that this is obvious if $\displaystyle Y$ is bounded, and if it isn't we can "erase" the parts that are far enough away from $\displaystyle X$ to be ignored.

So, we know that $\displaystyle X$ is compact and thus totally bounded (specifically bounded). Thus, $\displaystyle X\subseteq \left\{\bold{x}\in\mathbb{R}^n:\|\bold{x}\|\leqsla nt M\right\}$ for some constant $\displaystyle M$. Let $\displaystyle N=\left\{\bold{x}\in\mathbb{R}^n:\|\bold{x}\|\leqs lant M+d(X,Y)+1\right\}$. This is a compact subspace of $\displaystyle \mathbb{R}^n$ and so $\displaystyle G\cap N=K$ is compact (since the intersection of a compact set and a closed set is compact). Now, we next claim that $\displaystyle d(E,G)=d(E,K)$, but I leave this for you to prove. Thus, we must merely show that $\displaystyle d(E,G)=d(E,K)=d(e,k)$ for some $\displaystyle e\in E,k\in K$. To do this we note that since $\displaystyle G,K$ are compact that $\displaystyle G\times K$ is compact by Tychonoff's theorem. Also, it is clear that $\displaystyle d:G\times K\mapsto\mathbb{R}$ is continuous and so it assumes it's minimum on $\displaystyle G\times K$ and the conclusion follows.