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If $G$ is an open set and $K$ a compact set with $K\subset G$, show that there is a $\delta>0$ such that $\{x:\textrm{dist}(x,K)<\delta\}\subset G.$

PCeltide
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    There even exists a $\delta > 0$ s.t. ${x: \text{dist}(x, K) \leq \delta} \subseteq G$. This follows since every metric space is in particular normal. More generally, given some closed $K$ and open $G$ s.t. $K \subseteq G$, then there exists an open $V$ s.t. $K \subseteq V \subseteq \overline{V} \subseteq G$.This is then also equivalent to being normal. – G. Chiusole Mar 15 '20 at 15:10

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Suppose this was not true, and let $(X,d)$ be the metric space. Then for all $\delta$, there exists a $x\in X\backslash G$, such that $\textrm{dist}(x,K)<\delta$.

Let $\delta_n$ be decreasing sequence of postive real numbers converging to $0$.

Then there exists a sequence $x_n$ such that each $x_n\in X\backslash G$ and $\textrm{dist}(x_n,K)<\delta_n.$

As $K$ is compact and thus closed, for each $n$ consider $y_n\in K$ such that $$d(x_n,y_n)=\min\{d(x_n,y):y\in K\}.$$

By compactness of $K$, the sequence $y_n$ has a convergent subsequence $y_{n_k}$. Let $x\in X$ such that $y_{n_k}\rightarrow x$; as $K$ is closed $x\in K$.

Now, observe that $d(x_n,y_{n_k})\rightarrow 0$. Indeed, this follows from the fact that $\textrm{dist}(x_n,K)<\delta_n\rightarrow 0$ and the definition of $y_n$.

So, $$d(x_n,x)\leq d(x,y_{n_k})+d(x_n,y_{n_k})\rightarrow 0,$$ and thus $$d(x_n,x)\rightarrow 0.$$

But, $x\in K\subset G,$ and $G$ is open and so, there exists an $r>0$ such that $B(x,r)\subset G.$ So, there must occur a $N$ such that $x_n\in G,$ for all $n\geq N,$ a contradiction to our assumption.

Hence our claim must be true.

PCeltide
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