Compact features

Question

Consider this problem:

Let $X$ be a metric space, $U$ be open, $K$ compact and $K\subset U$, show that there exists a $r>0$ such that $B(k,r)\subset U$ $\forall k\in K$

Here $B(k,r)=\{x\in X ;d(x,k)<r\}$, and note that $r$ does not depend on $k$, there is a $r$ that works for all $k$.

I saw a proof by contradiction using the fact that every sequence in $K$ have a subsequence that converges to a point on $K$. But when I'm working with compact sets, I like to investigate the problem using other features, like every continuous function defined on $K$ assumes a maximum and minimum, or every open covering of $K$ admits a finite cover. So I tried without success, for the former I defined the function $F:K\to \mathbb{R}$ by $F(k)=d(k,X-U)$, it's continuous and I tried to show that $r=\min_{k\in K}F$ works. For the latter I noticed that $k\in U$ $\forall k\in K$ and because U is open exists a $r_k$ such that $B_k(k,r_k)\subset U$ and so $\bigcup_{k\in K}B_k$ is a open covering of $K$ so I can extract a finite one, but then I got stuck.

I would like to understand and finalize the two proofs. Thanks for help!

Answer 1

You claim that $\exists r\ \forall k \in K: B(k, r) \subset U$. Let us assume, that the opposite is true:

$$\forall r\ \exists k \in K: B(k, r) \not\subset U.$$

This is very powerful and it actually shows the power of the reductio ad absurdum approach.

Let $r_n := \frac1n$. Then there is $x_n$ such that $B(x_n, r_n) \not\subset U$. At the same time there are $x^\star$ and $\{n_k\} \in {\bf N}$, such that $x_{n_k} \to x^\star$ (as K is compact and $\{x_n\} \in K$).

The limit means precisly that $\forall \varepsilon > 0\ \exists n_0\ \forall n \ge n_0: \rho(x_{n_k}, x^\star) < \varepsilon$, where $\rho$ is the metric.

Because $K \subset U$, we have that $x^\star \in U$, which is an open set. Therefore $\exists r > 0$ such that $B(x^\star, r) \in U$. Setting $\varepsilon := r$, we have that $\exists n_0 \in {\bf N}$ such that $\forall n > n_0: \rho(x_{n_k}, x^\star) < r$. At the same time, $B(x_{n_k}, \frac1n) \not\subset U$.

As $r$ does not change with $x_{n_k}$ while $r_n \to 0$, $B(x_{n_k}, r_n)$ will eventually be a subset of $B(x^\star, r)$. But that is a subset of $U$, which is a contradiction.

(This proof could probably be made clearer, but unfortunately I do not have the time to make it so right now. Hope it helps!)