Does the open cover of a compact always contain an r-neighborhood of the compact

Question

Suppose I have $O$ constructed from a finite number of open balls, with $O$ a cover of a compact $C$.

Does $O$ necessarily contain a $\epsilon$-neighborhood of $C$?

I think so, with a potential $\epsilon$ defined as

$\epsilon = \min_{x \in \bar{C}} \epsilon_x$

with, for a constant $k > 0$,

$\epsilon_x = \max( \epsilon : \epsilon \le k, B(x,\epsilon) \in O)$

I tend to work with $\mathbb{R}^k$ but more general results would be welcome too.

Answer 1

If a metric space is compact and $\cal U$ is an open cover of it, there is some $\epsilon > 0$ such that for all elements $x$ of the space, there is some $U\in \cal U$ so that $B_\epsilon(x)\subseteq U.$ This is the so-called Lebesgue-number of the cover.

Answer 2

In order to define what an $\epsilon$-nbhd of a set is, you need some notion of distance.

Any metric space is a Hausdorff space. If $Y$ is a compact subspace of a Hausdorff space and $x \not \in Y$, then there are disjoint open sets $U$ and $V$ that contain $x$ and $Y$ respectively.

Using all those results, we can finally define our $\epsilon$. Consider all the points $x$ in the complement of the cover $O$. The quantity: $$\epsilon_m = \min_x \{ d(C, x) \} $$ Is always well defined and greater than zero. If it were zero, there would be an $x$ with zero distance between itself and $C$, but this contradicts the fact that there are two disjoint open sets that separate them.

You can check for yourself that all the numbers in the interval $(0, \epsilon_m)$ can be used to define a $\epsilon$-neighbourhood of $C$ contained in $O$.

The problem with this approach is that it will not work with topological spaces that are not Hausdorff spaces.