Uniform radius of points in a compact set that is in an open set Let $K$ be a compact set and let $U$ be an open set that contains $K$. Prove that there is an $\epsilon > 0$ such that if $k \in K$, then the interval $(k - \epsilon, k + \epsilon)$ is contained in $U$
Why can't we just take the minimum over all $\epsilon$-ball of points in $K$? I also try to solve it using the fact that $K$ has a finite open covering, but don't know to how to proceed.
 A: Here's a solution using the open cover definition of compactness; I'll write it so that it makes sense in general metric spaces, but if you're only interested in the real line then read "ball of radius $r$ around point $p$" as "interval $(p-r,p+r)$. For each point $x\in K$, let $\delta_x>0$ be so small that the open ball of radius $\delta_x$ around $x$ is included in $U$.  (You can do that because $U$ is open.)  Now shrink all these balls; consider, for each $x\in K$ the open ball $B_x$ with center $x$ and radius $\delta_x/2$. These open balls cover $K$, and therefore, by compactness, finitely many of them, say $B_{x_1},\dots,B_{x_n}$, cover $K$. Let $\epsilon$ be the smallest of $\delta_{x_1}/2,\dots,\delta_{x_n}/2$.  I claim this $\epsilon$ does what the problem requires.  To prove that, consider any $k\in K$ and any $y$ within a distance $\epsilon$ of $k$; I need to show that $y\in U$.  Well, since $k\in K$, we know that $k$ is in one of the balls $B_{x_1},\dots,B_{x_n}$, say it's in $B_{x_i}$.  Since that ball has radius $\delta_{x_i}/2$, the distance from $x_i$ to $k$ is less than $\delta_{x_i}/2$.  Also, the distance from $k$ to $y$ is less than $\epsilon\leq\delta_{x_i}/2$.  Putting those two distance facts together, you get that the distance from $x_i$ to $y$ is less than $\delta_{x_i}$.  That puts $y$ into the open ball with center $x_i$ and radius $\delta_{x_i}$, and that ball is included in $U$ by our original choice of $\delta_{x_i}$.  Therefore, $y\in U$ as required.
Let me also say something about your question "Why can't we just take the minimum over all $\epsilon$...?"  The minimum of infinitely many positive numbers might not exist; consider for example the "minimum" of $\{1/n:n=0,1,2,\dots\}$.  That's why it was important to use compactness, to get to a point where I just needed to take the minimum of finitely many positive numbers $\delta_{x_1}/2,\dots,\delta_{x_n}/2$.  That minimum exists and is positive, and that's what makes the rest of the proof succeed.
A: Since $U$ is open and contains $K$, for each point $p \in K$ there exists a positive number $r_{p}>0$ so that the open interval $(p-2r_{p}, \, p+2r_{p})$ is contained in $U$. Clearly the family of open intervals $\mathscr{F} = \{(p-r_{p},\,p+r_{p})\}_{p \in K}$ covers $K$. Since $K$ is compact there exists a finite refinement of $\mathscr{F}$ that covers $K$ (Heine-Borel). We denote the finite subcover by $\{(p_i-r_{i},\,p_i+r_i)\}_{i=1}^N$. We may set $\lambda = \min \{r_{1}, \,  \ldots, \,r_N \}$. Let $x \in K $. We can find $i \in \{1, \ldots, N \}$ so that $x \in (p_i - r_i,\,p_i + r_i)$ . We will show that $(x-\lambda, \, x+\lambda) \subseteq (p_i-2r_i,\,p_i+2r_i)$. Let $y \in (x-\lambda, \, x+\lambda)$. Then $y \in (p_i - r_i - \lambda, \, p_i + r_i + \lambda )$. Since $\lambda \leq r_i$, we observe that $(p_i - r_i - \lambda, \, p_i + r_i + \lambda ) \subseteq(p_i - 2r_i, \, p_i + 2r_i)$. Hence the desired result.
A: Let $C \subseteq \mathbb R$ be compact. Let $O \supset C$ be open. Assume that there does not exist an $\epsilon > 0$ such that for all $x \in C$ $B_\epsilon(x) \subseteq O$. Then for each $n \in \mathbb N$ we may choose $x_n$ such that $B_{1/n}(x_n) \not\subseteq O$, and thus $y_n \in B_{1/n}(x_n) \setminus O$. Since the sequence $y_n$ is bounded, we may extract a convergent subsequence $y_{n(k)}$. This sequence has the same limit as $x_{n(k)}$, and thus the limit of $y_{n(k)}$ lies in $C$, which contradicts the closedness of $\mathbb R \setminus O$.
