# Every cover covers sets with $\mathrm{diam}(A) < \lambda$ [duplicate]

Possible Duplicate:
Proof of the Lebesgue number lemma

Let $(X, d)$ be a metric space and $K \subset X$ a compact set.

Now I have to show that for all open covers $\mathcal U$, there is an $\lambda > 0$ so that for every subset $A \subset K$ with $\mathrm{diam}(A) < \lambda$ there is an $U \in \mathcal U$ so that $A \subset U$.

If I picture the set $K$ as a closed interval on the real numbers, I can imagine why this works. Since every open cover composed of several $U$ will overlap (since they are open). And if I take $\lambda$ to be the smallest overlap, any set that is that small will always be completely in either one of the two sets that overlap.

But how can I show that generally?

• We defined “compact” as: For every family of open covers, you can take a finite number of patches that still cover the whole set.

• Duplicate

## marked as duplicate by Rudy the Reindeer, t.b., robjohn♦, Davide Giraudo, Henning MakholmApr 23 '12 at 20:33

Let $O_1,\ldots,O_N\in \mathcal U$ such that $K\subset \bigcup_{j=1}^NO_j$. Then define $f_j(x):=\inf\{d(x,y),y\in \complement O_j\}$. Then $f_j$ is continuous, and so is $f\colon x\mapsto \max_{1\leq j\leq N}f_j(x)$. Since $O_J^c$ is closed and $f>0$ for all $x$, we can find $\alpha>0$ such that $f(x)\geq \alpha$ for each $x\in K$.
Now, if $x\in X$, then $f(x)=f_j(x)$ for some $j$, so $d(x,O_j^c)\geq\alpha$ and $B(x,\alpha)\subset O_j$.