# 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

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• This could help for some inspiration... – t.b. Apr 23 '12 at 18:31
• In your course, did you take sequential compactness or ('open cover') compactness as your definition of compactness? – T. Eskin Apr 23 '12 at 18:43
• We defined compact so that there is an finite open subcover. – Martin Ueding Apr 23 '12 at 18:49
• Alright. The reason why I asked was that some elementary metric topology textbooks first define compactness with convergence of subsequences and then show that it coincides with other definition of compactness. In particular, Lebesgue covering lemma is used to show the implication from sequential compactness to compactness, and the proof of the covering lemma is quite nice also for the sequential compact case. – T. Eskin Apr 23 '12 at 18:53
• @t.b.: Thanks, that helped a lot! – Martin Ueding Apr 23 '12 at 19:14

## 1 Answer

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 $B(x,\alpha)\subset O_j$.