# Show that any subspace of a compact space can be covered with one open subspace.

Here's the problem I'm dealing with:

Let $$(X,d)$$ be a compact metric space and let $$(U_{\lambda})_{\lambda \in \Lambda}$$ be an open cover of $$X$$. Show that there exist $$\delta >0$$ such that for all $$A \subset X$$ with diam$$A < \delta$$, there exist $$\lambda \in \Lambda$$ such that $$A \subset U_{\lambda}$$.

I am not sure how to solve this one. I tried to take an open ball of radius $$r= \delta + \epsilon$$, for a small non-negative number $$\epsilon$$ which contains $$A$$, i.e. $$A \subset B(x,r)$$ for some $$x \in A$$. Then I consider a finite sub cover of $$X$$ : $$(U_{\lambda})_{\lambda =1}^{n}$$. Now this is also a cover of $$A$$, so I guess that I can find some open sets in my finite sub cover such that my ball is the union of these elements, i.e. $$B(x,r)= \cup_{i=1}^k U_i$$.

Is it clear that $$\cup_{i=1}^k U_i$$ is again contained in one of the $$U_{\lambda}$$ of the cover ?

Thanks for your help.

• – Kavi Rama Murthy Apr 15 at 11:55
• Oh I see. Thank you. – Alain Apr 15 at 11:57
• @KaviRamaMurthy Reading the proof, I don't really understand why the function is defined like that. Also do you understand why the minimum has to be strictly positive ? Thanks again – Alain Apr 15 at 12:28
• If $f(x)=0$ then $d(x,C_i)=0$ for all $i$ which implies $x \in C_i=A_i^{c}$ for each $i$. This means $x$ does not belong to any $A_i$, a contradiction. – Kavi Rama Murthy Apr 15 at 12:35
• thank you for your help. – Alain Apr 15 at 12:45