# Nested descreasing intersection of nonempty covering compact sets is nonempty (check proof)

Proof that the nested descreasing intersection of nonempty covering compact sets is nonempty.

Let $$A_{1} \supset A_{2} \supset \cdots \supset A_{n} \supset \cdots$$ be a nested decreasing sequence of compacts. Suppose that $$\displaystyle \bigcap A_{n} = \emptyset$$. Take $$U_{n} = A_{n}^{c}$$, then $$\bigcup U_{n} = \bigcup A_{n}^{c} = \left(\bigcap A_{n}\right)^{c} = A_{1}.$$ Here, I'm thinking of $$A_{1}$$ as the main metric space. Since $$\{U_{n}\}$$ is an open covering of $$A_{1}$$, we can extract a finite subcovering, that is, $$A_{\alpha_{1}}^{c}\cup A_{\alpha_{2}}^{c} \cup \cdots \cup A_{\alpha_{m}}^{c} \supset A_{1}$$ or $$(A_{1}\setminus A_{\alpha_{1}})\cup(A_{1}\setminus A_{\alpha_{2}})\cup\cdots\cup(A_{1}\setminus A_{\alpha_{m}}) \supset A_{1}.$$ But, this is true only if $$A_{\alpha_{i}} = \emptyset$$ for some $$i$$, a contradiction.

Is correct?

• Why is $\{U_n\}$ an open covering? Compact subsets of topological spaces are not necessarily closed. See $X=\{0,1\}$ with the trivial topology, and the compact subset $\{0\}$. Dec 28, 2018 at 22:10
• @SmileyCraft I'm pretty sure that, at least in metric spaces (see the tag), compact subsets are closed. Dec 28, 2018 at 22:11
• @Noble Mushtak I saw the tag, but what confuses me is that the definition of compactness Lucas uses is very topological. I've never seen it in a real analysis course. Dec 28, 2018 at 22:12
• @SmileyCraft I'm reading Rudin's Analysis book right now, and they define a compact subset to be such that any open cover of the set has a finite subcover. This is the same definition Lucas seems to be using as well. Dec 28, 2018 at 22:14
• @SmileyCraft, thats true! In fact, the book Im using define first compact as sequentially compact. But after, the author present the covering compact definition and in this question is requested to use the second definition. Dec 28, 2018 at 22:16

In my opinion, you should add a justification of the assertion according to which that$$(A_1\setminus A_{\alpha_1})\cup(A_1\setminus A_{\alpha_2})\cup\cdots\cup(A_1\setminus A_{\alpha_m})\supset A_1\tag1$$can only occur if $$A_1=\emptyset$$. This happens because$$(1)\iff A_{\alpha_1}\cap\cdots\cap A_{\alpha_m}=\emptyset,$$but$$A_{\alpha_1}\cap\cdots\cap A_{\alpha_m}=A_{\max\{\alpha_1,\ldots,\alpha_m\}}\neq\emptyset.$$