I am looking to prove that the infinite intersections of compact sets is compact. I'd like some feedback on my proof.
Note: This course isn't a formal course in topology and we specifically work in $\mathbb{R}^n$. I have done some reading and there seems to be some discussion about this question in other spaces.
Proof.
Let $A_1$ and $A_2$ be arbitrary sets. We need to show that $A_1 \cap A_2$ is compact.
Let $O$ be the open cover $A_1 \cup A_2$ which completely covers $A_1 \cap A_2$ since if $x \in A_1 \cap A_2$ then $x \in A_1$ and $x \in A_2$, which is completely covered by $O$.
Since $A_1$ and $A_2$ are compact, then there exists a finite subcover $O_1 \subset O$ and $O_2 \subset O$ respectively which cover the respective sets. So $O_1 \cup O_2$ cover $A_1 \cup A_2$. Therefore $O_1 \cap O_2$ covers $A_1 \cap A_2$. (Perhaps I can justify this better?)
Since the sets were arbitrary, we have that the infinite intersection is compact. QED.
Let me know what you think. Thanks!
Best Answer
The following are my messages (with a few updates) in the chat with MH1993, I moved them to here because I'm not sure if he can access it.
Well, in my opinion, you should start with an open cover $O$ of $A_1\cap A_2$ instead of taking an open cover for $A_1\cup A_2$ from the beginning. Of course, you can induce an open cover for $A_1\cup A_2$, say $O'=O\cup\{\mathbb{R}^n\setminus A\cap B\}$, because $A_1\cap A_2$ is closed.
Also, when you claim that $O_1\cap O_2$ covers $A_1\cap A_2$, I believe we may get some problems with this. For if $x\in A_1\cap A_2$, can we always find a single $U\in O_1\cap O_2$ such that $x\in U$? It could be the case that none of the open sets in $O_1$ belong to $O_2$, vice-versa.
Finally, as far as I can see, by showing that $A_1\cap A_2$ is compact you can conclude by induction that the intersection of finitely many compact sets is compact, but you can't conclude that the intersection of infinitely many compact sets is compact. For example, one cannot conclude that since "the product of any two compact sets is compact" then "the arbitrary product of compact sets is compact": the former is true in ZF while the later is equivalent to the axiom of choice. Maybe there is a way to prove what you want in the way you want, but I don't know how to do it.
However, in any case, you need to use the fact that if $K\subset\mathbb{R}^n$ is compact, then $K$ is closed. Hence any intersection of compact sets is closed and bounded. So, Heine-Borel Theorem handles the arbitrary case without many efforts.