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Using the fact that:

S is compact: every open cover has a finite subcover.

Prove:

Given $A,B \subset \Bbb R^n$ are compact sets, then $A\cap B$ is compact.


Here is my attempt, using the fact that $A$ follows the above Heine-Borel criterion above and $A \cap B \subset A$:

Since $A$ is compact, every open cover of $A$ has a finite subcover. We wish to show that every open cover in $A \cap B$ has a finite subcover.

Since $A \cap B \subset A$, then every open cover of A must be an open cover of $A \cap B$ (from the definition of an open cover shown below).

A collection of sets ${U_\alpha}$ is an open cover os $S$ if $S$ is contained in $\bigcup U_\alpha$.

Since $A$ is compact, we know that every open cover has a finite subcover.

Therefore, since $A \cap B \subset A$ and $A$ has a finite subcover for every open cover, $A \cap B$ has a finite subcover for every open cover.

Is this the correct way to approach this problem?

Thanks!

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  • $\begingroup$ I understand that $A\cap B$ is closed and bounded so $A\cap B$ is compact, but is there a way to show $A \cap B$ is compact using fact that every open cover has a finite subcover? $\endgroup$
    – Raoul Duke
    Commented Dec 19, 2019 at 2:23
  • $\begingroup$ Thanks for the feedback George. I appreciate it. So, I could complete the proof by showing the same for $B$ as I did for $A$? Therefore "covering" all covers of $A\cap B$ $\endgroup$
    – Raoul Duke
    Commented Dec 19, 2019 at 2:35
  • $\begingroup$ @GeorgeDewhirst: No "Heine-Borel" as he states is something about open covers, not about "closed and bounded". That thing about open covers works in situations other than Euclidean space. So stating the problem this way is prompting Raul to come up with a proof applicable in spaces other than $\mathbb R^n$, even if he does not know it. $\endgroup$
    – GEdgar
    Commented Dec 19, 2019 at 2:43

2 Answers 2

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A red flag here is that you never used the fact that $B$ is compact. And you have to be more careful about what exactly you mean by a "cover."

You are given that $A$ and $B$ are compact subsets of $\mathbb R^n$ and you want to prove that $A\cap B$ is compact, using only the definition. So, give $A\cap B$ the subspace topology, and let $\{U_{\alpha}\}_{\alpha\in \Lambda}$ be an open cover of $A\cap B$. By definition of the subspace topology, there are opens $\{O_{\alpha}\}_{\alpha\in \Lambda}$ in $\mathbb R^n$ such that $O_{\alpha}\cap(A\cap B)=U_{\alpha}$. Then, and here we use the fact that $A\cap B$ is closed (because it is the intersection of compact, hence closed sets), the sets $\{O_{\alpha}\}_{\alpha\in \Lambda}\cup \mathbb R^n\setminus A\cap B$ form an open cover of $A$, and now we can extract a finite subcover which covers $A$, hence covers $A\cap B$. And since $\mathbb R^n\setminus A\cap B$ is $\textit{not}$ a covering element of $A\cap B$ so it must be that the finite cover consists of elements of $\{O_{\alpha}\}_{\alpha\in \Lambda}$ alone. And now, unwinding the definitions, we get a finite subcover of the original cover $\{U_{\alpha}\}_{\alpha\in \Lambda}$.

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  • $\begingroup$ Hi @Matematleta, thanks for the answer. Could you elaborate on "there are opens $\{O_{\alpha}\}_{\alpha\in \Lambda}$ in $\mathbb R^n$ such that $O_{\alpha}\cap(A\cap B)=U_{\alpha}$." $O_{\alpha}$ is an open set in $\mathbb R^n$. Why is it that the intersection with said open set and $A \cap B$ equal to$U_{\alpha}$? $\endgroup$
    – Raoul Duke
    Commented Dec 19, 2019 at 3:05
  • $\begingroup$ It's the definition of the subspace topology. $\endgroup$ Commented Dec 19, 2019 at 3:08
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In a very general approach you can consider the following result:

Proposition: Let $(X, \mathscr{T})$ be an arbitrary topological space, $K \subseteq X$ be compact (with the relative topology) and $F \subseteq X$ a closed subset. Then $K\cap F$ is also compact.

Proof: As the intersection with an absolutely closed subset, $K \cap F$ will be closed relatively to the subspace topology on $K$; the result immediately follows from the general proposition that closed subsets of compact spaces are compact. $\Box$

This general result applies indeed to your case since any compact subset in a Hausdorff space is necessarily closed.

Try to see if you can understand and prove the proposition mentioned at the end, let me know if you need more details.

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