Let $A,B$ be compact subsets of $X$. Prove that $A \cap B$ is compact.

Question

Let $A,B$ be compact subsets of $X$. Prove that $A \cap B$ is compact.

Attempt: Suppose by contrapositive, that $A \cup B$ is compact. Then let $V$ be an open cover of $A \cup B$. Then let $A$ be compact, then $ V$ has a finite subcover of $A$ . But suppose $B$ is not compact. Then $V$ does not contain a finite subcover of $B$. Thus their union will not be have a finite subcover of $ V$ that converse $A \cup B$.

Can someone please help me? I already proved if $A, B$ are compact , then their union is compact. So I was trying to use a similar argument. Any feedback/help would really help. Thanks

Answer 1

There is a problem with your contrapositive.

Hint: show that a closed part of a compact set is compact.

Answer 2

If $X$ is Hausdorff (in particular all metrci spaces are Hausdorff), every compact subset is closed.

So, $A,B$ are closed $\Rightarrow$ $A \cap B$ is closed in $A$. But $A$ is compact, so $A\cap B$ is a closed subset of a compact set, hence it is compact.

Answer 3

In any topological space if you suppose that A and B are compact then it holds that A can be written as a finite cover of open sets and so can B (definition of compactness). So if you intersect open sets you still get open sets therefore that should be a finite cover of open sets of = (A intersection B) and again according to defenition the intersection is compact.

ok, this happens only in hauusdorf spaces as I have seen and it's incorrect for every space. I've seen a counter example: (intersection of two compacts isn't compact) Y-with the discrete topology Y is infinite and X is taken to be X=Y uninon {c1} union {c2}, where {c1} and {c2} are two arbitary points. The topology on X is defined to be all the open sets in Y. Now can anyone understand this counter example? It doesn't make sense...