[Math] Intersection of Closed and Compact Set is Closed

compactnessgeneral-topology

Fact

The intersection of a closed set $A$ and a compact set $B$ is closed.

Background

I wanted to prove that the intersection of closed and compact sets is compact. There might be other ways, but I am not interested in them for now and am looking to prove the above statement.

Relevant Facts

I don't have much idea what to use. $A^c$ is open. Every cover $\cup_{i\ge1}U_i$ of $B$ contains a subcover $\cup_{i=1}^kU_i$. I might want to prove $A\cap B^c$ is open. Limit points may be another points to consider but I don't get any idea in that direction.

I also don't have that $B$ is closed otherwise it would be trivial.

Some Clarification

I am aware that compact sets are not closed in arbitrary topological spaces. The reason why I was motivated to believe tha intersection of closed and compact set is compact is true is: See Matt E.'s Answer Here.

So if $K$ is compact in an arbitrary topological space (which is just to say that
it is a compact topological space when given its induced topology) and
$L$ is closed then $K \cap L$ is a closed subset of $K$ in its induced topology

Is it true if $A$ and $B$ are not whole spaces?

I would appreciate any hints.

Best Answer

It is not always true. The following example witnesses it.

Example 1: Let $X= \mathbb R$ with finite complement topology. Note that $X$ is $T_1$, not $T_2$. It is not difficult to prove that $X$ is compact. Now let $A=X$; $B=\{0,1,2,\dots,n,\dots\}$. Note that $A$ of course is closed and $B$ is compact, however it is not closed. Therefore $A \cap B=B$ is not closed.

However it is true if $A$ and $B$ belong to a Hausdorff space.

Theorem 2: Every compact subspace of a Hausdorff space $X$ is a closed subspace of $X$.

Proof: Let $A$ be a compact subspace of $X$. For every $x\in X\setminus A$ there exists an open set $V\subset X$ such that $x \in V$ and $A \cap V=\emptyset$, so that $X \setminus A$ is an open subset of $X$.

So in a Hausdorff space, $B$ is always closed as it is compact, and hence $A \cap B$ is closed, as the intersection of two closed set is closed.

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