Intersection of Compact Sets Is Not Compact

Question

What is an example of a topological space $X$ such that $C,K\subseteq X$; $C$ is closed; $K$ is compact; and $C\cap K$ is not compact?

I know that $X$ can be neither Hausdorff nor finite.

I am interested in this question because I recently read the following definition (in a Rudin book):

If $\left(X,\tau\right)$ is a topological space and $\infty\not\in X$, then $\left(X_\infty,\tau_\infty\right)$, where $X_\infty=X\cup\left\{\infty\right\}$ and every $U\in\tau_\infty$ is such that $U\in\tau$ or $U^c\subseteq X$ is compact, is a topological space.

I believe that this definition requires that $U^c\subseteq X$ be compact and closed.

Edit: The first question was my attempt to show that if $U,V\in\tau_\infty$ are such that $U\in\tau$ and $V^c\subseteq X$ is compact, then $\left(U\cup V\right)^c=U^c\cap V^c$ is not compact. However, this holds, as the comments show. A correct counter-example would be to show that if $U,V\in\tau_\infty$ are such that $U^c,V^c\subseteq X$ are compact, then $\left(U\cup V\right)^c=U^c\cap V^c$ is not compact, as Rob Arthan shows in his answer.

$C \cap K$ is a (relatively) closed subset of $K$, hence compact. — Daniel Fischer, Aug 09 '18 at 21:42
For any open cover ${\cal U}$ of $C$, adding $K\setminus C$ to ${\cal U}$ gives an open cover of $K$. It must therefore have a finite subcover; removing $K\setminus C$ from it gives a cover of $C$. — anomaly, Aug 09 '18 at 21:46
And yes, if you take a non-closed (quasi)compact $K \subset X$, let $U = X_{\infty}\setminus K$, then $U \cap X \notin \tau_{\infty}$. Does Rudin only consider Hausdorff spaces there? — Daniel Fischer, Aug 09 '18 at 21:46
Possible duplicate of Is the intersection of a closed set and a compact set always compact? — blub, Aug 09 '18 at 21:48
@zzuussee: this is about topological spaces not metric spaces and so not a duplicate of the question you cite. — Rob Arthan, Aug 09 '18 at 21:49
@RobArthan I was referring to the answer given in the corresponding question as well — blub, Aug 09 '18 at 21:51

Rob Arthan · Accepted Answer · 2018-08-09T22:54:10.870

I'm reading this question as asking about the OP's belief in the last sentence as well as just about the sentence with the question mark at the beginning (which has been addressed in the comments: the intersection $C \cap K$ of a closed set $C$ and a compact set $K$ is always compact).

In a space that isn't Hausdorff, compact sets aren't necessarily closed under intersections. E.g., take $(X, \tau)$ to be the line with two origins: then (using a notation that I hope is obvious), $A = [0_a, 1]$ and $B = [0_b, 1]$ are both compact but $A \cap B = (0_a, 1] = (0_b, 1]$ is not compact. So in Rudin's definition, you do indeed need to require $U^{c}$ to be compact and closed for the proposed set of open sets $\tau_{\infty}$ to satisfy the axioms of a topological space.

N.b., with the new MSE theme you can barely see it, but the words "line with two origins" above are a link to the Wikipedia page on that subject.

Perhaps, now that I finally did bin my answer, it's worth just saying that there is indeed no example of what OP asks for (as in the comments to their question). — Matt, Aug 09 '18 at 22:52
@Matt: thanks for the reminder. I've edited my answer to address that. — Rob Arthan, Aug 09 '18 at 22:55

Intersection of Compact Sets Is Not Compact

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