Talk:Intersection of Closed Set with Compact Subspace is Compact/Proof 2

$K$ instead of $H$

I believe it should be:

Hence $\family {U_\alpha} \cup S \setminus H$ is an open cover of $K$.

Rather than "open cover of $H$". This is because $S\setminus H$ by definition excludes $H$.

And:

It follows by definition that a finite subcover:

$\set {U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n}, S \setminus H}$

of $K$ exists.

Rather than "of $H$ exists". This is because $K$ is compact, yielding the finite subcover.

Well spotted. Good call. --prime mover (talk) 20:03, 15 January 2022 (UTC)

All corrections are appreciated :) BTW, you should sign talk page messages with ~~~~. Caliburn (talk) 20:17, 15 January 2022 (UTC)