Baby Rudin Theorem 2.35 Proof Explanation

Question

The theorem states: Closed subsets of compact sets are compact.
The proof is:

Suppose $F\subset K\subset X$, $F$ is closed (relative to $X$), and $K$ is compact. let $\{V_\alpha\}$ be an open cover of $F$. If $F^c$ is adjoined to $\{V_\alpha\}$, we obtain an open cover $\Omega$ of $K$. Since $K$ is compact, there is a finite subcollection $\Phi$ of $\Omega$ which covers $K$, and hence $F$. If $F^c$ is a member of $\Phi$, we may remove it from $\Phi$ and still retain an open cover of $F$. We have thus shown that a finite subcollection of $\{V_\alpha\}$ covers $F$.

My question is are the two "if"s always true? Specifically:

If $F^c$ is adjoined to $\{V_\alpha\}$…
If $F^c$ is a member of $\Phi$…

I will admit that I still do not quite understand the concept of open covers and subcovers, so any clarifications on that will be very appreciated!

Answer 1

If $F^c$ is adjoined to $\{V_\alpha\}$, we obtain an open cover $\Omega$ of $K$.

The $V_\alpha$ are open sets, by definition. $F^c$ is also open (because $F$ is closed). So the $V_\alpha$ with the additional set $F^c$ is a collection of open sets. To see that they form a cover of $X$, it suffices to show that any point of $x$ lies in some set in this collection of open sets. If $x \in K$, then $x \in V_\alpha$ for some $V_\alpha$ (since $\{V_\alpha\}$ is a cover of $K$). Otherwise if $x \in K^c$, then $x \in F^c$ because $F \subset K$.

If $F^c$ is a member of $\Phi$, we may remove it from $\Phi$ and still retain an open cover of $F$.

$\Phi$ is a finite sub-collection of the collection $\Omega$ that covers $K$.

• If $F^c$ is not a member of $\Phi$, then $\Phi$ covers $F$ (because it covers $K$), and is a finite sub-collection of the $V_\alpha$.
• If $F^c$ is a member of $\Phi$, the claim is that if we remove $F^c$ from $\Phi$, the remaining collection of finitely many open sets would still be an open cover of $F$. But no point of $F$ lies in $F^c$, so $F^c$ is unnecessary for covering $F$.