There's a mistake:
so we obtain an open cover $K^c \cup V$ of $K$, and because $K$ is compact
You don't know that $K$ is compact. You are trying to prove that $K$ is compact. You only know that $K$ is closed and $F$ is compact. In fact, the mere fact that you don't use $F$ anywhere in your proof should be cause for alarm.
Also, you are confusing covers (which are collections of sets) with their coverage (i.e., union of their component sets).
So, when you write that you take a finite subcover $\Omega$ of $K$, you then say that $K\subset \Omega$, which is not true. What is true is that $$K\subset\bigcup_{A\in\Omega} A$$ which is different!
I advise you to restart the proof again, and here's a couple points to start you off:
- You need to prove that $K$ is compact
- That means you need to prove every open cover of $K$ has a finite subcover
- That means you take some cover, $\mathcal V$, such that $K$ is covered by $\mathcal V$ (you were on point until here).
- Now you need to prove there exists a finite subcover of $\mathcal V$.
And a little hint:
- You know $F$ is compact, so any open cover of $F$ has a finite subcover.
- Can you add one more set to $\mathcal V$ such that the resulting cover will be a cover for $F$?
Aftert your edit:
You still have that misstake of cover vs coverage. So, instead of saying that $K^c\cup V$ is an open cover of $F$, you should say that $\{K^c\}\cup \{V_a\}$ is an open cover of $F$.
I am even more concerned that I think you are confused a bit about what an open cover is. When you say
Now, consider the union $K^c \cup V$. Since $K \subset V$, it follows that $X =K^c \cup K \subset K^c \cup V$, meaning that $X = K^c \cup V$, so $F \subset K^c \cup V$ and $X$ is an open subset of itself
I'm thinking "yes, all he said is true, but it's irrelevant".
An open cover is a collection of open sets, not a collection of sets whose union is open. So, you don't need to prove that $$K^c\cup V$$ is open (even though it is), you need to prove that every element of $$\{K^c\}\cup\{V_a\}$$ is open (which is not hard, it's just that if you don't do that, the proof is incorrect).
Well you can start by redefining the concept "compact" by stating that a space $X$ is compact if for every family $(F_{\alpha})_{\alpha\in A}$ of closed sets that has the so-called "finite intersection property" the intersection $\bigcap_{\alpha\in A}F_{\alpha}$ is not empty.
Here family $(F_{\alpha})_{\alpha\in A}$ has by definition this "finite intersection property" if for every finite $B\subseteq A$ the intersection $\bigcap_{\alpha\in B}F_{\alpha}$ is not empty.
Using this underlying definition of "compact" (equivalent with the definition that states that open covers must have finite subcovers) it is enough to prove here that in metric spaces every compact subspace $K$ is closed (so that the family of compact sets $(K_{\alpha})$ can be recognized as a family of closed sets).
edit:
The following statements are equivalent:
- (1) Every collection of open sets that cover $X$ has a finite subcover.
- (2) Every collection of closed sets that has the finite intersection propery has a non-empty intersection.
(1)$\implies$(2)
Let $(F_{\alpha})_{\alpha\in A}$ be a collection of closed sets that has the finite intersection property.
Now define $U_{\alpha}=F_{\alpha}^{\complement}$ for $\alpha\in A$ and observe that for every finite $B\subseteq A$ we have $$\bigcup_{\alpha\in B}U_{\alpha}=\bigcup_{\alpha\in B}F_{\alpha}^{\complement}=(\bigcap_{\alpha\in B}F_{\alpha})^{\complement}\neq X$$
This tells us that the collection $(U_{\alpha})_{\alpha\in A}$ has no finite subcover of $X$. It is a collection of open sets, so this allows us to conclude that $(U_{\alpha})_{\alpha\in A}$ does not cover $X$.
That means that: $$\bigcap_{\alpha\in A}F_{\alpha}=\bigcap_{\alpha\in A}U_{\alpha}^{\complement}=(\bigcup_{\alpha\in A}U_{\alpha})^{\complement}\neq\varnothing$$as was to be shown.
(2)$\implies$(1)
Let $(U_{\alpha})_{\alpha\in A}$ be a collection of open sets that cover $X$.
Now define $F_{\alpha}=U_{\alpha}^{\complement}$ for $\alpha\in A$ and observe that: $$\bigcap_{\alpha\in A}F_{\alpha}=\bigcap_{\alpha\in A}U_{\alpha}^{\complement}=(\bigcup_{\alpha\in A}U_{\alpha})^{\complement}=\varnothing$$
That implies the existence of a finite $B\subseteq A$ such that $\bigcap_{\alpha\in B}F_{\alpha}=\varnothing$ (non-existence of such finite $B$ should contradict that closed family $(F_{\alpha})_{\alpha\in A}$ has the finite intersection property).
Then $$\bigcup_{\alpha\in B}U_{\alpha}=\bigcup_{\alpha\in B}F_{\alpha}^{\complement}=(\bigcap_{\alpha\in B}F_{\alpha})^{\complement}=\varnothing^{\complement}=X$$
Proved is now that cover $(U_{\alpha})_{\alpha\in A}$ has a finite subcover $(U_{\alpha})_{\alpha\in B}$ and we are ready.
Best Answer
$K$ can be an improper subset in the definition. Some authors use $\subseteq$ and $\subset$ for subset and proper subset respectively, while others use $\subset$ and $\subsetneq$ respectively. Rudin uses the latter.
An open set obviously forms an open cover of itself, but a set is compact if every open cover has a finite subcover. You need to understand the notation for subset and definition of compact to avoid confusion.