I was trying to understand the proof of the following theorem :
" A space X is compact if and only if every collection of closed subsets of X satisfying the finite intersection property has non-empty intersection."
The standard proofs of this theorem that I am seeing are same as the one covered in the question below :
Finite Intersection Property implies compactness?
As in the question above I could understand the proof for compactness implying non empty intersection of collection of closed sets having finite intersection property however I am not clear about the proof for the other direction .
Coming to the second proof as provided in the first answer above which goes as follows :
Suppose K has the finite intersection property. To prove that K is compact let {Ui}i∈I be a collection of open sets that cover K. We claim that this collection contains a finite subcollection of sets that also cover K.
Suppose that $K \neq\bigcup_{j \in J} U_j$ where $J\subset I$ is finite. Taking compliments gives $K^c \neq \bigcap U_j^c$, which by hypothesis is non-empty – since $U_i$ is open, $U_i^c$ is closed. Since $K$ has the fip we thus have that
$ \emptyset \neq \bigcap_{i \in I} U_i^c = \left( \bigcup_{i \in I} U_i \right)^c$. This contradicts $U_i$ being an open cover for $K$.
Here I am unable to see the significance of the part of this proof with the finite set $ J $ which is contained in $ I $ . Could we not have directly started with the following part ?
$ \emptyset \neq \bigcap_{i \in I} U_i^c = \left( \bigcup_{i \in I} U_i \right)^c$
Most importantly we obtained a contradiction to {Ui}i∈I being an open cover of the set in question . I can't see how does that lead to compactness ? How does it mandate the existence of an open subcover for all such open covers of this set ?
I saw a similar proof in the following blog :
but here as well I could not see how the contradiction leads to existence of a subcover .
Best Answer
A family of sets with the finite intersection property is said to be centred; for convenience I will use that term.
Dan Ma’s proof is not by contradiction. He wants to prove that if every centred family of closed sets in $X$ has non-empty intersection, then $X$ is compact. To do this, he proves the contrapositive: if $X$ is not compact, then $X$ has a centred family of closed sets whose intersection is empty. This is logically equivalent to the desired implication.
The argument itself is straightforward. Suppose that $X$ is not compact; then it has an open cover $\mathscr{U}$ with no finite subcover. For each $U\in\mathscr{U}$ let $F_U=X\setminus U$, and let $\mathscr{F}=\{F_U:U\in\mathscr{U}\}$; clearly $\mathscr{F}$ is a family of closed sets. Let $\mathscr{F}_0$ be any finite subset of $\mathscr{F}$. There is a finite $\mathscr{U}_0\subseteq\mathscr{U}$ such that $\mathscr{F}_0=\{F_U:U\in\mathscr{U}_0\}$. Then
$$\bigcap\mathscr{F}_0=\bigcap_{U\in\mathscr{U}_0}F_U=\bigcap_{U\in\mathscr{U}_0}(X\setminus U)=X\setminus\bigcup\mathscr{U}_0\,.$$
$\mathscr{U}$ has no finite subcover, so $\bigcup\mathscr{U}_0\ne X$, and therefore
$$\bigcap\mathscr{F}_0=X\setminus\bigcup\mathscr{U}_0\ne\varnothing\,.$$
Thus, $\mathscr{F}$ is centred: every finite subset of $\mathscr{F}$ has non-empty intersection. But
$$\bigcap\mathscr{F}=\bigcap_{U\in\mathscr{U}}(X\setminus U)=X\setminus\bigcup\mathscr{U}=\varnothing\,,$$
since $\mathscr{U}$ is a cover of $X$, so $\mathscr{F}$ is a centred family of closed sets in $X$ whose intersection is empty.
The proof that you copied into your question uses essentially the same idea but does organize it as a proof by contradiction. I’ll try to present it a bit more clearly. We start with an arbitrary open cover $\mathscr{U}=\{U_i:i\in I\}$ of a compact space $K$, and we suppose, to get a contradiction, that it has no finite subcover. Then for each finite $J\subseteq I$ we know that $\bigcup_{j\in J}U_j\ne K$. Now for each $i\in I$ let $F_i=K\setminus U_i$; then $\mathscr{F}=\{F_i:i\in I\}$ is a family of closed sets in $K$, and for each finite $J\subseteq I$ we have
$$\bigcap_{j\in J}F_j=\bigcap_{j\in J}(K\setminus U_j)=K\setminus\bigcup_{j\in J}U_j\ne\varnothing\,,$$
so $\mathscr{F}$ is centred. We are assuming that every centred family of closed sets in $K$ has non-empty intersection, so we conclude that $\bigcap\mathscr{F}=\bigcap_{i\in I}F_i\ne\varnothing$. But then
$$\bigcup\mathscr{U}=\bigcup_{i\in I}U_i=\bigcup_{i\in I}(K\setminus F_i)=K\setminus\bigcap_{i\in I}F_i\ne K\,,$$
contradicting the fact that $\mathscr{U}$ is a cover of $K$. This contradiction shows that there must in fact be a finite $J\subseteq I$ such that $\bigcup\{U_j:j\in J\}=K$, i.e., such that $\{U_j:j\in J\}$ is a finite subcover.