Intersection of compact subsets of metric space / seeking alternative proof

Question

Theorem 2.36 of Baby Rudin states:

If $\{K_\alpha\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_\alpha\}$ is nonempty then $\cap K_\alpha$ is nonempty.

Rudin proves this by contradiction:

Choose $K_1 \in K_\alpha$, assume no point of $K_1$ belongs to every $K_\alpha$. Then the sets $K_\alpha^c$ form an open cover of $K_1$. Since $K_1$ is compact there are finitely many indices $\alpha_1, \dots, \alpha_n$ such that $K_1 \subset K_{\alpha_1}^c \cup \dots \cup K_{\alpha_n}^c$ which means $K_1 \cap K_{\alpha_1} \cap \dots \cap K_{\alpha_n} = \emptyset$. Hence we have an empty intersection of a finite subcollection of $K_\alpha$ which contradicts our assumption.

My question is this: is there a way to prove this directly instead of by contradiction? So to derive the fact that $\cap K_\alpha$ is nonempty directly from an assumption that every finite subcollection is nonempty.

To be clear, I understand Rudin's proof, but I am looking for alternative proof that is done directly as opposed to by contradiction. Someone marked this as a duplicate, but I do not see any existing question that addresses this specific concern.

Answer 1

Fix $K_1 \in \{ K_\alpha\}$. For any finite choice of indices $\alpha_1 , \dots, \alpha_n$,

$K_1 \cap \bigcap\limits_{i=1}^nK_{\alpha_i} \neq \emptyset \implies K_1 \not\subset \bigcup\limits_{I=1}^n K_{\alpha_i}^c$ which means that the open sets $\{K_\alpha^c\}$ do not have a finite subcover which covers $K_1$. Since $K_1$ is compact this means these sets do not cover $K_1$, hence there exists $p \in K_1$ such that $p \notin K_\alpha^c$ for all $\alpha \in X$. Thus $p \in K_\alpha$ for each $\alpha \in X \implies p \in \bigcap K_\alpha$ and specifically $\bigcap K_\alpha \neq \emptyset$.

Answer 2

Ordering the nonempty sets $K_\alpha$ and letting $L_i$ be the intersection of all the $K_\alpha$ for $\alpha\leq i$ we obtain a nested decreasing sequence of nonempty compact sets $L_i$ where $i$ runs through $\mathbb N$. Now we can apply Cantor's intersection theorem to conclude that the intersection is nonempty.

Cantor's intersection theorem does admit more direct proofs e.g, for $X=\mathbb R$. For instance, one can define $a_i$ to be the infimum of $L_i$, producing a monotone increasing sequence $(a_i)$. One can similarly define $b_i$ to be the supremum of $L_i$ to get a monotone decreasing sequence $(b_i)$. Both sequences are therefore bounded, and a point in the intersection of all the $L_i$ can be exhibited as the limit of $(a_i)$ or alternatively of $(b_i)$.

Answer 3

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).

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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.

Answer 4

In metric spaces, compactness is equivalent to sequential compactness - i.e. every infinite sequence in a compact set has a convergent subsequence. Picking $ x_i \in K_i $ for each i, we get an infinite sequence of points in $ K_1$. By definition of sequential compactness this must converge to a point. By construction of our sequence, this must be in the intersection of all the K's.