Family of Compact Sets as subset of $X$ wherein the finite intersection is non-empty, implies the infinite intersection is non-empty

Question

I have the following problem:

Let $X$ be a Hausdorff Space, and $J \neq \varnothing$, such that $\{K_{j}:j \in J\}$ is a family of compact sets, whereby any finite subset $J_{0} \subseteq J$ shows that $\cap_{j \in J_{0}}K_{j} \neq \varnothing$. Prove that $\cap_{j \in J}K_{j} \neq \varnothing$.

My idea: Let's assume that $\cap_{j \in J}K_{j} = \varnothing$. We now look at $K_{j}^c:=X-K_{j}$. It is clear that $K_{j}^c$ is open as the $K_{j}$ is a compact subset in a Hausdorff Space. We can now say that $X=\cup_{j \in J_{0}}K_{j}^{c}$.

This is where my question comes in: There was no mention made of the compactness of $X$. Is there any way of proving the above without compactness of $X$?

Using compactness, I would say: $\exists N \in \mathbb N: \cup_{j=1}^{N}K_{j}^{c}=X$, and that therefore $\cap_{j=1}^{N}K_{j}^{c}=\varnothing$ which is a contradiction. q.e.d

Answer 1

Your start of the proof will actually show the following:

Lemma : Let $X$ be a compact space and $F_i ,i \in I$ a family of closed subsets of $X$ that has the FIP (every finite subcollection has non-empty intersection). Then $\bigcap_{i \in I} F_i \neq \emptyset$.

The proof goes as you started: suppose for a contradiction that $\bigcap_{i \in I} F_i = \emptyset$. Then by define $U_i = X\setminus F_i$, open in $X$, for all $i \in I$, and the empty intersection assumption implies that $\{U_i: i \in I\}$ is a cover of $X$. So by compactness of $X$ we have a finite subcover $X = U_{i_1} \cup \ldots U_{i_n}$ for finitely many $i_1, \ldots , i_n \in I$. It follows (de Morgan) that

$$\bigcap_{j=1}^n F_{i_j} = \bigcap_{j=1}^n (X\setminus U_{i_j}) = X\setminus \bigcup_{j=1}^n U_{i_j} = \emptyset$$

whcih contradicts the fact that the family $F_i, i \in I$ has the FIP. So this contradiction shows that the intersection is non-empty.

Now you're almost where you want to be: we have $X$ merely Hausdorff and a family of $K_j, j \in J$ of compact subsets with the FIP. Fix one $j_0 \in J$ and apply the above lemma to a new $X = K_{j_0}$ (compact) and $F_j = K_j \cap K_{j_0}$ for $j \in J$. This family consists of closed sets because all $K_j$ are closed sets (as $X$ is Hausdorff!) and so these intersections are closed in $K_{j_0}$ by definition of the subspace topology. It still has the FIP, because any finite intersection of them is just a finite intersection of the original family too.

That concludes the proof.

Answer 2

In the proof, we will be using the following result:

Let $X$ be a Hausdorff space, let $Y$ be a compact subspace of $X$, and let $x \in X$ such that $x \not\in Y$. Then there exist disjoint open sets $U$ and $V$ in $X$ such that $x \in U$ and $Y \subset V$.

For a proof of this result, see Lemma 26.4 in the book Topology by James R. Munkres, 2nd edition.

Now for our proof.

Suppose that $$ \bigcap_{j \in J} K_j = \emptyset. \tag{1} $$

Fix a (non-empty) set $K$ in your collection.

Then, for each point $x \in K$, there exists a set $K_x$ in your collection such that $x \not\in K_x$, because otherwise (1) would not hold. For each $x \in K$, pick such a set $K_x$ from your collection.

As $X$ is a Hausdorff space, so for each point $x \in K$ and for the set $K_x$ not containing that point $x$, there exist disjoint open sets $U_x$ and $V_x$ such that $$ x \in U_x \ \qquad \mbox{ and } \ \qquad K_x \subset V_x. \tag{2} $$ In this way, we obtain an open covering $$ \left\{ \ U_x \ \colon \ x \in K \ \right\} $$ of the compact set $K$; so there exist finitely many points $x_1, \ldots, x_n \in K$ such that $$ K \subset \bigcup_{r=1}^n U_{x_r}. \tag{3} $$ Let us put $$ U \colon= \bigcup_{r=1}^n U_{x_r} \qquad \mbox{ and } \qquad U \colon= \bigcap_{r=1}^n V_{x_r}. \tag{Definition A} $$ Then we find that $$ K \subset U. \tag{4} $$ And, also $$ \bigcap_{r = 1}^n K_{x_r} \subset V. \tag{5} $$ Moreover, if $x \in V$, then $x \in V_{x_r}$ for each $r = 1, \ldots, n$, and so $x \not\in U_{x_r}$ for each $r = 1, \ldots, n$, because $U_{x_r}$ and $V_{x_r}$ are disjoint sets, by our construction, and therefore $x \not\in U$. Thus we have shown that $$ U \cap V = \emptyset. \tag{6} $$

Now from (2), (4), (5), and (6) above we find that $$ K \cap K_{x_1} \cap \cdots \cap K_{x_n} \subset U \cap V = \emptyset, $$ and so $$ K \cap K_{x_1} \cap \cdots \cap K_{x_n} = \emptyset, $$ also.

Thus we have shown that if our collection has empty intersection, then some finite sub-collection of our collection also has empty intersection. This is the contrapositve of (and hence equivalent to) what you were trying to prove.

This completes our desired proof.

Please feel free to contact me should you have any confusion.