If $\bigcap\limits_\alpha K_\alpha=\emptyset,$ then there exists $\alpha_1,\dots,\alpha_n$ such that $\bigcap\limits_{k=1}^n K_{\alpha_k}=\emptyset$

Question

Let $\{K_\alpha\}$ be a collection of compact sets of a Hausdorff space. If $\displaystyle\bigcap_\alpha K_\alpha=\emptyset,$ then there exists $\alpha_1,\dots,\alpha_n$ such that $\displaystyle\bigcap_{k=1}^n K_{\alpha_k}=\emptyset.$

There is a hint by martini

Hint. Let $L_i = \bigcap_{j=1}^i K_j$, then the $K_1 \setminus L_i$ form an open cover of $K_1$. Now use $K_1$'s compactness to find an $i$ such that $K_1 = K_1 \setminus L_i$ (note that the $K_1 \setminus L_i$ are an increasing sequence), hence $L_i = \emptyset$.

I don't understand this

Now use $K_1$'s compactness to find an $i$ such that $K_1 = K_1 \setminus L_i$ (note that the $K_1 \setminus L_i$ are an increasing sequence), hence $L_i = \emptyset$.

why are $L_i$ increasing?

and why

hence $L_i = \emptyset$?

At the moment I have

I have $K_1\cap L_i^c=(K_1\cap K_1^c)\cup\dots\cup(K_1\cap K_j^c)$, all are open in $K_1$ then they cover $K_1,$ $\displaystyle K_1=\bigcup_{j=1}^i K_j^c=\big(\bigcap_{j=1}^i K_j\big)^c$

Could someone help to understand please?

thanks in advance.

Answer 1

martini's hint does not help you. It is only applies to the case that you have a countable cover $\{ K_i \}$. Nevertless, the sequence $K_1 \setminus L_i$ is increasing because the sequence $L_i$ is obviously decreasing. Now $\{ K_1 \setminus L_i \}$ is a cover of $K_1$, hence it has a finite subcover. This means (because the sequence $K_1 \setminus L_i$ is increasing ) that there exists $i$ such that $K_1 = K_1 \setminus L_i$. Since each $L_i \subset K_1$, we conclude that $L_i = \emptyset$.

So what can be done in the general situation? Pick any $K_{\alpha_0}$. Then the family $L_\alpha = K_{\alpha_0} \setminus K_\alpha$ is an open cover of $K_{\alpha_0}$. Here we invoke that the compact $K_\alpha$ are closed in the Hausdorff space $X$. Hence there exists a finite subcover $L_{\alpha_1}, \dots, L_{\alpha_n}$ of $K_{\alpha_0}$. This means $K_{\alpha_0} = \bigcup_{i=1}^n L_{\alpha_i} = K_{\alpha_0} \setminus \bigcap_{i=0}^n K_{\alpha_i}$, in other words $\bigcap_{i=0}^n K_{\alpha_i} = \emptyset$.

Answer 2

$K_{\alpha_0}$ (pick any $\alpha_0$, the index set $A$ is irrelevant) is compact.

Then $L_\alpha:=K_\alpha \cap K_{\alpha_0}$, for $\alpha \in A$ are all closed subsets of $K_{\alpha_0}$ (by Hausdorffness all compact sets are closed and so intersecting them with $K_{\alpha_0}$ gives us closed subsets of $K_{\alpha_0}$ again. If one of these sets is empty, say $L_\alpha$ is, we are done because we can take $\alpha_0$ and $\alpha$ as the finite subset of the index set we have to find. So we can assume they're all non-empty, but still $$\bigcap_{\alpha \in A} L_\alpha = \bigcap_{\alpha \in A} K_\alpha = \emptyset$$

and so the (open in $K_{\alpha_0}$) sets $U_\alpha = K_{\alpha_0} \setminus L_\alpha$ ( where $\alpha \in A$) form an open cover of $K_{\alpha_0}$: if not, some point would not be covered by any $U_\alpha$ and this point would thus be in all sets $K_{\alpha_0}\setminus U_\alpha = L_\alpha$ contrary to this intersection being empty.

As $K_{\alpha_0}$ is compact, finitely many $\alpha_1, \ldots \alpha_n$ exist such that $U_{\alpha_i}, i=1, \ldots n$ cover $K_{\alpha_0}$ and then it's equally clear that $K_{\alpha_0},K_{\alpha_1}, \ldots, K_{\alpha_n}$ have empty intersection and we are done: suppose $p$ would be in this intersection, then we'd have a point in $K_{\alpha_0}$ that is also in none of the $U_{\alpha_i}, i=1, \ldots n$, while we know this to be a finite cover of $K_{\alpha_0}$.