Intersection of nested, non-empty compact subsets of a hausdorff topological space

general-topology

So I feel like I'm on the verge of solving this problem, but honestly don't know the crux. The question is:

Let $(X,T)$ be a topological space and let this space be hausdorff. Consider the infinite sequence $K_1⊇K_2⊇,…,⊇K_n⊇..$ of non-empty, compact subsets of $X$. Prove that their infinite intersection is non-empty.

So far I have written:

Assume that this intersection is empty. Consider $U_n = X$ $\backslash K_n$. (Some of the following statements I will write without proofs).
Firstly, it is known that if X is hausdorff, then any compact subset of $X$ is closed. Additionally, for some $Z⊆K⊆X$, $Z$ too is compact.

This implies that for all $n$, $K_n$ is closed. It is also known that the intersection of an arbitrary family of closed subsets of $X$ is also closed. Therefore the intersection of all $K_n$ is closed too in $X$. (Now I have just noticed when writing this, by assumption the intersection was the empty set which is an open set, so can the proof end here or did I do something wrong?).

By definition, the compliment of a closed set is open. Then, $U_n$ is open in $X$. Define $Ω_n = K_1 $ \ $K_n$, and $Ω_n$ is an open set. $Ω_n$ then defines an open cover for $K_1$. As K is compact, there exists a finite sub-cover $(Ω_{n_1}, Ω_{n_2},…, Ω_{n_m})$.

I honestly don't know where to go from here at all, and in fact I'm not sure what I have written is correct. Any ideas?

Best Answer

You're very close, you've got all the right pieces.

I think this might be a little easier if you phrase it as a proof by contradiction. So, something like: `Assume to the contrary that $\bigcap K_n = \emptyset$, then the collection of $\Omega_n$ is an open cover of $K_1$'.

You've got the finite subcover $\{\Omega_{n_1}, \dots, \Omega_{n_m} \}$, and with Daniel's comment, you can make the following observation (hover over to get the hint, if you still need it):

If w order things so that $n_1 < n_2 < \dots, n_m$, you have that $\Omega_{n_1} \subseteq \Omega_{n_1} \subseteq \dots \Omega_{n_m}$. Since $\{\Omega_{n_1}, \dots, \Omega_{n_m} \}$ is a cover of $K_1$, this means that the lone set $\Omega_{n_m}$ is a cover of $K_1$.

From that observation, you can complete the proof:

But $\Omega_{n_m} = K_1 - K_{n_m}$. If $\Omega_{n_m}$ covers $K_1$, then $K_{n_m}$ must be empty! This contradicts the assumption that it is not empty, completing the proof.

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