Question:
If ${Sn}$ is a sequence of nonempty compact sets with $S_{n+1} ⊂ S_{n}$ for
every $n$, show that ${\bigcap}_{n=1}^∞
S_{n} \neq ∅$.
My Attempt
Choose $S_{k}$ from sequence ${Sn}$. Assume ${\bigcap}_{n=1}^∞ S_{n} = ∅$.
Such that no points inside $S_{k}$ are in unions of all the other $S_{n}$
Set $T_{\alpha} = S_{\alpha}^{c}$.
$S_{k}$ is contained in the open cover formed by several $T_{\alpha}$
Since $S_{k}$ is compact, there exists a finite subcover of several $T_{\alpha}$ that covers $S_{k}$.
This means $S_{k} ⊂ T_{\alpha_1}{\bigcap}T_{\alpha_2}{\bigcap}T_{\alpha_3}{\bigcap}….{\bigcap}T_{\alpha_n}$
This also implies $S_{k}{\bigcap}S_{\alpha_1}{\bigcap}S_{\alpha_2}{\bigcap}S_{\alpha_3}{\bigcap}….{\bigcap}S_{\alpha_n} = ∅$
However, the condition says, $S_{n+1} ⊂ S_{n}$ which means $S_{k}{\bigcap}S_{\alpha_1}{\bigcap}S_{\alpha_2}{\bigcap}S_{\alpha_3}{\bigcap}….{\bigcap}S_{\alpha_n} = ∅$ is not possible, thus making contradiction.
Thus ${\bigcap}_{n=1}^∞
S_{n} \neq ∅$.
I'm not sure whether this would be the legitimate proof for this question, especially the last part about the contradiction. Could anyone verify or suggest a better method to address this question?
Thanks in advance!
Best Answer
This is actually not true in general you need that the the compact sets are also closed. A simple counter example is the reals with the topology that has all sets of the form $(x,\infty)$ Any set of the form $[y,\infty)$ is going to be compact but it's not closed since the only closed sets are of the form $(-\infty,x]$. So taking the sequence of sets $([n,\infty) )_{n\in\mathbb{N}}$ then the intersection is empty even thought they are all compact. In a $T_2$ space like the real line with its standard topology all compact sets are closed. Your statement is true in this case.
Edit: I see that there are no general topology tags on your question. Just ignore this part if you haven't taken any introductory courses on point-set topology. The following proof works for the Real numbers because of the Heine-Borel Theorem which tells us compact sets are closed.
Proof: Given a sequence $S_n$ of decreasing non empty closed compact sets we can assume by taking a subsequence that it's strictly decreasing. assume that $\bigcap_{n} S_n=\emptyset$ then one defines an sequence of open sets as the complements $A_n=S_n^c$ for $n \geq 2$ Using Demorgan $\bigcup A_n=\bigcup S_n^c=(\bigcap S_n)^c=X$ So it's an open cover of the entire topological space $X$ so it must also cover $S_1$ so there must be a finite open subcover of $S_1$ since it's compact. But $A_n\subseteq A_{n+1}$ for all $n$ so if there is a finite subsequence there must be an $n$ such that $A_n$ covers $S_1$ Since $S_n\subseteq S_1$ and $A_n=S_n^c$ we get $S_n=S_1\cap S_n=S_1\setminus S_n^c =S_1\setminus A_n=\emptyset$. But this contradicts the fact that all of the $S_n$ were non empty so the intersection can't be non empty.