So I am looking over some topology, and am stuck on this problem.
I have that $(K_n)_{n\in \Bbb{N}}$ is a nested sequence of non-empty compact sets in a metric sets. I have managed to prove the first part of the problem, that $\bigcap K_n \ne \varnothing $. From here I want to prove that if an open set $U$ contains $\bigcap _{n \in \Bbb{N}}K_n$ then there exists an $n_0$ such that $K_n \subseteq U$ for all $n \geq n_0$.
Hoever, I am stuck on this. I was thinking that I might want to reason along the lines that as $K_{n+1} \subseteq K_n$ for every $n$, as $n \rightarrow \infty$ there must only be a small ball, say of radius $ \epsilon \gt 0$, about a point, say $z_0$ which is in $\bigcap K_n$: $B(z_0, \epsilon)$ in $K_{n+1}$. Thus, as U is open, it is a neighbourhood of all of its points, so we can see that for $K_{n+1}$ such that it can be expressed as a ball about the point in the intersection, $K_{n+1} \in U$. I don't know if this is correct, and I don't know if, to assert this I need to prove somehow that there is only one point on the intersection, or is it sufficient to just see that the the $K_n$s must constantly be decreasing in diameter as they are nested, meaning there can only be one point in the intersection?
Thanks
Suppose otherwise. That is, suppose that$$(\forall n\in\mathbb N):K_n\varsubsetneq U.$$Then, for each $n\in\mathbb N$, consider the non-empty set $L_n=K_n\cap U^\complement$. The set $L_n$ is a closed subset of $K_n$ and therefore it is compact. Furthermore,$$(\forall n\in\mathbb N):L_{n+1}\subset L_n.$$So, $\bigcap_{n\in\mathbb N}L_n\neq\emptyset$. Take $x\in\bigcap_{n\in\mathbb N}L_n$. Then $x\in\bigcap_{n\in\mathbb N}K_n$, but $x\notin U$, which is a contradiction.