(This is not a duplicate of the question in the link below. I want some more explanation but the question is made 7 years ago, so I don't expect to be answered if I just make a comment there.)
Let us say a set is an ordinal if it is well-ordered and transitive under the relation $\in$. $\omega_1$ is the first uncountable ordinal. I give the order topology on $\omega_1$. The statement I want to prove is:
For each positive integer $n$, let $K_n$ be a compact subset of $\omega_1$. Assume $K_n$ has uncountably many elements. Then the intersection $K=\bigcap_{n=1}^\infty K_n$ is an uncountable set.
I found a proof for the above proposition. It is in the link:
The proof that $K$ is uncountable(Math stack exchange)
In the answer, azarel says that:
If $\xi<\omega_1$, then there is an ordinal $\gamma$ in the set $K_0 \cap K_1$ such that $\xi<\gamma$.
But I don't see why this is enough. All I can think is that I can choose $\gamma<\omega_1$, so $K_0 \cap K_1$ has at least countably many elements. How to proceed to the uncountability?
Best Answer
Suppose $K_0\cap K_1$ were countable. Then the ordinal $\xi=\bigcup K_0\cap K_1$ is a countable union of countable sets, so it is countable. Thus $\xi<\omega_1$, so there exists $\gamma\in K_0\cap K_1$ such that $\xi<\gamma$. But this is a contradiction, since $\gamma\in K_0\cap K_1$ means $\gamma\subseteq\xi$ so $\gamma\leq \xi$.
The point here is that $\omega_1$ has uncountable cofinality: you can't "reach the end of it" with a countable set, since the supremum of a countable set of countable ordinals will still be a countable ordinal and thus strictly below $\omega_1$. So, any countable subset of $\omega_1$ is bounded in $\omega_1$.