Ordinal $\omega_1$ is not compact

Question

Any ordinal number can be turned into a topological space by using the order topology. The topological space $\omega_1$ is sequentially compact but not compact.

Why is $\omega_1$ not compact ?

Answer 1

Something is wrong here. $[0,\omega]$ is compact for any ordinal $\omega$.

Let $U$ be an open cover of $[0,\omega]$, and let $\theta$ be the smallest ordinal such that $[\theta,\omega]$ is covered by a finite subcover of $U$. Such a $\theta$ exists because there is clearly a set of all such $\theta$ that do the job, and so we are justified in defining $\theta$ to be the smallest one.

Now if $\theta$ is a successor, we can write $\theta=\eta+1$, then there is an element of $U$ that contains $\eta$, and so we can toss that into our finite cover of $[\theta,\omega]$ produces a finite cover of $[\eta,\omega]$, but this contradicts that $\theta$ was the smallest. Thus $\theta$ is not a successor.

On the other hand, we will also show now that $\theta$ is not a limit ordinal. Suppose for the sake of contradiction that it was. Then every open set that contains $\theta$ must also contain some $\zeta<\theta$, and therefore our finite subcover of $[\theta,\omega]$ is already a cover of $[\zeta,\omega]$. Again a contradiction.

The only possibility remaining is that $\theta$ is neither a successor nor a limit, but this is just saying that $\theta=0$.

After writing this, I realized you might have meant not $[0, \omega_1]$ but in fact $[0, \omega_1)$. It is easier I think to see that this space cannot be compact. This is simply because we can cover this space by an uncountable collection of disjoint intervals starting from $0$. Since the space doesn't contain its upper endpoint, and $\omega_1$ is a limit ordinal, no finite subcover can suffice.

Either way, hope this helps you straighten everything out, and welcome to MSE!

Answer 2

It is not compact since the collection $\{[0,\alpha):\alpha<\omega_1\}$ is an open cover with no finite subcover. Note that this isn't very surprising, since any limit ordinal is not compact by the exact same logic.

What is perhaps more novel is that $\omega_1$ is sequentially compact, whereas any countable limit ordinal is not since a cofinal $\omega$-sequence has no convergent subsequence. The novelty of $\omega_1$ is it that it has no cofinal $\omega$-sequences.