$\omega_1$ is not Lindelof

Question

I am looking to prove that $\omega_1$ is not Lindelof. Here is my proof so far:

I am attempting to reveal a contradiction.

Consider the collection $\mathscr{U} = \{(a,b]: a,b \in \omega_1$, $a<b$, $b$ is a limit ordinal$\}$. Clearly, $\mathscr{U}$ covers the space and is an open cover. Now, bounded sets in $\omega_1$ are countable. And the countable union of countable sets is once again countable. So suppose there exists a countable subcover $\mathscr{U}_0$. Then since $\mathscr{U}_0$ covers $\omega_1$ there exists some $(a, \omega_1] \in \mathscr{U}_0$.

But where do I go from here? The uncountability of $\omega_1$ is going to prevent a countable subcover, I understand, but putting this in proof formality isn't as intuitive.

Answer 1

Do it like this: set $\mathscr{U} = \{[0,a): a \in \omega_1\}$ and suppose $S=\{[0,a_n)\}_n$ is a countable subcover. Now, $A:=\sup S=\bigcup S$ is a countable ordinal (why?) and so $A\in \omega_1$ and so there is an integer $n$ such that $A\in [0,a_n).$ But then $A<a_n$ which is a contradiction.

Answer 2

First, your $\cal U$ forgets to cover $0$ (the first ordinal). Replace it by $$\mathcal U:=\{[0,\alpha)\mid\alpha\in\omega_1\}.$$ Secondly, there is no $(a,\omega_1]\in\mathcal U_0$ because $\omega_1\notin\omega_1=[0,\omega_1).$

Finally, to end your proof, note that $\mathcal U_0=\{[0,\alpha_n)\mid n\in\mathbb N\}<\sup_{n\in\mathbb N}\alpha_n\in\omega_1.$