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.
Best Answer
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 $\mathcal U_0=\{[0,\alpha_n)\mid n\in\mathbb N\}<\sup_{n\in\mathbb N}\alpha_n\in\omega_1.$