Let $\omega_1$ be the first uncountable ordinal. Let $L$ denote $\omega_1 \times [0,1)$ with the order topology and smallest element removed. How can we show this space is not second countable?

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    $\begingroup$ It has an uncountable discrete subset. $\endgroup$ – Daniel Fischer Feb 22 '14 at 18:40
  • $\begingroup$ @DanielFischer: To see that $\omega_1$ has an uncountable, discrete subset, I can just take $\{\alpha +1 : \alpha \in \omega_1\}$? $\endgroup$ – Mike F Dec 14 '15 at 21:36
  • $\begingroup$ @MikeF Yes, that works. $\endgroup$ – Daniel Fischer Dec 14 '15 at 21:39

If $A$ is a countable subset of $L$, say $A = \{(\alpha_n, t_n): n \in \omega \}$, then let $\alpha+1 \in \omega_1$ be an successor ordinal strictly larger than all $\alpha_n$, which can be done as all countable subsets of $\omega_1$ are bounded above. Then $(\alpha+1, \frac{1}{2})$ has a neighbourhood $\{\alpha+1\} \times (0,1)$ in $L$ that misses all members of $A$.

This shows that no countable subset of $L$ can be dense, i.e. $L$ is not separable. As a second countable space is always separable, $L$ does not have a countable base as well.


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