Is every Lindelöf subspace of the ordinal space $\omega_1$ countable?
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1So $\omega_1$ is given the usual order topology, and $Y \subseteq \omega_1$ is a Lindelöf subspace of $\omega_1$? – user642796 Mar 11 '13 at 07:23
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1Yes, ALL are yes. – Mar 11 '13 at 07:30
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I have completely re-worded the question. If you disagree, feel free to revert. – user642796 Mar 11 '13 at 07:33
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Good work and very appreciated it with your answer. – Mar 11 '13 at 07:42
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If $Y \subseteq \omega_1$ is uncountable, enumerate it is increasingly as $\{ \alpha_\xi : \xi < \omega_1 \}$. Then $$\mathcal{U} = \{ [ 0 , \alpha_\xi + 1 ) \cap Y : \xi < \omega_1 \}$$ is an open cover of $Y$ without a countable subcover.
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