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Using Urysohn's Lemma, it can be shown that a connected normal space $X$ (with more than one point) is uncountable. But then how can it be that a connected normal space might just be a single point? Is this immediate from Urysohn's Lemma?

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Trivially?${}{}$ –  Asaf Karagila Nov 11 '12 at 22:31
@AsafKaragila: Come again? –  Libertron Nov 11 '12 at 22:32
There is no way of decomposing a singleton into two non-empty disjoint sets... everything much hold vacuously. –  Asaf Karagila Nov 11 '12 at 22:33
Urysohn makes no (nonvacuous) statement about one-point spaces. But the very definition of normal is trivially verified for one-point space. –  Hagen von Eitzen Nov 11 '12 at 22:33
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up vote 5 down vote accepted

No, it’s immediate from the definitions of connectedness and normality. A one-point space is clearly not the union of two disjoint non-empty sets, open or otherwise, so it’s connected. A one-point space doesn’t contain two disjoint closed sets, so the defining condition of normality is vacuously satisfied.

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OK, now I got it. –  Libertron Nov 11 '12 at 22:33
@Sachin: Good. (By the way, what I wrote is basically just a longer version of what Asaf meant when he asked ‘Trivially?’.) –  Brian M. Scott Nov 11 '12 at 22:34
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