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Can $\Bbb N$ be topologized to be a compact Hausdorff space?

I guess it might be a topology that is strictly finer than cofinite topology and strictly coarser than the topology consists of all infinite subsets of $\Bbb N$. But is there such a topology?

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2 Answers 2

up vote 4 down vote accepted

Yes. Consider the ordinal $\omega+1$ (respect to the order topology.)

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Ah,right! Thank you. –  Metta World Peace Nov 5 '13 at 3:22

Sorry I can't comment, only answer. But isn't your question equivalent to asking whether there exists a denumerable compact Haussdorf space? An example would be a sequence in $\mathbb R$ with a limit that is also an element in the sequence, e.g. $\{1/n \mid n \in \mathbb N\} \cup \{0\}$. You then choose a bijection of $\mathbb N$ with this set and pull-back the topology.

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You're right.${}{}{}$ –  Metta World Peace Nov 5 '13 at 3:26

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