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I'm not sure if this is a true statement or not, but due to the compactness, it seems like it should be. My attempt at a proof involves supposing it has infinitely many components, and then taking a sequence such that each element in the sequence is in a unique component. Since compactness implies limit-pint compactness, we know this infinite subset of has a limit point. From here, it seems like this limit point should provide a contradiction to the construction, but I haven't been able to show any contradiction yet...

Essentially what I want to do is show that a component is closed and open, and if there are only finitely many, we get this for free.

Note that this is part of a larger problem, this is just what the guts of my proof comes down to, so if it isn't true, the direction of my proof is entirely wrong...

Thanks in advance!!

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    By the way, we know that components in a compact, Hausdorff space are disjoint since components and quasicomponents agree. Thus, my sequence isn't too ambiguous...2012-10-25
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    Note the compactness doesn't imply sequential compactness in general. See http://math.stackexchange.com/questions/220422/compact-space-which-is-not-sequentially-compact2012-10-25
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    That is certainly true, I wasn't suggesting limit-point compactness was equivalent to sequentially compact, merely that there had to exists a limit point of the sequence.2012-10-25

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