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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!!

notsure
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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... – notsure Oct 25 '12 at 15:53
  • Note the compactness doesn't imply sequential compactness in general. See http://math.stackexchange.com/questions/220422/compact-space-which-is-not-sequentially-compact – Jason DeVito - on hiatus Oct 25 '12 at 15:56
  • 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. – notsure Oct 25 '12 at 15:59

3 Answers3

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For a simpler example, note that $\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4}\ldots\}$ (with its natural topology as a subspace of $\Bbb{R}$ is compact, Hausdorff, and has $\aleph_0$ components.

Chris Eagle
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No, it’s not true. The middle-thirds Cantor set is totally disconnected, meaning that its components are all singletons. Since it has cardinality $2^\omega=\mathfrak c$, the same cardinality as $\Bbb R$, it has far more than finitely many components.

In fact it has an even stronger property: it’s zero-dimensional (in the sense of small inductive dimension, which is the usual sense when the term is used without further qualification), meaning that it has a base of clopen sets, and it’s $T_1$; every zero-dimensional $T_1$-space is totally disconnected, but not conversely.

Brian M. Scott
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It's not true, but under additional assumptions you can make it work. For example, a compact surface has finitely many connected components.

Maxim Gilula
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