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Is there some way to find the cardinality of set of all clopen subsets of a topological space, say, Cantor space, Baire space?

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Do you mean the cardinality of the set of all clopen subsets, or (as you wrote) the cardinality of specific clopen subsets? –  Chris Eagle Jan 30 '13 at 21:44
    
@ChrisEagle: Thank you for pointing that out. I've fixed it. –  Metta World Peace Jan 30 '13 at 21:46
    
Is it true that that every union of connected components is clopen? Is it also true that every connected component can be obtained in this way? If so, then the number of clopen subsets is 2 raised to the power of the number of connected components. –  goblin Jan 30 '13 at 21:58
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@user18921 In Cantor space or Baire space, the connected components themselves are singletons (so they're not open.) A space with this property is called totally disconnected. –  Trevor Wilson Jan 30 '13 at 22:09
    
Oh wow. I had no idea it was so complicated. –  goblin Jan 31 '13 at 6:41

1 Answer 1

up vote 3 down vote accepted

I don’t have an answer to the general question, but I can answer it for the specific spaces mentioned.

The clopen algebra of the Cantor space is the free Boolean algebra on $\omega$ generators, which has cardinality $\omega$.

For $n\in\omega$ let $B_n=\{n\}\times\omega^\omega$. Then $\bigcup_{n\in A}B_n$ is a clopen subset of $\omega^\omega$ for each $A\subseteq\omega$, so the Baire space has at least $2^\omega$ clopen sets. On the other hand, $\omega^\omega$ is second countable, so it has only $2^\omega$ open sets and therefore precisely $2^\omega$ clopen sets.

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Thank you for your answer. I don't follow why $\bigcup_{n \in A}B_n$ is closed, when $A$ is an infinite subset of $\omega$? –  Metta World Peace Feb 5 '13 at 3:28
    
@MettaWorldPeace: It’s just $A\times\omega^\omega$, a basic open set in the product. If you want to write it in excruciating detail, let $V_0=A$, and let $V_n=\omega$ for $n>0$; then $\bigcup_{n\in A}B_n=\prod_{n\in\omega}V_n$, and since only finitely many factors $-$ one! $-$ are restricted, this product is open. –  Brian M. Scott Feb 5 '13 at 3:31
    
I didn't see the topology, taking as a basis all sets of the form:$\{x \in \omega^{\omega}:s \subseteq x, s \in \omega^{<\omega}\}$ and the topology you're referring to ,the product topology, with $\omega$ discrete, are actually the same. Thank you very much for your patience. –  Metta World Peace Feb 5 '13 at 4:07
    
@MettaWorldPeace: You’re very welcome. –  Brian M. Scott Feb 5 '13 at 4:09

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