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Let $X$ be a topological space which, for my intents, may be assumed to be metrizable and compact if needed (let's say it's a closed subset of the unit cube or something like that).

I know that:

  1. If $X$ has only a finite number of connected components than each connected component is clopen. Generally, they are only closed.
  2. Each clopen set of $X$ is a union of connected components.

This means that if the Boolean algebra generated by connected components is finite, then the Boolean algebra of clopen sets is finite and the two algebras are equal.

However, if I know that the algebra of clopen sets is finite, can I get that the algebra generated by connected components is finite?

(It is possible to use the fact that in a compact Hausdorff space, connected components and quasi-connected components are the same, but I haven't been able to tie the ends together...)

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If the algebra of clopen sets is finite, then it's a powerset algebra of a finite collection of atomic (in the sense of boolean algebra) clopen sets. These are the connected components (they are connected because of atomicity and connected component cannot be obviously bigger than clopen set). So the algebras are the same since their finite generating sets of atoms are the same.

Summary: algebra of clopen sets is finite iff algebra generated (in $\mathcal{P}(X)$) by connected components is finite and in that case they are the same. This holds for arbitrary topological space.

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  • $\begingroup$ "connected component cannot be obviously bigger than clopen set". I don't understand this claim. In general a connected component is a closed set, not necessarily a clopen set. $\endgroup$ – Yoni Rozenshein Aug 8 '13 at 23:13
  • $\begingroup$ @YoniRozenshein: Connected component cannot be bigger than clopen set. If it was then it would contain non-trivial clopen set and so it wouldn't be connected. This claim is even weaker that the fact, that quasi-components are bigger than connected components. $\endgroup$ – user87690 Aug 9 '13 at 5:59
  • $\begingroup$ Great, understood now. Thanks! $\endgroup$ – Yoni Rozenshein Aug 12 '13 at 9:28

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