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I am aked to show that in a compact metric space we can find at most countably many subsets which are both: open and close. I would be grateful for your help.

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If this is of any help: The quasi-components coincide with the components, where a quasi-component of $x$ is the intersection of all clopen subsets containing $x$. They are the equivalence classes where two elements are equivalent if there is no separation between them, i.e. there is no clopen subset containing one point but not the other. I think this is true in each compact normal space. – Stefan Hamcke Apr 22 '13 at 19:11
@StefanH.: By the way, each compact Hausdorff space is normal (see, for instance, Ryszard Engelking, “General topology”, 3.1.9 in Russian edition). – Alex Ravsky Apr 23 '13 at 1:42
up vote 7 down vote accepted

It seems that we can show the claim as follows. It is well know and easy to prove that each metrizable compact space $X$ has a countable base. Fix such a base $\mathcal B$. Let $U$ be a clopen (that is, closed and open) subset of $X$. For each point $x\in X$ there exist a neighborhood $x\in U_x\in\mathcal B$. Since $U$ is compact, there exists a finite subset $Y$ of $U$ such that $U=\bigcup\{U_x:x\in Y\}$. Hence the cardinality of the family of all clopen subsets of $X$ is not greater than the cardinality of the family of all finite subsets of $\mathcal B$, which is countable.

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Why is $U_x$ compact? – Stefan Hamcke Apr 23 '13 at 11:24
It was a misprint. Corrected. Thanks. – Alex Ravsky Apr 23 '13 at 13:23
This should work in every compact second-countable space, shouldn't it? – Stefan Hamcke Apr 23 '13 at 14:07
It is almost the same, because each regular second-countable space is metrizable. – Alex Ravsky Apr 23 '13 at 15:00

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