I'm having trouble understanding why this fact is true. A lot of sites just assume it with out reason and it doesn't seem so direct to me.

Anyways, here is the theorem:

For any topological space $(X,T)$ with finite number of components, each component is clopen.

I only know the most basic definitions of components, basically that it is a maximal connected subset around a point.

Any help is appreciated.

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    $\begingroup$ Each component is closed. Hence, the complement of each component is open. But each component is itself a finite intersection of complements of components, hence is also open. $\endgroup$ – wckronholm Dec 5 '14 at 5:35
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    $\begingroup$ You should write that as an answer, it's exactly what OP is looking for, and that way it can be upvoted more substantially, and be accepted as an answer. $\endgroup$ – Alfred Yerger Dec 5 '14 at 7:03

A connected component $C$ of a space $X$ is always closed. This follows from the following fact:

Lemma: If $A$ is a connected subset of a topological space $X$, and $A\subseteq B\subseteq \overline A$, then $B$ is connected as well.

Now if $X$ has only finitely many components, then each component is complement of finitely many closed sets.


A minor variant: since the connected components of $X$ form a partition of $X$ and there is by assumption a finite number of them we can write $X=C_1\cup \cdots\cup C_n$. Since e.g. $C_1$ is closed, $X\setminus C_1= C_2\cup\cdots\cup C_n$ is open. On the other hand since each $C_i$ is closed the finite union (and this is what goes wrong for an infinite number of connected components) of closed sets $C_2\cup\cdots\cup C_n$ is closed. Hence $C_1$ is open as well. So we proved that $C_1$ is both closed and open, but this clearly works for any of the connected components $C_i$.


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