Openness and closeness in components.

Question

Problem Each connected component of a space $X$ is closed. If $X$ has only finitely many connected components, then each component of $X$ is also open.

I know that closure of a connected subspace of $X$ is connected but from this how we can say that each component of a space $X$ is closed? I didn't get this point.

Edit- The topologist’s since curve, $\overline{S}$, of Section $24$(Munkres Topology) is a space that has a single component(since it is connected) and two path-components. One path component is the curve $S$ and the other is the vertical interval $V=0×[-1,1]$. Note that $S$ open in $\overline{S}$ but not closed, While $V$ is closed but not open.

Someone explain dark part.

Any hint or help will be appreciable.

Thanks!

Answer 1

Yes, path-components need not be closed, as your example shows. But components are always closed, or the closure of a component would be a strictly larger connected set, contradicting its maximality.

And as $X$ is always a disjoint union of its components, if there are finitely many, each component's complement is the union of the finitely many other components, all of which are closed as we saw, and a finite union of closed sets is closed, so that complement is closed and the component is open.

Answer 2

Let $C$ be a connected component. $C\subseteq \overline{C}$ (closure of $C$) and $\overline{C}$ is also connected. Component is maximal connected set with respect to subset inclusion, so $\overline{C}=C$.