Finitely many connected components, prove interiors are also connected

Question

Show that in a space with finitely many connected components $C_i, i = 1, ..., n$ their interiors $Int(C_i)$ are also connected.

Is it true in general that the interior of a connected component is necessarily connected?

Don't have really any idea, how to do this...Any guidance would help.

Answer 1

A connected component is always closed, because the closure of a connected set is connected, and there cannot be a larger connected set containing the component.

If there are finitely many connected components, then they are also open, since $C_i $ is the complement of $\bigcup_{j\ne i} C_j$, which is closed. Hence, the interior of $C_i$ is $C_i$ itself.

In general, the interior of a connected component is not necessarily connected. I'll modify the example of Alex G. for this purpose. Let $X$ be the subset of $\mathbb R^2$ which is the union of three sets:

1. closed disk of radius $1$ centered at $(1,0)$
2. closed disk of radius $1$ centered at $(-1,0)$
3. the set $\{(0,1/n): n =1,2,3,\dots \}$

The union of two disks is a connected component. The point $(0,0)$ is not an interior point of this component. Removing this point, we are left with disconnected interior.