# Clopen iff union of open connected components

Wikipedia says, that in case all connected components are open, a subset is clopen iff it is a union of connected components.

While the first implication has been shown in this post, I'm trying to prove the opposite direction.

Let $$(X,\tau)$$ be a topological space with the assumptions from above. Further let $$A \subseteq X$$ be the union of a family $$(U_i)_{i \in I}$$ of connected components.

Of course, $$A$$ is open (arbitrary union of open sets). But why is $$A$$ closed? I'm thinking proof by contradiction, by showing that for $$A \subsetneq \bar A$$ one $$U_{i_0}$$ can't be maximal but I haven't had success so far.

Problem: Take $$x \in \bar A \setminus A$$ and its respective connected component $$U_x$$. The assumptions are not violated, if $$U_x = \{x\}$$. Following this path I get, that $$\bar A$$ is clopen.

Any hints?

• The complement of $A$ is also a union of connected components... – kccu Jan 8 '16 at 17:36
• @kccu: This does the trick. Thanks! – el_tenedor Jan 8 '16 at 17:39

A set $X$ is a disjoint union $(U_i)_{i\in I}$ where $U_i$ is a connected component, suppose that $Y\subset X, Y=\cup U_{j,j\in J\subset I}$ write $I=J\cup L$, the complement of $Y$ is $Z=\cup_{l\in L}U_l$. $Z$ is open, thus $Y$ is closed.