proof that components of open sets in a locally connected space are open

Theorem.

A topological space $X$ is locally connected if and only if each component of an open setis open.

Proof.

First, suppose that $X$ is locally connected and that $U$ is an open set of $X$.Let $p\in C$, where $C$ is a component of $U$.Since $X$ is locally connected there is an open connected set, say $V$ with$p\in V\subset U$. Since $C$ is a component of $U$ it must be that $V\subset C$.Hence, $C$ is open.For the converse, suppose that each component of each open set is open. Let $p\in X$.Let $U$ be an open set containing $p$. Let $C$ be the component of $U$ whichcontains $p$. Then $C$ is open and connected, so $X$ is locally connected.

∎

As a corollary, we have that the components of a locally connected space are bothopen and closed.