Locally path-connected implies that the components are open If $X$ is a locally path-connected space, then its connected components are open. 
I am trying to prove this, but for some reason it doesn't seem right to me, knowing that components are always closed. If the statement is true, wouldn't it be the case the components are the whole space $X$?

Here's another proof. It uses two facts:


*

*the union of any non-disjoint connected subsets is connected

*path connected sets are connected


Suppose $C$ is a component of the locally path connected space $X$, and suppose it's not open. Then it must contain a boundary point, call it $b$. Since the space is locally path connected, we can find a neighborhood $V$ of $b$ that is path-connected.
But path-connected subsets are always connected. Also, $C$ and $V$ share the point $b$, so $C \cup V$ is a connected subset that is strictly larger than $C$ due to the fact that $V$, being a neighborhood of a boundary point of $C$, intersects $X - C$. This contradicts that $C$ is a maximal connected subset (since that is what components are).
So in fact $C$ must be open.
