[Math] Locally path-connected implies that the components are open

connectednessgeneral-topologypath-connected

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 $X$?

Best Answer

Hints:

1) If $X$ is locally path-connected, then path components of $X$ are open
2) If $X$ is locally path-connected, then path components and connected components coincide