Locally path-connected implies that the components are open

Question

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 $X$?

Answer 1

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

Answer 2

To prove that the connected components are open, it is sufficient that each point has a connected neighborhood.

If, moreover, each point has a neighborhood base of connected sets (we say $X$ is locally connected), then one can even show that every open subset of $X$ has only open connected components. Actually, the following are equivalent:

• $X$ is locally connected
• The components of each open subset are open
• Each point $x\in X$ has a local base of open connected sets

The same holds if we replace connected by path connected and component by path component, and the proofs are the same.
Also note that path connected implies connected.