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 $X$?
 A: 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
A: 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.
A: Consider a connected component $C$ of $X$.
(*) Given any point $x \in C$ there must exist an open set $V \subset X$ that is path-connected such that $x \in V$

We say X is locally path connected if for every point $x \in X$ and every open neighborhood $U \subset X$ of $x$, there exists a open set $V \subset X$ that is path-connected and $x \in V \subset U$.

To get (*) just consider $U = X$ in the definition above.
Now, Since $V$ is path-connected then it must bet connected. And since $V \cap C \neq \emptyset$ ($x \in V\cap C$) this means $V \subset C$.
$C$ is open.
A: 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.
