Prove that if a space $X$ is locally path connected then each path component of each open subset of $X$ is open.

Question

I am having difficulty proving the following statement.

If a space $X$ is locally path connected, then each path component of each open subset of $X$ is open.

In particular, while I have been able to argue that each path component of $X$ is open, I am having difficulty extending this to the open subsets of $X$. My proof of the former is as follows.

First, let $X$ be locally path connected, let $A$ be a path component of $X$, and let $x\in A$ be an arbitrary point. Because $X$ is locally path connected, there exists a path connected open subset $U\subset X$ containing $x$. Then $x\in A\cap U$, so $A\cup U$ is path connected. Since $A$ is maximal, this implies that $U\subset A$, i.e. that $x$ is an interior point of $A$. Since the same argument holds for every point $x\in A$, we have that $A$ is open. Thus, each path component of $X$ is open.

Would anyone here be willing to push me in the right direction? In case there is ambiguity, the text I am following defines a locally path connected space to be one that has a basis of path connected open sets.

If $X$ is locally path connected and $U$ an open subset, then $U$ is locally path-connected! — Hagen von Eitzen, May 18 '20 at 18:52
@HagenvonEitzen, thank you. I suspected as much, but it is not obvious to me why it holds. I thought I might be able to construct a basis for $U$ by taking intersections with $U$ of the basis elements for $X$, but I am not guaranteed that these intersections will be path connected. — seafood258, May 18 '20 at 20:32
Does this answer your question? Definition of locally pathwise connected See the second answer. — Paul Frost, May 18 '20 at 21:42

score 1 · Accepted Answer · answered May 18 '20 at 21:46

If $O$ is open in $X$ and $X$ is locally path-connected, so is $O$ (in its subspace topology): let $X \in O$ and $U \subseteq O$ open in $U$, which however just means that $U$ is also open in $X$ (as $U=U'\cap O$ with $U'$ open in $X$ by the definition of the subspace topology on $O$, but then $U$ is a finite intersection of open sets of $X$, so indeed open in $X$!). As $X$ is locally path-connected it has a local base of path-connected neighbourhoods so one of these sits inside $U$ and we're done, as this is still a path-connected neighbourhood of $x$ in $O$ too (as $O$ is open).

So you can apply your argument got $X$ to $O$ too: its path-components are also open (in $O$ and thus in $X$ too).

Prove that if a space $X$ is locally path connected then each path component of each open subset of $X$ is open.

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