Theorem. If $X$ is a topological space, each path component of $X$ lies in a component of $X$. If $X$ is locally path connected, then the components and the path components of $X$ are the same.
Proof. Let $C$ be a component of $X$; let $x$ be a point of $C$; let $P$ be the path component of $X$ containing $x$. Since $P$ is connected, $P\subseteq C$. We wish to show that if $X$ is locally path connected, $P=C$. Suppose that $P\subset C$. Let $Q$ denote the union of all the path components of $X$ that are different from $P$ and intersect $C$: each of them necessarily lies in $C$, then $Q\subseteq C$, therefore $P\cup Q\subseteq C.$ The rest of the proof is clear.
Question 1. Why so $C\subseteq P\cup Q$?
Proposition.(Exercise) Let $X$ be a locally path-connected topological space, then every open subset of $X$ is locally path connected.
Question 2. Any hints?
My attempt after hints Let's see if I understand: let $O\subseteq X$ open.
Claim: for each $x\in O$ and for each ngd $U$ (open in $O$) of $x$, exists a ngb $V$ of $x$ (open in $O$) path-connected such that $V\subseteq U$.
Since $O$ is open in $X$, for each $x\in O$ exists a ngb $U$ of $x$ such that $U\subseteq O$, moreover since $X$ is path-connected and $U$ is a ngb of $x$, exists $V$ ngb of $x$ path connected such that $V\subseteq U$. Observe that both $U$ and $V$ are open in $O$, therefore we have proved that for each $x\in O$ and for each ngh $U$(open in $O$) of $x$ exists a ngb $V$(open in $O$) of $x$ path-connected such that $V\subseteq U$, then $O$ is path connected.
Quite right?
Thanks!
Best Answer
Answering an alternative question 1: If $P \subseteq C$ and $Q \subseteq C$, simple set theory/logic tells us $P \cup Q \subseteq C$ ($x$ is in either $P$ or $Q$ and in both cases it's in $C$). I see no claim in your proof of the reverse inclusion.
If $X$ is locally path-connected and $O$ is open, then let $x \in O$ and $x \in U$, where $U$ is open in $O$. Then standard facts tell us that $U$ is also open in $X$ so there is a path-connected neighbourhood $P$ of $x$ with $x \in P \subseteq U$, as $X$ is locally path-connected. So $O$ is locally path-connected too.