A question about topological space locally path connected.

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!

• On 1) if $c\in C$ then there is a path component $R$ containing $c$ as element. It has a non-empty intersection with $C$ (since $R$ and $C$ both contain $c$). If $R=P$ then $c\in P$, and if $c\notin P$ then $c\in R\subseteq Q$. – drhab Apr 7 at 15:05
• Hint for 2): if $U$ is an open subset of $X$ and $V$ is an open subset of $U$, then $V$ is an open subset of $X$ as well ($V = V' \cap U$ for some $V'$ open in $X$, and intersection of opens is open). – Pel de Pinda Apr 7 at 15:12

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.
• The path-connected ngb $P$ of $x$ is also open in $O$, because $P\subseteq U\subseteq O$, where $U$ is an open set of $X$? – Jack J. Apr 7 at 17:16
• @JackJ. Yes, open in open is open, so a neighbourhood of $x$ inside an open set $O$ is also a neighbourhood of $x$ in the subspace topology of $O$. – Henno Brandsma Apr 7 at 17:57
For every $$x \in C$$ the locally path connected hypotesis tell us that exists an open path connected set $$U$$ such that $$x \in U$$. Therefore the union of all the elements in $$Q$$ is exactly $$C$$.
The proposition: let $$A$$ be an open subset of $$X$$. For every point $$x \in A$$ exists a neighborhood $$U \subset A$$ and a neighborhood $$V \subset X$$ that is path connected. The intersection is an open path connected neighborhood contained in $$A$$.