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?


  • 2
    $\begingroup$ 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 $c\in R\subseteq Q$. $\endgroup$ – drhab Apr 7 at 15:05
  • $\begingroup$ 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). $\endgroup$ – 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.

  • $\begingroup$ @ Henno Brandsna. Thanks for your answer. I changed my question by proposing a resolution attempt based on your suggestion. $\endgroup$ – Jack J. Apr 7 at 16:48
  • $\begingroup$ 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 $X$? $\endgroup$ – Jack J. Apr 7 at 17:16
  • $\begingroup$ @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 $O$. $\endgroup$ – 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 $A$.


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