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I'm trying to figure out whether my proof is correct for a question I'm trying to tackle in Topology by James R. Munkres.

Task: Let $X$ be locally path connected. Show that every connected open set in $X$ is path connected.

My attempt at a proof: Well, every open subset of the locally path connected space $X$ is locally path connected. In addition, the path components and components of $X$ are the same in view of Theorem $25.5$ (p. $162$), which states that if a space $X$ is locally path connected, then the components and the path components of $X$ are the same. Altogether, this implies that then every connected open set in $X$ is path connected.

Am I right, or do I need to make any changes? Please provide your input, thanks.

Libertron
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    Could you please edit in the formulation of Theorem 25.5. Unfortunately not everyone has the book in question. – sxd Mar 16 '13 at 17:15

2 Answers2

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That's a typical connectedness argument. As pointed out by muzzlator, fix a point $x$ in $\Omega$ your open connected subset. Then consider $\Omega_x$ the set of points in $\Omega$ which are path connected to $x$ within $\Omega$. This is nonempty as $x$ belongs to it. This is open in $\Omega$ by local path connectedness. And by local path connectedness again, it is easily seen that the complement $\Omega\setminus\Omega_x$ is open in $\Omega$. So $\Omega_x$ is a nonempty open/closed subset of $\Omega$. Thus $\Omega_x=\Omega$. Note that the fact that $\Omega$ is open is implicitly used in both steps.

Julien
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    In OP's solution, why is every open subset of a locally path connected space locally path connected? @Libertron – hengxin Feb 11 '14 at 05:20
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Here's an alternative way. Let $S$ be the set of all paths reachable from a point $x$. This is an open set due to the local path connectedness. Use this to form a disconnection of $X$ if $S$ wasn't the same as $X$.

muzzlator
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  • Interesting, but was my attempt incorrect? – Libertron Mar 16 '13 at 17:34
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    Your approach is fine if that theorem is to be used. That theorem was probably proved with the idea I was using. You take the equivalence classes of the "Is path connected to" relation, use local path connectedness to show the equivalence classes are all open and then you have the set written as a disjoint union of open sets. – muzzlator Mar 16 '13 at 17:37
  • "the set of all paths reachable" -> "the set of all points reachable by a path", of course. – LSpice Dec 01 '22 at 16:32