Intereting Posts

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What does it mean for something to hold “up to isomorphism”?

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.

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

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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.

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 $X$.

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