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I would appreciate help clarifying:

On a metric space, connected does not imply path connected (e.g. topologist's sine curve). But I saw a theorem that if an open subset of the complex field is connected, then it is polygonally connected.

I can see that the topologist's sine curve is closed. What is it in general that allows an open set that is connected to imply path connected, and what prevents a connected closed set from being path connected.

Thanks from a self-studier

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    Open and connected in $\mathbb{R}^n$.2011-08-11

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The key property is that open subsets of $\mathbb{R}^n$ are locally path-connected, whereas closed subsets need not be. There is a discussion in this previous question; from it you can quickly extract a proof of your desired property (using the fact that open balls are polygonally connected, which should be clear).