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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

Qiaochu Yuan
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  • Why don't you look at (what I think is ) the standard argument showing that connected in $\mathbb R^n$ implies path-connected? The method consists of showing that, in a connected set C in $\mathbb R^n $, the set of points S that can be joined by a path to a fixed point $x_o$is both open and closed , forcing S to be the whole space, since C is connected and so C cannot have non-trivial subsets that are both closed and open. – gary Aug 10 '11 at 18:30
  • Open and connected in $\mathbb{R}^n$. – leo Aug 11 '11 at 05:47

1 Answers1

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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).

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Qiaochu Yuan
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