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It seems to me that closure of a path connected set may not be path connected. For example Topologist's sine curve $S={(x,sin(\frac{1}{x}):0< x \le 1 }$ is path connected but it's closure is not path connected.

Am I right?... Is there any other easy counter example?

Samiron Parui
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    For another example see https://math.stackexchange.com/questions/1005925/another-example-of-a-connected-but-non-path-connected-set?rq=1 – mfl Nov 13 '17 at 15:02
  • Thanks.this example is helpful. – Samiron Parui Nov 13 '17 at 16:22
  • @AlgTop I don't see that the inclusion would need to be a surjection. For example, when $X=[0,1)$ (as a subspace of $\mathbb R$), the closure $\bar X=[0,1]$ is path-connected but the inclusion is not surjective. – Andreas Blass Nov 13 '17 at 16:52
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    This example is not true; the topologist's sine curve is not path connected (Munkres p.157) – Dana Feb 28 '19 at 08:44
  • For the record, this is correct and answered here. To answer @Dana's comment: the topologist's sine curve is $S\cup{(0,0)}$, not $S$. – Léo S. Nov 05 '23 at 21:51

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