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My teacher taught me that a set of points is connected if and only if it is path connected. However,the Topologist's sine curve is connected but not path connected. Am i missing something here?

coconutmercury
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    This is only true if we put extra conditions on the space in question. For instance, a locally path connected space is connected iff it is path connected. Generally though there are connected spaces that are not path connected as you point out. – Alex G. Feb 12 '21 at 12:49
  • Please elaborate – coconutmercury Feb 12 '21 at 12:49
  • If the set is open then it is path connected –  Feb 12 '21 at 13:41
  • @AdityaDwivedi yes, for open subsets of the plane because these are locally path-connected. And the Topologist's sine curve is closed and not-open. – Henno Brandsma Feb 12 '21 at 14:45
  • @HennoBrandsma does open connected subset of $ \mathbb{R}^n$ is not neccesarily path connected –  Feb 13 '21 at 05:41
  • @AdityaDwivedi for an open subset of $\Bbb R^n$ we have that it’s connected iff it’s path-connected. That might be the fact your teacher was referring to. – Henno Brandsma Feb 13 '21 at 07:10

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No, your teacher is wrong. Even for planar subsets connected does not imply path-connected. In $\Bbb R$ this equivalence does hold, maybe you misheard him and did he mean that instead. Otherwise just plain wrong. Within the class of locally path-connected spaces the equivalence does hold, though. So e.g. in $\Bbb C$, the regions/domains (open connected subsets) are also path-connected (even with piecewise linear paths) which becomes important for path-integration theory.

Henno Brandsma
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In general, path-connected implies connected. The converse is not true as you point out with the topologist's sine curve. As Henno is correctly pointing out, the notion is equivalent in $\mathbb{R}$ since the topology in $\mathbb{R}$ is "so poor" that it is impossible to find a non-path-connected set that is connected. To be able to find one, you need to go on "bigger" topological spaces, in some sense.

I suggest that you go here to see how far you can generalize the equivalence, where it is noticed that in a locally path-connected space, connected open sets are path-connected. Every topological vector space over $\mathbb{R}$ or $\mathbb{C}$ is locally path-connected since the balanced neighbourhoods form a local base at 0.

Son Gohan
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