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Is the boundary of a simply connected set of the plane bounded and with non-empty interior a path-connected set?

Can I consider as counterexample the area between the x-axis and the topologist's sine curve?

If not, besides the counterexample I would also appreciate a similar (but true) result concerning to the path-connectedness of set boundaries.

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  • $\begingroup$ This is also a pretty good example: d2vlcm61l7u1fs.cloudfront.net/…. $\endgroup$ – Mike Earnest Jun 14 '18 at 19:55
  • $\begingroup$ Thank you Mike, I precisely meant something like that. In fact, to ensure breaking path-connectedness we should use "two topologist's sine curves" instead of one. Thank you again. $\endgroup$ – dudas Jun 14 '18 at 22:04
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Another easy counterexample: $\Bbb R\times[0,1]$ has two separated lines as boundary.

And the boundary of $\Bbb R^2\setminus\{\,(tn,tm)\mid n,m\in \Bbb Z, t\ge 1, (n.m)\ne 0\,\}$ even has infnitely many connected components.

But of course your example (or simply the complement of the topologists sine curve) has the advantage of having a more "interesting" reason for failed path-connectedness.

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I don't have a similar result for you, but the interval $[0,1]$ is simply connected, but its boundary $\{0,1\}$ is not path-connected since it is discrete with more than one point.

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  • $\begingroup$ Excuse me! I forgot saying with non-empty interior. $\endgroup$ – dudas Jun 14 '18 at 19:05
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    $\begingroup$ @dudas The interval has non-empty interior. Specifically $(0,1)$. Are you looking for an example in the plane? $\endgroup$ – D. Brogan Jun 14 '18 at 19:06
  • $\begingroup$ Yes, I mean in the plane $\endgroup$ – dudas Jun 14 '18 at 19:17

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