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Is the boundary of an open, regular, bounded, path-connected, and simply connected set a Jordan curve?

Trying to find weakest condition on an open bounded set to apply Carathéodory's theorem. My bounded open sets can be assumed to be pretty well-behaved, but I wonder if the above conditions are sufficient.

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This was addressed in a recent paper of R.L. Moore:

A Characterization of Jordan Regions by Properties Having no Reference to their Boundaries
Robert L. Moore
Proceedings of the National Academy of Sciences of the United States of America
Vol. 4, No. 12 (Dec. 15, 1918), pp. 364-370
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  • $\begingroup$ Thank you once again. Good to see such recent work address this. $\endgroup$ – user2059893 Jul 3 '15 at 2:28
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    $\begingroup$ You could have mentioned that the answer was "no". 1918 = recent work? $\endgroup$ – YCor Jul 26 '15 at 6:18
  • $\begingroup$ YCor: The question was rephrased, it used to be "when is...", so the answer doesn't exactly match now. $\endgroup$ – user2059893 Jul 27 '15 at 7:24
  • $\begingroup$ YCor: '1918=recent work' was sarcastic $\endgroup$ – user2059893 Jul 27 '15 at 7:25
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    $\begingroup$ Yeah I was wondering. Actually when I first saw the reference I actually read Igor's post and didn't notice the year and I really thought this was recent (and got surprised a bit that such results could be only done recently). $\endgroup$ – YCor Jul 28 '15 at 9:32
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The two answers already posted show that there are plenty of counterexamples (it ought to be a $G_\delta$ set in a suitable hyperspace), I just thought one might also consider the Warsow circle (or rather its "inside" which could easily be defined even if this continuum is not a Jordan curve) as a specfic (easy) counterexample that perhaps first comes to mind. Though I do not know what "regular" means in this context, is it that $U=\mathrm{Int}\ \overline U$ ? Or is it as in PDE that boundary points must be regular ? Or is it, as in this paper that the boundary is a "closed, embedded, codimension-$1$ smooth submanifold (without boundary) which is the common topological boundary of the open sets ..." ?

enter image description here

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No, the boundary of an open, regular, bounded, path-connected, and simply connected set in $\mathbb R^2$ can be quite complicated. It might contain NO simple curves. The construction is similar to the construction of pseudo-arc.

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To help clarify a well known characterization: If U is a connected open bounded simply connected planar set, then the boundary of U is a simple closed curve iff the boundary of U is locally path connected and contains no cut points.

Comments:

0) Definition. p is a cut point of the connected space X iff X\p is not connected.

1) Definition. By `boundary' of an open bounded planar set U, we mean the difference U closure minus U.

2) For necessity, observe that the unit circle X is locally path connected and X\p is connected for all p in X.

3) For sufficiency, every orientation preserving conformal homeomorphism ( Riemann map) f: int(D) --->U extends continuously to the respective closures iff the boundary of U is locally path connected. (Caratheodory). If the extension is not 1-1, a straightforward geometric construction yields a non cutpoint on the boundary of U.

4) Mirko's nice sketch of the `Warsaw circle' shows a non locally path connected boundary can have no cut points, and pac-man with a closed mouth shows a locally path connected boundary can have cut points. Thus two the mentioned conditions are independent.

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  • $\begingroup$ Is the following variant of the claim still true: "the boundary of U is a simple curve iff the boundary of U is locally path connected"? So the curve need not to be closed and the condition of having no cut point on the boundary is omitted. $\endgroup$ – Twi Oct 27 '16 at 12:17
  • $\begingroup$ And the word "simple" should be deleted as well ... sorry. $\endgroup$ – Twi Oct 27 '16 at 12:30

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