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A Jordan curve is a continuous closed curve in the plane with no self-intersections. My question is, is the interior of a Jordan curve always a Borel set?

If not, is the interior of a convex Jordan curve at least a Borel set. I know that convex sets need not be Borel sets, but maybe the combination implies Borel.

If not, does anyone know of a counterexample?

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    $\begingroup$ The interior is open – isn't every open set a Borel set? $\endgroup$ – Martin R May 28 at 8:40
  • $\begingroup$ What do you mean by interior? If it's in the topological sense then of course since every open set is a Borel set, as @MartinR points out. If you mean the set 'enclosed' by the Jordan curve then it's not so obvious. $\endgroup$ – Tony S.F. May 28 at 8:41
  • $\begingroup$ Your choice of the word "interior" is a tad unfortunate, as indicated in the comment of @TonyS.F. To avoid this, topologists sometimes refer simply to the "inside" and "outside" of a Jordan curve. $\endgroup$ – Lee Mosher May 28 at 12:22
  • $\begingroup$ @LeeMosher What is the difference between the interior of a Jordan curve and the “inside” of a Jordan curve? $\endgroup$ – Keshav Srinivasan May 28 at 15:40
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    $\begingroup$ I did not take Tony S.F.'s comment to be quite so trivial. The answer provided by @MartinR is pretty short, but not quite trivial. $\endgroup$ – Lee Mosher May 28 at 19:00
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  • The image of a Jordan curve is a compact set, so that its complement is open.

  • The interior of the Jordan curve is one of the two connected components of that complement, and therefore open as well.

  • Every open set is a Borel set.

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