I am currently going through the proof of the Jordan-Schoenflies Theorem by S.S. Cairns (http://eretrandre.org/rb/files/Cairns1951_193.pdf).

I find that because it is a somewhat old article there are some conventions that are unclear to me. I wanted to ask about them and hopefully get unstuck.

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The article claims this can be quickly established by familiar methods. I tried doing it and I found that using the corollary (A) of the Jordan-Schoenflies:

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I was able to deduce that either the interior of the $b_1$ is in the interior of $b_2$ or else their interiors are disjoint. I assume by hypothesis that $b_1$ and $b_2$ both have the Jordan-Schoenflies property. This claim holds because if part of $b_1$ lied in the interior of $b_2$ I can only exit through the simple arc $b$, which forces all of $b_1$ to be inside the interior of $b_2$ or in $b$, else we have a self-intersecting curve. As a consequence the interior of $b_1$ is also contained in the interior of $b_2$. The argument if part of $b_1$ lied in the exterior of $b_2$ is similar.

I can then map all of $E$ to itself bringing $b_1$ to a circle, which makes things look nicer. But even in this scenario I don't see how deleting $b$ I can prove Jordan-Schoenflies for $b_1+b_2-b'$. Would someone have a suggestion?

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I think crossing a polygon is not well-defined in this paper. If I pass the straight line $y=1$ through the vertex of a triangle $(1,0), (0,1), (0,0)$ it doesn't have an endpoint interior to it.

I suppose I can say look for each side of the polygon a small tubular neighborhood of it. Then I could define "a polygonal path crosses a polygon if it intersects one side of the polygon, and that in a neighborhood of such intersection the path has points on both sides of the tubular neighborhood of the side it intersects on." I was hoping there is an easier definition to work with and that makes this Lemma true.


In the middle of proving Lemma 3.1 he establishes (A) and (B), which is fine:

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but then he proceeds to prove (C):

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My question is about part (2). Here $\alpha$ is defined earlier by:

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He claims proving (C) presents no difficulty, but I was not able to do it yet. It's not clear to me how $\alpha$ as defined have to come from either of $\alpha_1$ or $\alpha_2$. I would appreciate any nudge towards showing this point (2) of (C), Lemma 3.1.

  • $\begingroup$ Theorem 2.1. Let $C$ be a closed disk with $p\subset C$. Fix a point $x\not\in C.$ For $x\ne y\not\in p$ let $[x,y]$ be the line-segment from $x$ to $y$. Now let $S(y)$ be the set of those $z\in p\cap [x,y]$ such that if $e$ is any edge of $p$ and if $z\in e \cap [x,y]$ then $e\cap [x,y]=\{z\}.$ Now for any $w\not\in p$ let $w\in E$ iff $(w=x$ or $S(w)$ has an even number of members) and let $w\in F$ iff $w\not\in E.$ It is not hard to show that $E,F$ are open & connected, with $F$ bounded $\endgroup$ Jul 25 at 0:48

I did not read his paper but "crossing" means "meeting transversally". Here two polygonal arcs $a, b$ in the plane are said to meet transversally (or "cross") at a point $p$ if there is a local PL homeomorphism from a neighborhood of $p$ to $R^2$ which sends $a$ to an interval in the x-axis and $b$ to an interval of the y-axis (and sends $p$ to the origin).

Instead of (or in addition to) following Cairns' paper, I would suggest the more recent

Thomassen, Carsten, The Jordan-Schönflies theorem and the classification of surfaces, Am. Math. Mon. 99, No. 2, 116-130 (1992). ZBL0773.57001.

  • $\begingroup$ I actually started my attempt at understanding the proof of the Jordan-Schoenflies with Thomassen's article, but I found it harder to tackle. For example, in Lemma 2.3 it assumes that for a Jordan curve the regions of the complement of C all have C as boundary. I got stuck then, because regions of R^2\C a priori just mean its connected components. It is probably a consequence of the curve being simple but I could not prove that and he doesn't provide a proof either. $\endgroup$ Jul 24 at 17:41
  • $\begingroup$ @viniciuscantocosta I suggest asking a separate question regarding this point. $\endgroup$ Jul 24 at 18:56
  • $\begingroup$ You're right, thanks for the suggestion. My main point with the comment was to point out I had tried looking into your suggestion before posting my question. $\endgroup$ Jul 24 at 19:07

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