# Around Jordans Curve Theorem I

There are more or less three theorems that are often called Jordan Curve Theorems, while there is a distinction between them. Let us denote by E the Euclidean plane, $$E=\mathbb{R}^2$$ and by $$E^n$$ the Euclidean n-dimensional space, $$E^n=\mathbb{R}^n$$.

The Jordan Curve Theorem. Let $$f\colon S^1\to E$$ be an embedding of the unit circle into the plane. Then the complement of $$f[S^1]$$ has two connected components, one bounded and one unbounded.

The Jordan-Schönflies Theorem. Let $$f\colon S^1\to E$$ be an embedding of the unit circle into the plane. Then the complement of $$f[S^1]$$ has two connected components, one bounded and one unbounded; and the complement of $$f[S^1]$$ is homeomorphic to the complement of $$S^1$$.

The Jordan-Brouwer Theorem. Let $$f\colon S^n\to E^{n+1}$$ be an embedding of the unit n-sphere into the n+1-dimensional space. Then the complement of $$f[S^n]$$ has two connected components, one bounded and one unbounded.

These theorems are closely related and seems weird why are they not combined into one. Probably, to underline that the following is wrong:

The Wrong Theorem. Let $$f\colon S^n\to E^{n+1}$$ be an embedding of the unit n-sphere into the n+1-dimensional space. Then the complement of $$f[S^n]$$ has two connected components, one bounded and one unbounded; and the complement of $$f[S^n]$$ is homeomorphic to the complement of $$S^n$$.

Thus, Jordan Curve Theorem (J.C.T) generalizes to the Jordan-Brower Theorem (J.B.T) (I will point out how to prove it using Brouwer’s Fixed Point Theorem), but the stronger theorem, The Jordan Schönflies Theorem (J.S.T.) does not generalize to the Wrong Theorem, because it is wrong. This video on YouTube gives a counter example, the Alexander Sphere. (For some reason I cannot embed it here.)

It is intuitively quite obvious that after the limit process of the video is done (one can formalize it using countable union or countable intersection depending on the preferred way to do that), the surface is an embedded sphere, which satisfies J.S.T.. On the other hand the fundamental group of the complement is not trivial. Which means that you can flip a rope around one of the horns of the surface and if you glue the loose ends of that rope together, you cannot (homotopically) disconnect the rope from the surface. Thus the complement cannot be homeomorphic to the complement of $$S^n$$. Why? Because the homeomorphism would somehow map the flipped rope to one of the connected components of the complement of $$S^n$$, where it can be undone (freed from the surface) and shrinked to a point. This would give (by the homeomorphism’s inverse) a way to disconnect the rope from the Alexander Sphere.

Note that the important consequence — which is in fact equivalent to J.C.T. — is that one cannot get out of the inside of a circle no matter how complexly the circle is situated in the plain, without crossing the circle itself. A Klein-bottle-effect is impossible. This can be viewed as a generalization of the Bolzano-Darboux Theorem (see discussion) which states, roughly speaking, that from the negative half of the plane one cannot get to the positive without crossing the x-axis. Well, we can as a more general question: suppose there is a continuous mapping

$$f\colon \mathbb{R}\to E, f(x)=(f_1(x),f_2(x))$$

such that $$\lim_{x\to\infty} f_1(x)=\infty$$, $$\lim_{x\to-\infty} f_1(x)=-\infty$$ and for all $$x, -1<f_2(x)<1.$$

(this corresponds to the x-axis). Then one cannot get from $$(0,-1)$$ to $$(0,1)$$ without crossing the image of $$\mathbb{R}$$ under $$f.$$ $$(*)$$
It is easily seen that J.C.T. implies $$(*)$$, but requires some tedious-to-go-through technicalities in order to show the converse. These technicalities can be found in

The Jordan Curve Theorem Via the Brouwer Fixed Point Theorem
Ryuji Maehara
The American Mathematical Monthly, Vol. 91, No. 10 (Dec., 1984), pp. 641-643

For a proof of $$(*)$$ see Theorem (3) in my previous post.

The reader may verify that that Theorem is equivalent to the statement $$(*)$$.

Challenge: Can you generalize this to $$\mathbb{R}^n\,?$$