# How to deduce Jordan Curve Theorem from Schönflies Theorem

Recently I started reading Ethan Bloch's "A First Course in Geometric Topology and Differential Geometry" and I came upon this exercise to deduce the Jordan Curve Theorem from the Schönflies Theorem:

Schönflies Theorem: Let $$C\subseteq \mathbb{R}^2$$ be a 1-sphere (homeomorphic image of $$S^1$$). Then there is a homeomorphism $$H: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ such that $$H(S^1)=C$$ and $$H$$ is the identity map outside a disk (homeomorphic image of $$D^2$$).

Jordan Curve Theorem: Let $$C\subseteq \mathbb{R}^2$$ be a 1-sphere. Then $$(1)$$ the set $$\mathbb{R}^2 \setminus C$$ has pricisely two components, one of which is bounded and one of which is unbounded and $$(2)$$ the union of $$C$$ and the bounded component is a disk, of which $$C$$ is the boundary.

My attempt so far:

(1) $$\mathbb{R}^2 \setminus C = H(\mathbb{R}^2 \setminus S^1) = H(\mathbb{R}^2 \setminus D^2)\cup H(intD^2)$$ (disjoined)

• $$H(intD^2) \subseteq H(D^2)$$ which is a continuous image of a compact set and therefore is compact (especially bounded). Hence $$H(intD^2)$$ is also bounded.
• Since the unbounded $$\mathbb{R}^2 = H(\mathbb{R}^2 \setminus D^2)\cup H(intD^2) \cup C$$ and $$C$$ is compact (therefore bounded) we must have that $$H(\mathbb{R}^2 \setminus D^2)$$ is unbounded.

(2) Since $$H^{-1}(C \cup H(intD^2)) = S^1 \cup intD^2 = D^2$$ we have that $$C \cup H(intD^2)$$ is a disk. Also, $$\partial[C \cup H(intD^2)] = \partial H(D^2) = H(\partial D^2) = H(S^1) = C$$.

Is there something wrong with my solution? I feel like I'm missing something because I didn't use the second part of Schönflies Theorem, but I can't find any gap in my reasoning. Is there something missing?

Thank you in advance!

• Your proof is correct. The second part of the Schönflies Theorem is a "bonus", in the literature most formulations of the theorem do not include it. – Paul Frost Nov 2 '18 at 17:18
• Thank you for your answer! – Sotiris Simos Nov 4 '18 at 3:44