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!

  • 1
    $\begingroup$ 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. $\endgroup$ – Paul Frost Nov 2 '18 at 17:18
  • $\begingroup$ Thank you for your answer! $\endgroup$ – Sotiris Simos Nov 4 '18 at 3:44

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