Jordan curve theorem

Section 2.11 Jordan curve theorem

Theorem 2.11.1. Nonseparation theorem.

If $C\subseteq S^2$ is an arc, then $C$ does not separate $S^2\text{.}$

Proof.

This proof is adapted from an argument given by Ryuji Maehara in the article The Jordan curve theorem via the Brouwer fixed point theorem ¹.

Assume by contradiction that $\R^2-C$ is not connected. This means that among the connected components of $\R^2-C\text{,}$ in addition to the unbounded connected component $U_\infty\text{,}$ there is a bounded component $U\text{.}$

Now observe that we have $\partial U=\overline{U}-U\subseteq C\text{.}$ This is because we have $\overline{U}\cap U'=\emptyset$ for any connected component $U'\ne U\text{,}$ since $U'$ is open and disjoint from $U\text{.}$

Since $C$ is bounded, we can pick an element $\boldu\in U$ and $R> 0$ such that $C\subseteq B_R(\boldu)\text{.}$ Since $\R^2-B_R(\boldu)$ is unbounded and connected, we must have $\R^2-B_R(\boldu)\subseteq U_\infty$ and hence $U\subseteq B_R(\boldu)\text{.}$ Let $B=\overline{B}_R(\boldu)$ (homeomorphic to $B^2$) and let $S=\partial B$ (homeomorphic to $S^1$). Observe that $S\subseteq U_\infty\text{.}$

Since $C$ is a closed subset of the normal space $B\text{,}$ the Tietze extension theorem implies that the identity function $\id\colon C\rightarrow C$ extends to retraction $r\colon B\rightarrow C\text{.}$ Define $f\colon B\rightarrow B-\{\boldu\}$ as follows:

\begin{equation*} f(\boldx)=\begin{cases} r(\boldx) \amp \text{if } \boldx\in \overline{U} \\ \boldx \amp \text{if } \boldx\in B-U. \end{cases} \end{equation*}

Since $\overline{U}\cap U^c=\partial U\subseteq C\text{,}$ and since $r\vert_C=\id\text{,}$ the function $f$ is well-defined and continuous by the closed pasting lemma. We cannot have $f(\boldx)=\boldu$ since $\boldr(\boldx)\in C$ for all $\boldx\in \overline{U}$ and $f(\boldx)=\boldx$ for all $\boldx\notin U\text{.}$

Finally, consider the map $a\circ p\circ f\colon B\rightarrow S\subseteq B\text{,}$ where $p\colon B-\{\boldu\}\rightarrow S$ is the usual retraction map of the ball onto the circle, and $a\colon S\rightarrow S$ is the antipodal map on $S\text{.}$

$Map without fixed point$

This map has no fixed point, contradicting Brouwer’s fixed point theorem. Indeed, since $a\circ p\circ f(B)\subseteq S\text{,}$ any candidate for a fixed point would have to be an element of $S\text{;}$ but for any $\boldx\in S\text{,}$ we have

\begin{align*} a(p(f(\boldx)))\amp =a(p(\boldx)) \amp (f\vert_S=\id) \\ \amp =a(\boldx) \amp (p\vert_S=\id)\\ \amp \ne \boldx\text{,} \end{align*}

since $a(\boldx)$ is the antipode of $\boldx\text{.}$

Since we have reached a contradiction, we conclude that $\R^2-C$ must be connected.

Theorem 2.11.2. Complementary Seifert-van Kampen.

Let $X=U\cup V\text{,}$ where $U$ and $V$ are open, and $U\cap V=A\cup B$ is a separation. Assume $a\in A$ and $b\in B\text{,}$ and that there are paths $\alpha\in P(U; a, b)$ and a path $\beta\in P(V; b, a)\text{.}$ Let $f=\alpha*\beta\in P(X; a,a)\text{.}$

The subgroup $\langle [f]\rangle\subseteq \pi_1(X, a)$ is infinite cyclic. Furthermore, if $\pi_1(X,a)\cong \Z\text{,}$ then $\pi_1(X,a)=\langle [f]\rangle\text{.}$
Suppose $a'\in A$ and we have paths $\gamma\in P(U; a,a')$ and $\delta\in P(V; a', a)\text{.}$ Setting $g=\gamma*\delta\text{,}$ we have $\langle [f]\rangle\cap \langle [g]\rangle=\{e\}\text{.}$

Proof.

Theorem 2.11.3. Jordan curve theorem.

Let $C_1$ and $C_2$ be closed connected subsets of $S^2$ whose intersection consists of two points. If neither $C_1$ nor $C_2$ separates $S^2\text{,}$ then $C_1\cup C_2$ separates $S^2$ into exactly two components.

As a consequence, if $C\subseteq S^2$ is simple closed curve, then $C$ separates $S^2$ into exactly two components, $W_1, W_2\text{.}$ Furthermore, we have $C=\partial W_1=\partial W_2\text{.}$