Possible corollary of Jordan's curve theorem

Question

Suppose $\gamma$ and $\phi$ are simple closed curves in $\mathbb{R}^2$, simple meaning that they have no self-intersections. Suppose that $\gamma$ and $\phi$ intersect in exactly one point and that $\phi$ minus the intersection point is contained in the interior of $\gamma$. How can we prove, possibly by using Jordan's curve theorem, that $\phi$ induces a decomposition of the interior of $\gamma$ in two connected components? I know the interior of $\gamma$ is homeomorphic to $\mathbb{R}^2$, but what is bothering me here is that a point of $\gamma$ (and exactly one) is on the boundary of the connected component. Thus, an homeomorphism which sends the interior of $\gamma$ to $\mathbb{R}^2$ cannot possibly send $\gamma$ to $S^1$, right?

Answer 1

The easiest way to argue is to quote the full Jordan-Schoenflies theorem which implies that the "interior" region $D$ of $\gamma$ (i.e. the bounded component of its complement) is homeomorphic to ${\mathbb R}^2$. (The actual Jordan-Schoenflies theorem is even stronger than this.)

Let $p$ denote the intersection point of the two curves. Then $\phi -\{p\}$ is homeomorphic to the real line and this homeomorphism defines a continuous injection ${\mathbb R}\to D$. Composing with the homeomorphism $h: D\to {\mathbb R}^2$, we obtain a continuous injection
$$ {\mathbb R} \stackrel{\eta}{\longrightarrow} \phi -\{p\} \stackrel{\zeta}{\longrightarrow} {\mathbb R}^2, f: {\mathbb R}\to {\mathbb R}^2. $$

Lemma 1. The map $f$ is proper.

Proof of properness. Let $K$ be a compact in ${\mathbb R}^2$. Its preimage $f^{-1}(K)$ in ${\mathbb R}$ is closed, we need to check that it is bounded to prove compactness. Suppose that this preimage is unbounded. Then there is a sequence $x_n\in f^{-1}(K)$ which diverges to $\infty$ in the 1-point compactification of the real line. But then its image sequence $y_n$ in $\phi$ converges to $p$, since we have a homeomorphism $$ \eta: {\mathbb R}\cup \{\infty\}=S^1 \to \phi. $$ Then $y_n$ contains no subsequences converging in $D$, which contradicts compactness of $K$. qed

Since the map $f$ is proper, by taking the one-point compactifications of both ${\mathbb R}$ and ${\mathbb R}^2$, we obtain a continuous map $$ F: S^1\to S^2. $$

Lemma 2. $F$ is injective.

Proof. We already have injectivity of the restriction of $F$ to the real line, $f: {\mathbb R}\to {\mathbb R}^2$; we also have that $F(\infty)=\infty$. Thus, for every $x\in {\mathbb R}$, ${\mathbb R}^2\ni F(x)\ne F(\infty)=\infty$. qed

Now, you quote Jordan curve theorem again and conclude that the image of $F$ (which is $h(\phi -\{p\}) \cup \{\infty\}$) separates $S^2$ in exactly two components. Applying $h^{-1}$, we see that $\phi -\{p\}$ separates $D$ in exactly two components as well, as required. qed

Addendum. If you know about compactly supported cohomology and Alexander duality, this argument can be streamlined and one can avoid the homeomorphism $h$ and the 1-points compactifications. This is how a proof would go if you were to ask a similar question about $n-1$-dimensional spheres embedded in ${\mathbb R}^n$ and intersecting in exactly one point. In this situation, the complement to a sphere need not be homeomorphic to ${\mathbb R}^n$ and one would have to use a purely homological argument.

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Edit. 1. Here is a criterion for properness one can use instead of the direct argument in Lemma 1: Let $X, Y$ be metrizable spaces, $f: X\to Y$ is a continuous map. Then $f$ is proper if and only if: For every sequence $x_n\in X$ that contains no convergent subsequences, the image sequence $f(x_n)$ contains no convergent subsequences either. I will write a proof later if you are interested (it's no longer needed for my solution).

2. As for the JCT (Jourdan Curve Theorem), the complement to (any) point in $S^2$ is homeomorphic to the plane. Hence, the planar JCT is equivalent to the JCT in the 2-sphere.

Answer 2

Moishe Kohan's answer is certainly more elegant, but after some fiddling I think this can be done without using the full Jordan-Schoenflies.

We will denote by $v$ the point where $\phi$ and $\gamma$ intersect, and use $int(\pi)$ and $ext(\pi)$ to denote the interior and exterior regions of the Jordan curve $\pi$.

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To begin, pick $s \in ext(\gamma)$ and $t \in int(\phi)$ such that the straight line $\overline{st}$ does not contain $v$. Note that $\overline{st}$ must cross both curves. Let $p$ be the point of $\overline{st} \cap \gamma$ that is closest to $t$, and let $q$ be the point of $\overline{pt} \cap \phi$ that is furthest from $t$ (these points exist by compactness). Observe that, except for its endpoints, the line segment $\overline{pq}$ must lie in $int(\gamma) \cap ext(\phi)$.

We have now have a subspace of $\mathbb{R}^2$ formed by the union of four Jordan curves: $\phi$, $\gamma$, and the two curves formed by following part of $\phi$, then $\overline{pq}$, then part of $\gamma$ (see the figure). Thus every point not on one of the curves must be in $ext(\gamma)$, or in the interior of one of the other curves. These interiors are all connected by the Jordan Curve Theorem. Further, a path can be drawn from the region $B$ to the region $C$ that only crosses $\overline{pq}$ (by compactness, and since $\overline{pq}$ lies on the boundary of both $B$ and $C$, we can take a small disk somewhere on $\overline{pq}$ that does not intersect $\gamma$ or $\phi$ and draw a path that crosses $\overline{pq}$ through this disk).

So we conclude that $B \cup C \cup \overline{pq} - \{p,q\}$ is one connected region of $\mathbb{R}^2 - (\gamma \cup \phi)$, and that this region and $A$ cover all the points of $int(\gamma)$.

Answer 3

Here's a solution using winding numbers combined with a theorem of Janiszewski (at very end). A useful reference is sections 3 and 4 of chapter 11 of Beardon's Complex Analysis. The more advanced Schoenflies theorem is from chapter 11, section 5 of said book, and outside the scope of this post.

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Summary of JCT implications:
When you prove the JCT for Jordan curve $\sigma$, then $\mathbb C-[\sigma]$ has two components, where the image of the curve, $[\sigma]$, is the boundary of each component. The two components are (1) an unbounded component which necessarily has winding number zero i.e. $n\big(\sigma, p\big)=0$ for any $p$ in unbounded component and (2) a single bounded component that has winding number $\pm 1$. By suitable choice of orientation (i.e. reversing the curve if needed) for purposes of this problem we can assume WLOG that the winding number is always 1 in the bounded component. Note if $[\phi]\cap [\gamma] = \big\{v\big\}$ then for $z\in [\phi]-\big\{v\big\}$ OP's statement that this is in the interior of $[\gamma]\implies n\big(\gamma,z\big)=1$.

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For convenience we may assume
$\gamma:[0,1]\longrightarrow \mathbb C$ and $\phi:[0,1]\longrightarrow \mathbb C$ and let $A:=[\gamma]$ and $B:=[\phi]$ and define the cycle $\Gamma:= \big(\gamma, \phi\big)$

Select some $p,p'\in \mathbb C-[\Gamma]=\mathbb C-\big(A\cup B\big)$ with the same winding numbers. Since for purposes of this question our Jordan curves only have winding numbers of $0$ or $1$ and there are two curves, there are only 3 possibilities

(a) $n\big(\Gamma,p\big)=n\big(\Gamma,p'\big)=0$ i.e. both in the unbounded component of $\mathbb C-A$ and $\mathbb C-B$
(b) $n\big(\Gamma,p\big)=n\big(\Gamma,p'\big)=1$ i.e. both in the bounded component of $\mathbb C-A$ and unbounded component of $\mathbb C-B$

(c) $n\big(\Gamma,p\big)=n\big(\Gamma,p'\big)=2$ i.e. both in the bounded components of $\mathbb C-A$ and $\mathbb C-B$

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clarification for (b):
Consider the other "possibility" of say $p$ in the bounded component of $\mathbb C-B$-- i.e. $n\big(\phi,p\big)=1$-- but $p$ is in the unbounded component of $\mathbb C-A$--i.e. $n\big(\gamma, p\big)=0$. This would imply the existence of a polygonal path $\lambda$ from $p$ to $w$ in $\mathbb C-A$ where $\vert w\vert \gt \max_{u\in B}\vert u\vert $ thus $n\big(\phi,p\big)=1$ and $n\big(\phi,w\big)=0$ so $[\lambda]$ and $B$ must have non-trivial intersection (otherwise the winding number would be constant for all points on $[\lambda]$) so there is some $z \in [\lambda]\cap B$ (but $z\not \in A$ since $[\lambda]\in \mathbb C-A$). Finally $n\big(\gamma, z\big)=1$ (ref 'Summary of JCT implications') and $z$ is path connected to $p$ in $\mathbb C-A$ (via $\lambda$) hence $1=n\big(\gamma, z\big)= n\big(\gamma, p\big)=0$ which is impossible.

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existence of (b) and (c)
For avoidance of doubt: while it is obvious that for reasons of compactness an unbounded component-- i.e. (a)-- must exist, it may not be obvious that (b) and (c) must exist, so to justify this:
$w:=\phi\big(\frac{1}{2}\big)$ (or if that is the intersection point then $w:=\phi\big(0\big)$). We have $\text{distance}\Big(\big\{w\big\}, A\Big) = 2\epsilon\gt 0$ since $A$ is compact and $\big\{w\big\}$ is closed.

Then the open ball $B(w,\epsilon)$ does not meet $A$ but since $w\in B$, which is the boundary for the two components of $\mathbb C-B$, there is some $p, p' \in B(w,\epsilon)$ such that $n\big(\phi, p\big)=0$ and $n\big(\phi, p'\big)=1$ and note that they are path connected to $w$ in $\mathbb C-A$ (since the entire open disc $B(w,\epsilon)$ exists in $\mathbb C-A$).

Thus
$n\big(\Gamma, p\big) = n\big(\gamma, p\big)+n\big(\phi, p\big) =n\big(\gamma, w\big)+n\big(\phi, p\big) = 1 + 0 = 1$
$n\big(\Gamma, p'\big) = n\big(\gamma, p'\big)+n\big(\phi, p'\big) = n\big(\gamma, w\big)+n\big(\phi, p'\big) = 1 + 1 = 2$

so cases (b) and (c) must exist as well.

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Finally, since $A$ and $B$ are compact and $A\cap B$ is connected then in each case we can conclude that $p, p'$ are in the same component of $\mathbb C-\big(A\cup B\big)$, due to a theorem attributed to Janiszewski, which is Theorem 11.3.4 in Beardon:

Let $A$ and $B$ be compact subsets of $\mathbb C_\infty$ such that $A\cap B$ is connected. If $z_1$ and $z_2$ lie in the same component of $\mathbb C_\infty -A$ and $\mathbb C_\infty -B$, then they lie in the same component of $\mathbb C_\infty -(A\cup B)$.

Put differently, $n\big(\Gamma,p\big)=n\big(\Gamma,p'\big)$ is obviously a necessary condition and due to Janiszewski we conclude it is also a sufficient condition for $p$ and $p'$ to be in the same component of $\mathbb C-[\Gamma]=\mathbb C-\big(A\cup B\big)$ and since the winding number for $\Gamma$ takes on exactly 3 values, there are exactly 3 connected components.