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?

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    $\begingroup$ "Without loss of generality, suppose that ϕ minus the intersection point is contained in the interior of γ" You do lose generality here, actually. If you have two externally tangent circles, for instance, then you can't contain either curve in the interior of the other. $\endgroup$
    – user147556
    Jul 4, 2021 at 19:34
  • $\begingroup$ Not too sure, but it may be helpful to use the full Jordan-Schoenflies theorem here? This states that a Jordan curve and its interior are homeomorphic to a disk $D^2$. $\endgroup$ Jul 4, 2021 at 19:46
  • $\begingroup$ @MichaelBarz But then one is contained in the exterior of the other, which is essentially the same thing. After all, one can always swap interior and exterior by means of stereographic projection, which is a homeomorphism, right? $\endgroup$
    – Lele99_DD
    Jul 4, 2021 at 20:05
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    $\begingroup$ @Lele99_DD Yes, you can do all this. It's often convenient to just work on the sphere $S^2$ instead of the plane whenever you are only discussing compact sets (like Jordan curves), as all your results are the same but you don't have to worry about the difference between bounded and unbounded regions. $\endgroup$ Jul 4, 2021 at 20:06
  • $\begingroup$ @BrandonduPreez We can assume all these things if you want, but they don't seem very helpful since they don't change the fact that $\phi$ intersects $\gamma$ exactly once. $\endgroup$
    – Lele99_DD
    Jul 4, 2021 at 20:08

3 Answers 3


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.

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).

  1. 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.
  • $\begingroup$ Could you please convince me on why the map $\mathbb{R} \to D$ is a proper embedding? And wait, how does the Jordan curve theorem work on $S^2$? I mean, I only knew it for $\mathbb{R}^2$. $\endgroup$
    – Lele99_DD
    Jul 6, 2021 at 17:31
  • $\begingroup$ @Lele99_DD: See the edit. $\endgroup$ Jul 6, 2021 at 17:54
  • $\begingroup$ Thank you. Sorry, I'm not familiar with the criterion for properness you mentioned, may I ask you a proof or even a reference? $\endgroup$
    – Lele99_DD
    Jul 6, 2021 at 19:07
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    $\begingroup$ @Lele99_DD: Homeomorphisms map compact sets to compact sets; $D\cup \{\phi\}$ is compact, so its image will be compact, hence, bounded. $\endgroup$ Jul 14, 2021 at 14:19
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    $\begingroup$ @Lele99_DD: Yes, exactly three components. The easiest way to solve such problems is to use Alexander duality; I am sure, there is also a "low-brow" proof. $\endgroup$ Jul 15, 2021 at 18:33

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 $\pi$.

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 $int(\gamma)$.

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  • $\begingroup$ I don't know if this is correct, but I might have found an easier way to prove this. Let $X$ be the connected component of $\mathbb{R}^2 \setminus \gamma$ which contains $\text{int}(\phi)$. I will just prove that $X \cap \text{ext}(\phi)$ and $\text{int}(\phi)$ are the connected components of $X \setminus \phi$. Obviously they are path-connected. Moreover, $\text{int}(\phi)$ is maximal, because any open path-connected set strictly containing $\text{int}(\phi)$ must intersect $\phi$, thus it would not be in $X \setminus \phi$. Analogously, one proves that $X \cap \text{ext}(\phi)$ is maximal. $\endgroup$
    – Lele99_DD
    Jul 5, 2021 at 13:51
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    $\begingroup$ The entire difficulty here is showing that $X\cap ext(\phi)$ is path connected, so I'm not sure how we can assume this without some work. $\endgroup$ Jul 5, 2021 at 19:03
  • $\begingroup$ Thank you, now I'm starting to better understand and appreciate your proof. $\endgroup$
    – Lele99_DD
    Jul 5, 2021 at 20:38
  • $\begingroup$ @Lele99_DD If you are interested in a general framework for these kinds of problems about configurations of Jordan curves in the plane, I recommend Thomassen's paper "The Jordan-Schonflies Theorem and the Classification of Surfaces". In particular, Theorem 3.3 lets you find similar results for any finite collection of Jordan curves that only have finitely many intersection points. $\endgroup$ Jul 5, 2021 at 21:06
  • $\begingroup$ I'm trying to formally justify why $\overline{st}$ must cross both curves. To do this, I think it is reasonable to prove that $t \in \text{int}(\gamma)$ and $s \in \text{ext}(\phi)$. I tried to reason by way of contradiction, but I've had no luck so far. If I assume $t \in \text{ext}(\gamma)$, then I guess one could try proving that $\text{int}(\phi)$ and $\text{ext}(\gamma)$ are path-connected components in $\mathbb{R}^2 \setminus \{\gamma,\phi\}$ (I only managed to do so with $\text{ext}(\gamma)$). This would imply $\text{int}(\phi) = \text{ext}(\gamma)$, thus $\phi=\gamma$, contradiction. $\endgroup$
    – Lele99_DD
    Jul 6, 2021 at 13:42

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.

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 $[\gamma]\implies n\big(\gamma,z\big)=1$.

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 $\mathbb C-B$

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.

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 $\mathbb C-A$).

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

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

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.


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