# If two closed plane curves are outside each other, can there be a point inside both of them?

I think this recent question (also here) has a quick answer if the conjecture below is true.

It looks "obviously" true, but I've learned to distrust my judgement in such matters. It also looks as if it should have an "obvious" proof, but I'm having no ideas.

## Definitions

Let $$[a, b]$$ be any compact interval in $$\mathbb{R}$$. For any continuous function $$\gamma \colon [a, b] \to \mathbb{C}$$, denote the connected, compact set of points $$\{\gamma(t) : a \leqslant t \leqslant b\}$$ by $$[\gamma]$$. Define $$r(t) = |\gamma(t)|$$ ($$a \leqslant t \leqslant b$$). By Theorem 7.2.1 of A. F. Beardon, Complex Analysis (Wiley, Chichester 1979), there exists a branch of $$\operatorname{Arg}\gamma$$ on $$[a, b]$$, i.e. a continuous function $$\theta \colon [a, b] \to \mathbb{R}$$ such that $$\gamma(t) = r(t)e^{i\theta(t)}$$ ($$a \leqslant t \leqslant b$$). Quoting from the same reference:

Definition 7.2.1 Let $$\gamma \colon [a, b] \to \mathbb{C}$$ be any curve and suppose that $$w \notin [\gamma]$$. We define the index $$n(\gamma, w)$$ of $$\gamma$$ about $$w$$ by $$n(\gamma, w) = \frac{\theta(b) - \theta(a)}{2\pi},$$ where $$\theta$$ is any branch of $$\operatorname{Arg}(\gamma - w)$$ on $$[a, b]$$. If $$\gamma$$ is closed then $$n(\gamma, w)$$ is an integer.

The index $$n(\gamma, w)$$ is sometimes called the winding number of $$\gamma$$ about $$w$$, for it represents the number of times that a point $$z$$ moves around $$w$$ as it moves from $$\gamma(a)$$ to $$\gamma(b)$$ along $$\gamma$$. [...]

The index can be used to clarify the difficult question of what is meant by the 'inside' and 'outside' of a closed curve $$\gamma$$. We shall say

(a) that $$z$$ is inside $$\gamma$$ if $$z \notin [\gamma]$$ and $$n(\gamma, z) \ne 0$$,

(b) that $$z$$ is on $$\gamma$$ if $$z \in [\gamma]$$, and

(c) that $$z$$ is outside $$\gamma$$ if $$z \notin [\gamma]$$ and $$n(\gamma, z) = 0$$.

[...] Observe that [...] the outside of $$\gamma$$, say $$O(\gamma)$$, is the union of those components of $$\mathbb{C} \setminus [\gamma]$$ on which the index is zero. Thus $$O(\gamma)$$ is an open set. Further [...] $$O(\gamma)$$ contains the complement of some closed disc. If we denote the inside of $$\gamma$$ by $$I(\gamma)$$, then $$\mathbb{C} \setminus O(\gamma) = [\gamma] \cup I(\gamma),$$ and so the set of points which lie inside or on $$\gamma$$ is a compact set.

## Conjecture

For closed curves $$\sigma$$ and $$\tau$$, if $$[\sigma] \subset O(\tau)$$ and $$[\tau] \subset O(\sigma)$$, then $$I(\sigma) \subset O(\tau)$$.

Because the premise $$[\tau] \subset O(\sigma)$$ implies $$I(\sigma) \cap [\tau] = \emptyset$$, the conclusion can be expressed symmetrically as $$I(\sigma) \cap I(\tau) = \emptyset$$ - whence the title of the question.

• Your title has me scratching my head. Do you have a picture of the situation you’re describing? – Lubin Feb 14 at 3:03
• Hint: (1) $I(\sigma)$ is contained in the same connected component of $R^2- [\tau]$ as $[\sigma]$. (2) Winding number with respect to $\tau$ is constant on connected components of $R^2- [\tau]$. – Moishe Kohan Feb 14 at 4:15
• @MoisheCohen I like the hint - thank you! (I'm still working on it. I just got out of bed, having posted the question at bedtime. I may need coffee, but I don't expect to need further hints.) – Calum Gilhooley Feb 14 at 12:32
• @Lubin I was hoping for smiles, rather than scratched heads! The puzzlingly paradoxical quality of the title is a consequence of the conjecture seeming so utterly "obvious". As the situation described seems in fact to be impossible - as I'd hoped - I can't draw a picture of it. That would need a stronger drug than coffee. :) – Calum Gilhooley Feb 14 at 12:37
• @MoisheCohen I wouldn't be surprised to learn that I've made an unnecessarily laboured use of your hint. In that case, I would be grateful (I mean, even more grateful!) if you would post an expanded version of it as an answer - which I would then almost certainly accept. – Calum Gilhooley Feb 14 at 19:54

(Thanks to Moishe Cohen for his very helpful hint. He is not to blame for the use I have made of it, however!)

Lemma If $$E$$ is a connected, closed subset of a connected, normal topological space $$X$$, and $$G$$ is a connected component of $$X \setminus E$$, then $$G \cup E$$ is connected.

Proof Define $$H = \overline{G} \cap E$$.

If $$H = \emptyset$$, then $$\overline{G}, E$$ are disjoint closed subsets of $$X$$, whence there exist disjoint open subsets $$U, V$$ of $$X$$ such that $$\overline{G} \subseteq U$$ and $$E \subseteq V$$. We cannot have $$U = \overline{G}$$, because then $$U$$ would be a closed, open, non-empty proper subset of $$X$$, so $$X$$ would be disconnected, contrary to hypothesis. On the other hand, we cannot have $$U \ne \overline{G}$$, because then $$G$$ would be properly contained in the open subset $$U$$ of $$X \setminus E$$, contrary to its definition.

Therefore, $$H \ne \emptyset$$.

We have $$G \subseteq G \cup H \subseteq \overline{G}$$, therefore $$G \cup H$$ is connected.

Also, $$(G \cup H) \cap E = H \ne \emptyset$$.

Since $$G \cup H$$ and $$E$$ are both connected, it follows that their union $$(G \cup H) \cup E = G \cup E$$ is connected. $$\square$$

Corollary The union of $$E$$ with any collection of connected components of $$X \setminus E$$ is connected. $$\square$$

Proposition If $$\sigma$$ is a closed curve in $$\mathbb{C}$$, then $$[\sigma] \cup I(\sigma)$$ is connected.

Proof Let $$z, w \in \mathbb{C}$$. If $$z \in I(\sigma)$$, then by definition $$n(\sigma, z) \ne 0$$. If $$z, w$$ are in the same connected component of $$\mathbb{C} \setminus [\sigma]$$, then $$n(\sigma, w) = n(\sigma, z) \ne 0$$, whence $$w \in I(\sigma)$$. Therefore, $$I(\sigma)$$ is a union of connected components of $$\mathbb{C} \setminus [\sigma]$$. The result now follows from the above corollary. $$\square$$

Theorem For closed curves $$\sigma$$ and $$\tau$$ in $$\mathbb{C}$$, if $$[\sigma] \subset O(\tau)$$ and $$[\tau] \subset O(\sigma)$$ then $$I(\sigma) \subset O(\tau)$$.

Proof Because $$[\tau] \subset O(\sigma)$$, we have $$I(\sigma) \subseteq \mathbb{C} \setminus [\tau]$$, therefore $$[\sigma] \cup I(\sigma) \subseteq \mathbb{C} \setminus [\tau]$$. It follows that the connected set $$[\sigma] \cup I(\sigma)$$ is contained in the same connected component of $$\mathbb{C} \setminus [\tau]$$ as $$[\sigma]$$. By hypothesis, this component is a subset of $$O(\tau)$$, whence $$I(\sigma) \subset O(\tau)$$. $$\square$$

• What you wrote is just fine (I have to say, when I wrote the hints, I did not think about the degree of generality in which the argument works and only thought about subsets of $R^n$). I think you should just accept your own answer. – Moishe Kohan Feb 14 at 23:52
• I wanted to make sure I understood. I hadn't grasped that $[\gamma]\cup I(\gamma)$ was connected. (I'm especially distrustful of my intuition at the moment, because I made several silly mistakes working on the problem in the questions referred to - fortunately I didn't post an answer, but I'm working on one now, with renewed confidence.) That's why your hint was so valuable. And it's been a good exercise for my rusty old brain (45 years after attending some topology lectures, although I've tried to top up my knowledge since), having to work the thing out, instead of having it all handed to me. – Calum Gilhooley Feb 15 at 0:26
• You are welcome. :) – Moishe Kohan Feb 15 at 0:29