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This is the original statement I want to prove:

Conjecture: Let $A$ and $B$ be open connected sets in $\mathbb R^2$ with the usual metric such that $A \cap B \neq \emptyset$. Then $ \partial A \cap \partial B = \emptyset \Longrightarrow$ $ A \cap B $ is connected.

It is a simple statement, but I cannot find a proof neither a counter example. Any ideas are welcome!


Update 0:

Some comments about this being a general topological result made me realize that an example of space where this doesn't hold is important as motivation.

Here is a simple example of such space:

Consider the circle $S^1 = \mathbb R / \mathbb Z$ and the intervals $A = (\frac{2}{8}, \frac{7}{8})/ \sim$ and $B = (\frac{-3}{8}, \frac{3}{8})/ \sim$ ($\sim$ is the equivalence relation defining the circle). Note that they satisfy the conditions but their intersection is not connected: $A \cap B =( (\frac{2}{8}, \frac{7}{8}) \cup (\frac{5}{8}, \frac{7}{8}))/ \sim$. This is illustrated below. enter image description here


Update 1:

I came with a weaker version of this statement that is sufficient for the application I need.

Weak Conjecture: Let $A,B$ be open connected sets in $\mathbb R^2$ such that $A \cap B \neq \emptyset$ and each connected component of $\partial A$ and $\partial B$ is the image of a proper embedding from $\mathbb R$ to $\mathbb R^2$. Then $ \partial A \cap \partial B = \emptyset \Longrightarrow$ $ A \cap B $ is connected.

$\ \ \ $Proof:

$\quad$ First, we will refer to images of proper embeddings from $\mathbb R$ to $\mathbb R^2$ as lines. Note that lines splits the plane into two connected components. In the case that a line $\phi$ is a connected component of the boundary of some connected set $U$, we can define $D(\phi)$ as the connected component of $\mathbb R^2 \setminus \phi$ which contains $U$. The other c.c. is denoted $E(\phi)$. If we denote $\Phi_U$ the collection of all c.c. of $\partial U$, we can write

$$ U = \bigcap_{\phi_i \in \Phi_U} D(\phi_i) \qquad \ \ \ \ \ U^c = \bigsqcup_{\phi_i \in \Phi_U} E(\phi_i)\;. $$ It is important to note that the union above is disjoint. Otherwise, we would get $i_1 \neq i_2$ such that $\phi_{i_1} \cap \phi_{i_2} \neq \emptyset$, absurd since they are different c.c. components of $\partial U$.

$\quad$ Now we begin the proof. Let $C : = A \cap B$ and suppose $C$ has at least two nonempty connected components $C_1$ and $C_2$. Note that $ C_2 \subset C_1^c = \bigsqcup_{\phi_{i} \in \Phi_{C_1}} E(\phi_i)$, therefore $\exists ! \ i_n$ s.t. $C_2 \in E(\phi_{i_n})$. Since $\partial A \cap \partial B = \emptyset $, we can suppose without loss of generality that $\phi_{i_n} \subset \partial A$. However, $C_1 \subset A$ and $C_2 \subset A$ are contained on different c.c. of $\mathbb R^2 \setminus \phi_{i_n}$, which would imply that $A$ is not connected, absurd.

$\qquad $Q.E.D.

The main question remains open for discussion. The understanding of the topology of $\partial U$ is all that we should need to conclude it.

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  • $\begingroup$ @FranklinPezzutiDyer For open sets, $\partial A=\overline A\setminus A$. $\endgroup$ Jul 1, 2020 at 15:43
  • $\begingroup$ @ FranklinPezzutiDyer $\partial A$ is the boundary of $A$. Points which are accumulated by points of $A$ and $ A^c$. $\endgroup$ Jul 1, 2020 at 15:44
  • $\begingroup$ @DonThousand I really enjoy the engagement. Continue thinking in the problem. $\endgroup$ Jul 1, 2020 at 15:53
  • $\begingroup$ @FranklinPezzutiDyer I really enjoy the engagement. Please continue thinking in the problem. $\endgroup$ Jul 1, 2020 at 15:53
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    $\begingroup$ I was editing my attempt sparingly during the day and I didn't notice your edit. $\endgroup$ Jul 4, 2020 at 22:12

2 Answers 2

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I'll try to sketch some ideas that might lead to a proof. The following is definitely incomplete, also too long for a comment. Did you consider anything along these lines?

Assume by way of contradiction that the hypotheses hold but $A\cap B$ is not connected. Then take two points $x,y\in A\cap B$ in different connected components. Since connected open sets in $\mathbb{R}^2$ are arcwise connected, we may find a path $P_A:[0,1]\to A$ from $x$ to $y$ inside $A$. Considering that $\partial A\cap \partial B=\emptyset$, we might assume the starts in $A\cap B$, traverses $A\setminus B$ and reaches $A\cap B$ in that order:

paths between points

(To see this, it should be enough to find a maximal subinterval of $P_A^{-1}(A\smallsetminus B)$, which must exist, otherwise, $B$ would accumulate along the path and we would have a common boundary point.) The point $a$ lies in the boundary of $A\cap B$: more precisely, in $A \cap \partial(A\cap B) \smallsetminus B$.

Similarly we can find a path $P_B:[0,1] \to B$ from $x$ to $y$ (though now I won't be confident that I can have the same property as above). At least, we can be sure that we have some point $b$ in $B \cap \partial(A\cap B) \smallsetminus A$.

Now, let us restrict ourselves to an open ball $V$ containing $x$, $y$, and both paths. It is easy to see, since $A$ and $B$ are open, that every point of $\partial (A\cap B)$ must be on $A\smallsetminus B$ or $B\smallsetminus A$, otherwise it would lie in $\partial A\cap \partial B$. So (using compactness) we can cover $\partial (A\cap B)\cap\overline V$ by finitely many separated balls contained in $A$ or in $B$.

If we focus on the balls containing $a$ and $b$, we have a situation as pictured below:

two balls

where red corresponds to $A$ and blue to $B$, and the left ball is completely included in $B$ and the right ball inside $A$. This looks rather strange, since $x$ lies in (the interior of) $A\cap B$, and therefore we should be able to find some point of $\partial (A\cap B)$ in the region between the two paths.

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  • $\begingroup$ This is a great attempt. I initially had some a similar idea, but I struggled with the development. Mainly with arguments regarding the boundary (for example, how to know if these $a$ and $b$ are in the same connected component of $\partial (A \cap B)$? This pushed me towards the weak version of the conjecture. But I guess the best way to achieve the full result is to follow some construction like yours. Thank you a lot for the engagement, any new development I contact you! $\endgroup$ Jul 5, 2020 at 0:43
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    $\begingroup$ Another idea was to make both paths simple (by using compactness and looking for balls included in $A$ resp. $B$), use Jordan curve theorem and somehow replicate the situation in a smaller place... At some point, the successive points $a$ and $b$ would converge to something in $\partial A \cap \partial B$. Anyway, thank you very much for such a great problem! $\endgroup$ Jul 5, 2020 at 1:30
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In the situation you describe, for both $A$ and $B$ there is an increasing sequence of regions $\Omega_{A}^k \subset A$, $\Omega_{B}^k \subset B$, each bounded by finitely many disjoint analytic Jordan curves (parameterized so that the regions are to the left), such that the closures of these regions form exhausting sequences of compact subsets for both $A$ and $B$ and $\overline{\Omega_{A}^{k}} \subset \Omega_{A}^{k+1}$, $\overline{\Omega_{B}^{k}} \subset \Omega_{B}^{k+1}$ both hold.

After a mobius transformation we may also assume that $A$ and $B$ are neighbourhoods of $\infty$, and each $\Omega_{A}^k$, $\Omega_{B}^k$ is the intersection of all the exterior regions of the analytic Jordan curves they are each bounded by. Since we are working in $\overline{\mathbb{C}}$, the boundaries of these regions converge uniformly to the boundaries of $A$ and $B$ respectively, since the latter are a positive distance apart, for large enough $k$ , $\Omega_{A}^k$ and $\Omega_{B}^k$ are $\cap_{j=1}^{s^{A}_k} \text{ext}(\gamma^{A}_{j})$ and $\cap_{j=1}^{s^{B}_{k}} \text{ext}(\gamma^{B}_{j})$, and $\gamma^{A}_{1},...,\gamma^{A}_{s^{A}_{k}},\gamma^{B}_{1},...,\gamma^{B}_{s^{B}_{k}}$ are disjoint analytic Jordan curves. Then we have that $\Omega_{A}^k \cap \Omega_{B}^k$ is path-connected as long as we know that whenever $\gamma_1,...,\gamma_t$ are disjoint Jordan curves, $\cap_{i=1}^t \text{ext}(\gamma_t)$ is connected, which I think one can show basically by the Jordan curve theorem (the Jordan curve theorem tells that for two disjoint Jordan curves, the interior regions are either nested or disjoint and separated a positive distance apart, this allows one to reduce to the case where all Jordan curves in consideration have interiors that are disjoint and a positive distance apart, hence if any path connecting two points in the intersection of the exteriors enters into one of the interior regions bounded by say $\gamma_1$, it does not enter any other interior region before it exits this interior region, which means that on some definite time interval we can correct our path so that it closely follows along $\gamma_1$ but remains in the intersection of all the exteriors, and do this whenever we need to correct (we will only need to do a finite number of corrections by compactness of the finite union of the closures of all the interior regions) ).

So $\Omega^{A}_{k} \cap \Omega^{B}_{k}$ is path connected for sufficiently large $k$, say for $k \geq K_0$. Finally, since $A \cap B = \lim_{k \geq K_0} \Omega^{A}_{k} \cap \Omega^{B}_{k}$ is an increasing sequence of path-connected open sets, $A \cap B$ is path connected.

Edit: More generally, the result I was using in the first paragraph could be formulated as follows:

Let $E \subset \mathbb{C}$ be compact, and let $\Omega$ be the unbounded component of $\overline{\mathbb{C}} \setminus E$. Then there exist an increasing sequence of regions $\Omega_k$ of the form described in the first paragraph increasing to $\Omega$.

In this case, $A$ and $B$ are connected open sets, WLOG we can assume they contain the point at $\infty$, then they are the unbounded components of the complements of $A^c$, $B^c$ respectively, which are compact subsets of the plane.

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