More specifically, let $B$ be a open ball and $C, D$ be open disjoint sets in $\mathbb{R}^n$, $n>1$. Suppose that $B\cap C\neq\emptyset$ and $B\cap D\neq\emptyset$, furthermore, $B\subset \bar{C}\cup\bar{D}$. Is there **at least** one path in $B\cap\partial C$?


Edit: for what i need, the statement actually can be a little less strong: is there some path $\varphi$ in $B$ such that $\varphi\cup\partial C$ has uncontable many points?



Is there at least one path in the common boundary of two open sets?
The answer to the second question is yes: there is an arc containing uncountable points of $B\cap\partial C$. It is enough to prove it in the case $n=2$.


Applying an affine transformation if necessary, we can suppose that $[0,1]^2\subseteq B$, $[0,1]\times\{0\}\subseteq C$ and $[0,1]\times\{1\}\subseteq D$. This implies that for any $x\in[0,1]$, there is some point of $\partial C$ in $\{x\}\times[0,1]$.


Now let $W$ be the set of finite strings of $0$ and $1$. Given a word $w\in W$, we write $w0$ and $w1$ for the words obtained by adding "$0$" or "$1$" at the end of $w$.


To each word $w\in W$ we will associate a rectangle $R\_w=[x\_w,x\_w']\times[y\_w,y\_w']\subseteq[0,1]^2$ such that:


* For any $w\in W$, the set $R\_w\cap\partial C$ has an uncountable projection onto the $x$-axis.
* If $w$ has length $n$, then $R\_w$ has diameter $\leq2^{-n}$.
* For any $w\in W$, $R\_{w0}$ and $R\_{w1}$ are contained in $R\_w$, and $x\_{w0}'<x\_{w1}$. So $R\_{w0}$ and $R\_{w1}$ are disjoint.


It is easy to see how to construct the rectangles inductively. Now let $\omega\in2^\mathbb{N}$ be an infinite word of ones and zeros, with $\omega\_n$ being the finite word formed by the first $n$ characters of $\omega$. Remember that $2^\mathbb{N}$ (with the product topology) is homeomorphic to the ternary Cantor set $X\subseteq[0,1]$ via the function $f:2^\mathbb{N}\to X;\omega=(x\_n)\_{n\in\mathbb{N}}\mapsto\sum\_{n\in\mathbb{N}}2x\_n3^{-n}$.


To each $\omega\in2^\mathbb{N}$ we associate the point $p\_\omega=\cap\_{n\in\mathbb{N}}R\_{\omega\_n}$. This defines an imbedding $f:X\to\partial C$, because $f$ is continuous from a compact space to a T2 space and $f$ is bijective: in fact, if $x\_1,x\_2\in X$ with $x\_1<x\_2$, then the $x(f(x\_1))<x(f(x\_2))$ (where $x(p)$ represents the $x$-coordinate of a point $p$).


We can extend this homeomorphism to an arc $F:[0,1]\to[0,1]^2$: to do this, we just have to define $F$ in the countable intervals $(p\_n,q\_n)$ of $[0,1]\setminus X$. We do this by interpolating linearly between $f(p\_n)$ and $f(q\_n)$, that is, $F(tp\_n+(1-t)q\_n)=tf(p\_n)+(1-t)f(q\_n)$. The continuity of $F$ can be deduced easily from the continuity of $f$, and $F$ is injective because different points of $[0,1]$ get sent to points of $[0,1]^2$ with different $x$-coordinates.


So this arc contains continuum many points of $B\cap\partial C$.


