# Is there at least one path in the common boundary of two open sets?

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?

• No. For $n=2$ you can arrange for the shared boundary of $B$ and $C$ to be a pseudo-arc. Such a space in particular doesn't contain any nontrivial paths. I'm sure one can perform a similar construction in higher dimensions. May 2, 2022 at 17:26
• For the second question, maybe you want to change 'path' by 'arc'? (Because there are space filling curves) May 3, 2022 at 0:23
• @SaúlRM yep, i believe that's better May 3, 2022 at 1:48

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 $$x_{w0}'. So $$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 $$x_1, then the $$x(f(x_1)) (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 $$B\cap\partial C$$.

To expand on the comment by @Wojowu: Higher-dimensional hereditarily indecomposable continua, Trans. AMS, 71 (1951), 267-273 R. H. Bing showed that between disjoint closed sets in $$\mathbb{R}^n$$ one can always find partitions all components of which are hereditarily indecomposable.