# Separating a certain planar region with an open set

I have a fairly specific question related to plane separation properties. I couldn't quite see how to use Phragmen–Brouwer properties to answer it because those kind of results generally apply to closed sets.

Let $$U$$ be an open connected subset of the closed unit square $$[0,1]^2$$ such that $$U\cap \partial I^2=(0,1)\times \{0\}$$. Suppose $$V\subseteq U$$ is open (but not necessarily connected), $$V\cap \partial I^2=(1/3,2/3)\times\{0\}$$, and that $$V$$ separates $$a=(1/4,0)$$ and $$b=(3/4,0)$$ in $$U$$. The last condition means that $$a$$ and $$b$$ lie in distinct connected components of $$U\backslash V$$.

I'd basically like to know if the component of $$V$$ that intersects the x-axis (call it $$V_0$$) limits on $$[0,1]^2\backslash U$$ and, alone, does the job of separating $$a$$ and $$b$$ in $$U$$.

Question 1: If $$V_0$$ is the connected component of $$V$$ that contains $$(1/3,2/3)\times\{0\}$$, must $$V_0$$ separate $$a$$ and $$b$$ in $$U$$?

Question 2: Must $$\overline{V_0}$$ meet $$[0,1]^2\backslash U$$?

• are you sure that by separating you mean "no path"? then take $U$ to be the open square with an extra segment, $V$ to be a thinner rectangle, but pierced by a topologist's sine curve. Perhaps a more interesting question is when separated means "no connected set"
– erz
Aug 13, 2020 at 23:51
• @erz whoops! You're right. That's the standard definition of "separated by" that I meant. Thanks. Aug 14, 2020 at 1:40
• I guess then you have to show that the outer boundary of $V_0$ is connected
– erz
Aug 14, 2020 at 10:05