Boundary of an open set in a Banach space

[Valerio Capraro]

Prove (or disprove) the following statement: the boundary of an open set in a Banach space does not contain open sets.

[dissonance]

I believe this claim to be true in an arbitrary topological space. Indeed, let \(U\) be an open subset of a topological space \(X\). We claim that the interior part of the boundary of \(U\) is empty, that is \(\big( \partial U \big)^\circ= \varnothing\). To see this write

\[X=U \cup \partial U \cup (U^C)^\circ \]

where \(U^C\) stands for the complement of \(U\). Suppose there exists \(x \in \big(\partial U\big)^\circ\). Since \(U\) and \(\partial U\) are disjoint, \(x \in (U^C)^\circ\). But \((U^C)^\circ\) is open and so \((U^C)^\circ\) and \(\partial U\) need be disjoint as well. We just got a contradiction.

What do you think?

[Valerio Capraro]

I think you are right, but I was thinking to another thing, while writing that one. So, let us state my original problem.

Let \( \displaystyle {\left({X},{d}\right)} \) be a metric space. Let \( \displaystyle {A} \) be an open and bounded subset of \( \displaystyle {X} \), \( \displaystyle {r}\gt{0} \), define \( \displaystyle {B}{\left({A},{r}\right)}={\left\lbrace{x}\in{X}:{d}{\left({x},{A}\right)}\lt{r}\right\rbrace} \) and \( \displaystyle {N}{\left({A},{r}\right)}={\left\lbrace{x}\in{X}:{d}{\left({x},{A}\right)}\leq{r}\right\rbrace} \).

1)Prove that if \( \displaystyle {X} \) is a Banach space (probably normed is enough), then \( \displaystyle {N}{\left({A},{r}\right)}\backslash{B}{\left({A},{r}\right)} \) cannot contain open sets.

2) Find an example of a metric space \( \displaystyle {X} \) and a subset \( \displaystyle {A} \) such that it does.

3) Find an example of metric space \( \displaystyle {X} \) and a subset \( \displaystyle {A} \) such that \( \displaystyle {N}{\left({A},{r}\right)}\backslash{B}{\left({A},{r}\right)} \) contains an homeomorphic copy of \( \displaystyle {A} \).