Cluster points and equality of sets

Question

I am trying to understand a proof of the following result:

If $A $ is a subset of a fixed set $V $ in $\mathbb R^n $ then the boundary of $A $ in $V$ is $V\cap \partial A $.

The proof runs as follows:

The boundary of $A $ in $V$ is $(V\cap\overline{A})\cap(V\cap\overline{V\setminus A})$ which is equal to $(V\cap\overline {A})\cap(V\cap\overline {\mathbb R^n\setminus A})$ which is equal to $V\cap (\overline {A}\cap\overline {A^c})$ which is equal to $V\cap \partial A $.

What I don't understand is why $(V\cap\overline {A})\cap(V\cap\overline {V\setminus A})$ is equal to $(V\cap\overline {A})\cap(V\cap\overline {\mathbb R^n\setminus A})$? How can I rigorously show that this is true. The direction which is problematic is to show the RHS set within the LHS set.

Just to clarify here are the definitions: A point $a\in V$ is said to be a cluster point of $A $ in $V $ if every ball centred at it intersects $A $. The set of all cluster points of $A $ in $V $ is denoted by $\overline {A^V}$ and we denote the boundary of $A $ in $V $ by $\overline {A^V}\cap \overline {(V\setminus A)^V}$.

Answer 1

This is not true without some additional conditions. Let's see what goes wrong if we try to prove it:

Consider a point $$x \in V \cap \overline{\mathbb{R}^n \setminus A}\,.$$ Then clearly $x \in V$, and every ball around $x$ intersects $\mathbb{R}^n \setminus A$. Now there are two possibilities:

1. every ball around $x$ intersects $V\setminus A$. Then $x \in V \cap \overline{V \setminus A}$ and it is clear that $x \in (V \cap \overline{A}) \cap (V \cap \overline{\mathbb{R}^n \setminus A})$ if and only if $x \in (V \cap \overline{A}) \cap (V \cap \overline{V \setminus A})$.
2. there is a ball, say $B$, around $x$ not intersecting $V\setminus A$. That is, $V \cap B \subset A$, or equivalently that $x$ lies in the interior of $A$ relative to $V$. Then clearly $x \in V \cap \overline{A}$, so $$x \in \bigl((V \cap \overline{A}) \cap (V \cap \overline{\mathbb{R}^n \setminus A})\bigr) \setminus \bigl((V \cap \overline{A}) \cap (V \cap \overline{V \setminus A})\bigr)\,.$$

Thus we have $\partial_V A = V \cap \partial A$ if and only if option 2 never happens.

If option 2 happens, then $x$ lies in the boundary of $V$, for every ball around $x$ intersects $$(\mathbb{R}^n \setminus A) \setminus (V\setminus A) = \mathbb{R}^n \setminus V\,.$$ Thus if option 2 happens, then $x \in \partial V \cap \operatorname{int}_V A$. Conversely, it is easy to see that for $x \in \partial V \cap \operatorname{int}_V A$ option 2 applies.

So we have $\partial_V A = V \cap \partial A$ if and only if $$\partial V \cap \operatorname{int}_V A = \varnothing\,. \tag{1}$$

A trivial example where $(1)$ doesn't hold is any non-open $V$ and $A = V$, then $\partial_V A = \varnothing$ and $V \cap \partial A \neq \varnothing$.

This can never happen, i.e. $(1)$ holds for all $A \subset V$, if and only if $V$ is open.

Note: nothing in this uses particular properties of $\mathbb{R}^n$. If every occurrence of "ball around" is replaced with "neighbourhood of", the conclusion holds in any topological space for subsets $A \subset V \subset X$.