Relative boundary of $A$ in $V \subset \mathbf{R}^n$ I am studying Multidimentional Real Analysis by Duistermaat and Kolk.
Proposition 1.2.17 states that, given $A \subset V \subset \mathbf{R}^n$, we have $\partial_V A = V \cap \partial A$. Here, $\partial A$ is boundary of $A$ in $\mathbf{R}^n$  and $\partial_V A$
is a boundary of $A$ in relative topology of $V$.
This statement seems to be wrong. For example, if $V = [-1, 1]$ and $A = [0, 1]$, then
$\partial_V A = \{0\} \ne \{0, 1\} = V \cap \partial A \ .$
If I am right, then what is the most general condition on $V$ that makes $\partial_V A = V \cap \partial A$ true? I think it is true if $V$ is open. Of course, in general we have, $\partial_V A \subset V \cap \partial A$. Is it worth going on with the book or should I switch to a different one?
The boundary of $A$ is defined as follows: $\partial A = \overline{A} \cap \overline{A^c}$ where bar represents closure. Similarly, the boundary of $A$ in $V$ is defined as follows: $\partial_V A = \overline{A}^V \cap \overline{V \backslash A}^V$ where bar with $V$ represents relative closure.
 A: The equality $\partial_V A = \partial A \cap V$ holds for all $A \subset V$ if and only if $V$ is open. This is not specific to subspaces of $\mathbb{R}^n$, it holds for all subspaces $V$ of a topological space $X$.
For the necessity, note that $Y \subset X$ is open if and only if $Y \cap \partial Y = \varnothing$ :
$$Y\cap\partial Y  = Y\cap \bigl(\overline{Y} \setminus \overset{\Large\circ}{Y}\bigr)
= Y \cap \bigl(\overline{Y}\cap \bigl(X\setminus \overset{\Large\circ}{Y}\bigr)\bigr) = Y \cap\bigl(X\setminus \overset{\Large\circ}{Y}\bigr) = Y \setminus \overset{\Large\circ}{Y}\,.$$
Now take $A = V$. We have
$$\partial_V V = \partial V \cap V \iff \partial V \cap V = \varnothing \iff V \text{ is open.}$$
The sufficiency follows from
$$\partial A \cap V = \partial_V A \cup (A \cap \partial V)\,. \tag{$\ast$}$$
If $V$ is open, and $A \subset V$, then $A \cap \partial V \subset V \cap \partial V = \varnothing$, so $\partial A \cap V = \partial_V A$ follows from $(\ast)$.
Now let's prove $(\ast)$. Suppose $p \in \partial_V A$, and let $U$ be an arbitrary $X$-neighbourhood of $p$. Then $W = U \cap V$ is a $V$-neighbourhood of $p$, and by either the definition or a characterisation of the boundary it follows that $W \cap A \neq \varnothing$ and $W \setminus A \neq \varnothing$. But $U \supset W$, hence a fortiori $U \cap A \neq \varnothing$ and $U \setminus A \neq \varnothing$. Since $U$ was arbitrary it follows that $p \in \partial A$. Since trivially $p \in V$, the inclusion $\partial_V A \subset \partial A \cap V$ is proved.
Next, suppose $p \in A \cap \partial V$ and let again $U$ be an arbitrary $X$-neighbourhood of $p$. Since $p \in U$ it follows that $U \cap A \neq \varnothing$, and since $p \in \partial V$ it follows that $U \setminus A \supset U \setminus V \neq \varnothing$. By the arbitrariness of $U$, $p \in \partial A$. Trivially $p \in A \subset V$, and thus we have proved the inclusion $A \cap \partial V \subset \partial A \cap V$. Together with the previous step,
$$\partial_V A \cup (A \cap \partial V) \subset \partial A \cap V\,.$$
Finally, suppose $p \in \bigl(\partial A \cap V\bigr) \setminus \partial_V A$. We need to show $p \in A \cap \partial V$. Since $p \notin \partial_V A$, there is a $V$-neighbourhood $W$ of $p$ such that either $W \subset A$ or $W \cap A = \varnothing$. By definition of the subspace topology, there is an $X$-neighbourhood $U$ of $p$ such that $W = U \cap V$. Since $p \in \partial A$, we have
$$\varnothing \neq U \cap A = U \cap (V \cap A) = (U\cap V) \cap A = W \cap A\,.$$
Thus we must have $W \subset A$, in particular $p \in W \subset A$. It remains to see $p \in \partial V$. If it weren't so, $p$ would be an interior point of $V$. But then
$$A \supset W = U \cap V \supset \overset{\Large\circ}{U} \cap \overset{\Large\circ}{V} \ni p$$
would be an $X$-neighbourhood of $p$, i.e. $p \in \overset{\Large\circ}{A}$, contrary to the assumption $p \in \partial A$.
