# 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.

• I very much suspect that $V$ is assumed to be open. That may be a general assumption in the book and not stated explicitly every time a $V$ occurs. Apr 25, 2020 at 20:00
• I don't think so. The authors define relative topology one page before, and explicitly state certain facts are true only if $V$ is open. Apr 25, 2020 at 20:50
• Well, "$\partial_V A = \partial A \cap V$ for all $A \subset V$" is one of the facts that are only true for open $V$. Note that $V$ is open if and only if $V \cap \partial V = \varnothing$. (That holds for all topological spaces.) Now choose $A = V$. Then $\partial_V A = \varnothing$. And thus $\partial_V A = \partial A \cap V \iff \partial A \cap V = \varnothing$, which means "$V$ is open" is a necessary condition. Apr 25, 2020 at 20:59
• Thank you! This helps a lot! Please post this as an answer, maybe with more details, and I will accept it. Apr 25, 2020 at 23:41

## 1 Answer

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 $$p \in \partial A$$.