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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.

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  • $\begingroup$ 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. $\endgroup$ Apr 25, 2020 at 20:00
  • $\begingroup$ 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. $\endgroup$
    – Yerbolat
    Apr 25, 2020 at 20:50
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    $\begingroup$ 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. $\endgroup$ Apr 25, 2020 at 20:59
  • $\begingroup$ Thank you! This helps a lot! Please post this as an answer, maybe with more details, and I will accept it. $\endgroup$
    – Yerbolat
    Apr 25, 2020 at 23:41

1 Answer 1

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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 $p \in \partial A$.

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