1

If we have an arbitrary metric space $(M,d)$ with $A \subseteq (M,d)$ then the boundary of $A$ is defined the following way for any $\epsilon > 0$

$\partial A :=$ $\{x \in M: B_{\epsilon}(x) \cap A \neq \emptyset, B_{\epsilon}(x) \cap A^{c} \neq \emptyset\}$.

However, I understand we can rewrite this definition: let $int(A)$ be the interior of $A$ and $cl(A)$ be the closure of $A$. Then we can rewrite $\partial A :=$ $\{x \in M: B_{\epsilon}(x) \cap A \neq \emptyset, B_{\epsilon}(x) \cap A^{c} \neq \emptyset\}$ as $$\partial A := cl(A) \cap cl(A^{c}).$$

While most of my colleagues in my program are just taking this at face value, I wanted to see why this is true; here is my reasoning

Since $\partial A :=$ $\{x \in M: B_{\epsilon}(x) \cap A \neq \emptyset, B_{\epsilon}(x) \cap A^{c} \neq \emptyset\}$, so $x \in \partial A$ if and only if no ball $B_{\epsilon}(x)$ is subset of either $A$ or $A^{c}$, which is true if and only if $x \notin int(A)$ and $x \notin int(A^{c})$. Therefore, $x \in (int(A))^{c}$ and $x \in (int(A^{c}))^{c}$. So we can write $$\partial A := (int(A))^{c} \cap (int(A^{c}))^{c}.$$ Now, clearly this can be simplified - But this is where I start to have questions. Is it true that $int(A) = (cl(A))^{c}$? and if so, wouldn't that imply that $(int(A))^{c} = cl(A)$? If this is the case, then we could arrive at the following result by applying the same logic to ($int(A^{c}))^{c}$: $$\partial A := cl(A) \cap cl(A^{c}).$$ Please correct me if I am wrong or if I am thinking about something backwards.

Taylor Rendon
  • 2,276

2 Answers2

1

Your definition of the boundary is slightly incorrect it should be: $$ \partial A := \{x \in M : \mbox{for all}~\epsilon > 0, B_{\epsilon}(x) \cap A \neq 0, B_{\epsilon}(x) \cap A^c \neq 0\}. $$ In other words, the quantifier on $\epsilon$ should be universal.

To see the equivalent characterization, namely that $\partial A = \mathrm{cl}~A \cap \mathrm{cl}~A^c$, we as usual check the two inclusions.

First inclusion: Let $\epsilon > 0$. Suppose that $x \in \mathrm{cl}~A$ and $x \in \mathrm{cl}~A^c$. Then $B_{\epsilon}(x) \cap A$ and $B_{\epsilon}(x) \cap A^c$ are nonempty, and therefore: $$ \partial A \supset \mathrm{cl}~A ~\cap\mathrm{cl}~A^c. $$

Second inclusion: Conversely, suppose that $\epsilon_n \downarrow 0$, and for some $x$, there are $y_n \in B_{\epsilon_n}(x) \cap A$ and $z_n \in B_{\epsilon_n}(x) \cap A^c$. Then by construction $x \in \mathrm{cl}~A$ and $x \in \mathrm{cl}~A^c$ (in fact $y_n \to x$ and $z_n \to x$). Consequently, $$ \partial A \subset \mathrm{cl}~A ~\cap\mathrm{cl}~A^c. $$

Drew Brady
  • 3,399
1

Yes, $$\partial A = \overline{A} \cap \overline{A^\complement}\tag{0}$$ which is quite clear from the corrected definition (with quantifier)

$$\partial A = \{x \in M\mid \forall r>0: (B_r(x) \cap A \neq \emptyset) \land (B_r(x) \cap A^\complement \neq \emptyset)\,\}\tag{1}$$

using $$\overline{B} = \{x \in M \mid \forall r>0: B_r(x) \cap B \neq \emptyset\}$$

which holds too for all $B \subseteq M$. The $\forall$ quantifier commutes over $\land$.

Also, another way of finding the boundary is seeing it as:

$$\partial A = \overline{A} \setminus \operatorname{int}(A)$$ which is clear from

the previous equation $(0)$ plus the standard fact

$$\operatorname{int}(A)^\complement = \overline{A^\complement}$$

Henno Brandsma
  • 242,131
  • Thank you for your response. With the corrected definition in mind, are the steps I have taken correct? – Taylor Rendon Sep 12 '20 at 23:18
  • 1
    @TaylorRendon not quite, others already pointed out flaws. Your reasoning can be corrected : Given any $A$, $M$ splits into three disjoint subsets $\operatorname{int} A$ (if some ball around $x$ stays in $A$), $\operatorname{int} A^\complement$ (if some ball around $x$ misses $A$ entirely, so this is the same set as $\overline{A}^\complement$, and the $x$ that remain always intersect both $A$ and $A^\complement$ and so are in $\partial A$. Together, $\operatorname{int} A$ and $\partial A$ form a partition of $\overline{A}$. – Henno Brandsma Sep 12 '20 at 23:23