Proof that $\overline A= Int(A) \cup \partial (A)$

general-topologymetric-spacesreal-analysis

I want to prove that $\overline A= Int(A) \cup \partial (A)$, My attempt is as follows.

For the first contention. Be $x\in \overline A$, then for all $r>0$ $B(x; r) \cap A \neq \emptyset$ and let $x \in Ext(A)=Int(M-A)$ i.e. exist $r>0$ such that $B(x;r) \subseteq A^{c}$ it follows that $A^{c}$ is open and $A$ closed.

How $B(x;r) \subseteq A^{c}$ and $B(x; r) \cap A \neq \emptyset$ then $A^{c} \cap A \neq \emptyset$ which is not true, so $x$ is not an exterior point then subtract that $x$ is an interior point or point on the boundary of $A$.

Now, I'm having a little trouble seeing the other implication, any suggestions?

By the way, I am not using sequences in the process of the proof, only the definitions of closure, interior and boundary using open balls.

Best Answer

When working with terms like closure/boundary/interior one should always give the definition one is working with as there are so many different and in some cases not even necessarily equivalent definitions out there. Let $A\subset\mathbb{R^n}$. I'll assume you use the following definitions as you mentioned open balls.

  • We define the closure of $A$, denoted $\overline{A}$, by $\overline{A}:=\left\lbrace x\in\mathbb{R}^n\ \big|\ B_r(x)\cap A\neq\emptyset\text{ for all }r>0\right\rbrace$
  • We define the interior of $A$, denoted $\overset{\circ}{A}$, by $\overset{\circ}{A}:=\left\lbrace x\in\mathbb{R}^n\ \big|\ \exists r>0:B_r(x)\subset A\right\rbrace$
  • We define the boundary of $A$, denoted $\partial A$, by $\partial A:=\left\lbrace x\in\mathbb{R}^n\ \big|\ \forall r>0:\left(B_r(x)\cap A\neq\emptyset\text{ and }B_r(x)\cap A^c\neq\emptyset\right)\right\rbrace$

We want to prove that $\overline{A}=\overset{\circ}{A}\cup\partial A$.
"$\subseteq$"
Let $x\in\overline{A}$, then by definition $B_r(x)\cap A\neq\emptyset$ for all $r>0$. If $B_r(x)\cap A^c\neq\emptyset$ for all $r>0$, then $x\in\partial A$ by definition. If $B_r(x)\cap A^c\neq\emptyset$ is not true for all $r>0$, then there is some $r_1>0$ with $B_{r_1}(x)\cap A^c=\emptyset$. But this means that $B_{r_1}(x)\subset A$ and thus $x\in\overset{\circ}{A}$.
"$\supseteq$"
Let $x\in\overset{\circ}{A}\cup\partial A$. If $x\in\overset{\circ}{A}$, then there exists some $r>0$ with $B_r(x)\subset A$, which immediately implies that $B_r(x)\cap A\neq\emptyset$ for all $r>0$ and thus $x\in\overline{A}$.
If $x\in\partial A$, then for all $r>0$ we have $B_r(x)\cap A\neq\emptyset$ and $B_r(x)\cap A^c\neq\emptyset$, which is a stronger statement than $B_r(x)\cap A\neq\emptyset$ for all $r>0$, which is equivalent to $x\in\overline{A}$. Thus we have proven the other inclusion as well.

Related Question