# Prove that $\partial A = \overline{A} \cap \overline{(X-A)}$

Definitions.

• $$\overline{A} = A \cup A'$$, where $$A'$$ is the set of limits point of $$A$$.
• (Boundary point) $$x$$ is a boundary point of $$A$$ if open set $$U$$ containing $$x$$, $$U$$ contain a point in $$A$$ and a point in $$X-A$$.
• $$\partial A = [\text{set of all boundary points of } A]$$.

Problem. Prove that $$\partial A = \overline{A} \cap \overline{(X-A)}$$.

Proof. $$\partial A = \overline{A} \cap \overline{(X-A)}$$

Let $$x \in \overline{A} \cap \overline{(X-A)}$$

$$x\in \overline{A}$$ and $$x \in \overline{X-A}$$

by how we define $$\overline{A}$$

$$x \in (A \cup A')$$ and $$x \in (X-A) \cup (X-A)'$$

$$x \in A$$ or $$x \in A'$$ and $$x \in (X-A)$$ or $$x \in (X-A)'$$

since $$x$$ cannot belong to $$A$$ and $$X-A$$ at the same time

either $$x \in A$$ and $$x \in (X-A)'$$ or

$$x \in A'$$ and $$x \in (X-A)$$

if $$x \in A'$$ and $$x \in (X-A)$$

by definition of limit point

there is an open set $$N$$ containing $$x$$ such that $$N$$ has another point in $$A$$ other than $$x$$.

then we can say, since $$x \in (X-A)$$ and $$N$$ contain another point in $$A$$ (This define boundary point)

hence $$\overline{A} \cap \overline{(X-A)} \subseteq\partial A$$

(i) Is this correct?

(ii) Can anyone help with the converse

if $$x \in \partial A$$

...???

• (i) is correct, however there is a little void in the definition of boundary point: x is a boundary point of A if for any open set U containing x, U contain a point in A and a point in X\A. – Masacroso Oct 12 '18 at 1:54
• It's equivalent, since N contain another point not equal x – Niang Moore Oct 12 '18 at 2:06
• no, you were right, my bad, your second definition is almost right. You only need to add for every open set containing $x$. – Masacroso Oct 12 '18 at 2:12
• okay, thanks. Any hint on the converse? – Niang Moore Oct 12 '18 at 2:17

Note that if $$x\in\partial A$$ then $$x$$ is a limit point of $$A$$ and also a limit point of $$X-A$$, that is, for any chosen open set $$U$$ such that $$x\in U$$ then $$(U-\{x\})\cap A\neq\emptyset$$ by definition of boundary point, so $$x\in A'$$. Consequently $$x\in\overline A$$.

Also we find that $$(U-\{x\})\cap(X-A)\neq\emptyset$$ by the same reason, so $$x\in(X-A)'$$ also. Then $$x\in\overline{(X-A)}$$ too.

Finally note that, by the definition of intersection of sets, we find that

$$x\in \overline A\text{ and }x\in\overline{(X-A)}\implies x\in \overline A\cap\overline{(X-A)}$$

Lemma. For each $$x$$, the followings are equivalent:

1. $$x \in \overline{A}$$.
2. For any open set $$U$$ containint $$x$$, $$U \cap A \neq \varnothing$$.

It is easily proved by consider two cases, $$x \in A$$ and $$x \notin A$$. Now using this,

\begin{align*} x \in \partial A &\quad\Longleftrightarrow\quad \forall U \text{ open s.t } x \in U \ : \quad U\cap A \neq \varnothing \text{ and } U \cap (X-A) \neq \varnothing \\ &\quad\Longleftrightarrow\quad \begin{cases} \forall U \text{ open s.t } x \in U \ : \quad U\cap A \neq \varnothing, \\ \forall U \text{ open s.t } x \in U \ : \quad U\cap (X-A) \neq \varnothing \end{cases} \\ &\quad\Longleftrightarrow\quad x \in \overline{A} \text{ and } x \in \overline{X-A} \\ &\quad\Longleftrightarrow\quad x \in \overline{A} \cap \overline{X-A}. \end{align*}

• thanks, this looks cleaner than mine. What of the converse? – Niang Moore Oct 12 '18 at 2:24
• @NiangMoore, $\Leftrightarrow$ means that both sides are equivalent. So my answer automatically proves both directions. – Sangchul Lee Oct 12 '18 at 2:25

x in closure A and x in closure X-A iff
x an adherance pt of A and x an adherance pt of X-A iff
for all open U nhood x, A $$\cap$$ U and A $$\cap$$ X-A aren't empty.

The use of limit points makes for a grotesque proof.
It also obsecures how simple the proposition is.