• $\overline{A} = A \cup A'$, 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 $x \in \partial A$


  • $\begingroup$ (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. $\endgroup$ – Masacroso Oct 12 '18 at 1:54
  • $\begingroup$ It's equivalent, since N contain another point not equal x $\endgroup$ – Niang Moore Oct 12 '18 at 2:06
  • $\begingroup$ no, you were right, my bad, your second definition is almost right. You only need to add for every open set containing $x$. $\endgroup$ – Masacroso Oct 12 '18 at 2:12
  • $\begingroup$ okay, thanks. Any hint on the converse? $\endgroup$ – 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

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


I understand your overall logic, but your written argument is not easy to read. For a more concise answer, notice that

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*}

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


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