proposition about boundary points of subset of topological space Let $(X,\mathcal T)$ be a topological space and let $A$ be a subset of $X$. Then:


*

*$A$ is closed if and only if $\partial A$ is a subset of $A$

*$A$ is open if and only if $A\cap\partial A=\emptyset$ 

*$\partial A=\emptyset$  if $A$ is both open and closed


Here $\partial A$ denotes the boundary of $A$.
 A: Define $\partial A := \overline{A} \cap \overline{X \setminus A}$. 
(i) $A$ closed $\Rightarrow \partial A \subseteq A$:  we have $A =   \overline{A}$ so
$$ \partial A = \overline{A} \cap \overline{X \setminus A} = A \cap \overline{X \setminus A} \subseteq A. $$
$A$ closed $\Leftarrow \partial A \subseteq A$: We will show that $\overline{A} \subseteq A$, so that $\overline{A} = A$, giving equality and hence closedness. Suppose for a contradiction that we can find an $a \in \overline{A} \setminus A$. As $a \notin A$, $a \in X\setminus A \subseteq \overline{X \setminus A}.$ On the other hand we know that $a \in \overline{A}$. So $a \in \partial A \subseteq A$, a contradiction. 
(ii) Suppose $A \cap \partial A = \emptyset$. We want to show that $A$ is open. So we will show that $A \subseteq A^{\circ}$. If this isn't true, we can pick $x \in A \setminus A^{\circ}$. Note that $A^{\circ} = X \setminus \overline{X \setminus A}$. So we see that $x \in \overline{X \setminus A}$. But $ x \in \subseteq A \subseteq \overline{A}$. So $x \in \partial A$. Hence $x \in A \cap \partial A = \emptyset$, a contradiction. 
Suppose $A$ is open. Then  $$A \cap \partial A = A \cap \overline{A} \cap \overline{X \setminus A} = A \cap \overline{A} \cap (X \setminus A^{\circ}) = A^{\circ} \cap (X \setminus A^{\circ}) = \emptyset.$$
(iii) If $A$ is both open and closed, then:
Because $A$ is open, then by part (ii) we have $A \cap \partial A = \emptyset$.
But as $A$ is closed, we know that $X \setminus A$ is open. So again by part (ii) $(X \setminus A) \cap \partial A = \emptyset$.
But then $$\partial A = ((X \setminus A) \cap \partial A ) \cup (A \cap \partial A) = \emptyset \cup \emptyset = \emptyset.$$
