How to prove boundary of a subset is closed in $X$? Suppose $A\subseteq X$. Prove that the boundary $\partial A$ of $A$ is closed in $X$.
My knowledge:


*

*$A^{\circ}$ is the interior 

*$A^{\circ}\subseteq A \subseteq \overline{A}\subseteq X$


My proof was as follows:
To show $\partial A = \overline{A} \setminus A^{\circ}$ is closed, we have to show that the complement $( \partial A) ^C = X\setminus{}\partial A =X \setminus (\overline{A} \setminus A^{\circ})$ is open in $X$. This is the set $A^{\circ}\cup X \setminus(\overline{A})$
Then I claim that $A^{\circ}$ is open by definion ($a\in A^{\circ} \implies  \exists \epsilon>0: B_\epsilon(a)\subseteq A$. As this is true for all $a$, by definition of open sets, $A^{\circ}$ is open.
My next claim is that $X \setminus \overline{A}$ is open. This is true because the complement is $\overline{A}$ is closed in $X$, hence $X \setminus \overline{A}$ is open in $X$.
My concluding claims are: We have a union of two open sets in $X$, By a proposition in my textbook, this set is open in $X$. Therefore the complement of that set is closed, which is we had to show. 
What about this ?

Your proof looks correct to me. The interior is indeed open by definition (and the closure is closed by definition). And the fact that a union of open sets is open is one of the axioms for a topology. You may wish to note that $\partial A$ is often defined as $\overline{A} \cap \overline{X - A}$. Since finite intersections of closed sets are closed, it follows that $\partial A$ is closed in $X$.
