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 ?
Best Answer
From your definition, directly, $$ \partial A=\overline{A}\setminus \mathring{A}=\overline{A}\cap (X\setminus \mathring{A}) $$ is the intersection of two closed sets. Hence it is closed.
No need to prove that the complement is open, it just makes it longer and more complicated.
Also, keep in mind that a set $S$ is open in $X$ if and only if its complement $X\setminus S$ is closed in $X$. This should be pavlovian.