A is closed in X$\Longleftrightarrow$ $\partial(A)\subset A$

Question

I would like to prove that

let A a subset of a metric space

A is closed in X $\Longleftarrow$ $\partial(A)\subset A$

Please I need a suggestion. I tried to prove that $A^´\subset A$...

Answer 1

$\partial(A)=\bar A-\text{int}(A)$, we deduce that $\bar A=\partial(A)\cup \text{int}(A)$ and $\bar A=A$ implies that $\partial(A)\cup A=A$, we deduce that $\partial(A)\subset A$. On the other hand $\partial(A)\subset A$ implies that $\bar A=\partial(A)\cup \text{int}(A)\subset A$ since $\text{int}(A)\subset A$, we deduce that $\bar A=A$ since by definition $A \subset \bar A$.

Answer 2

Hint: $\bar{A}=int(A)\cup\partial(A)$.

Answer 3

Let $\overline{A}$ denote the closure of $A$.

If $A$ is closed, $A=\overline{A}$. Hence, $\partial{A}=\overline{A}\cap X\backslash intA=A\cap X\backslash intA\subseteq A$.

Note, $\overline{A}=A\cup \partial{A}$. Thus if $\partial{A}\subseteq A$, then $\overline{A}\subseteq A$.