Definition 1:- Let $(X,d)$ be a metric space. $A$ be a subset of $X$. Then $\text{Boundary}(A)=\{x\in X:$open ball centered at $x$ intersects both $A$ and $A^c\}$
Definition 2:- $\overline A=A\cup A'$, $A'$ is the set of all limit points of $A$.
Using these two definitions
My aim is to prove
If $A$ is closed iff $A$ contains its boundary.
Let $A$ is closed $\implies \overline {A}=A\implies \text{Boundary}(A)=\overline A \cap \overline {X\setminus A}=A \cap \overline {X\setminus A}\subseteq A.$
Conversaly $A$ contains its boundary. That is $\text{Boundary}(A)\subseteq A.$ Let $x\in \overline A$ then we need to prove that $x\in A$. $x\in \overline A \implies $ every open set contains $x$ intersects $A$. How do I complete the proof?
Best Answer
Take $x\in\overline A$. Could we not have $x\in A$? In that case, $x\in A^\complement$ and therefore $x\in\overline{A^\complement}$. But $x\in\overline A$ together with $x\in\overline{A^\complement}$ means that $x\in\operatorname{Bd}A$. And therefore $x\in A$, since $A\supset\operatorname{Bd}A$. A contradiction was reached, which follows from assuming that $x\notin A$. So, $x\in A$.