# If $\text{Boundary}(A)\subseteq A.$, then $\overline A=A$?

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?

• Show that $A^c$ is open. – Surb Feb 7 at 17:17
• How do I prove? Let $x\in A^C$. $x$ not a boundary point. So There exists a open ball which is disjoint from either $A$ or $A^C$. Right? – Unknown x Feb 7 at 17:20
• Suppose $A^c$ not open, i.e. there is $x\in A^c$ s.t. for all $U\ni x$ open $A\cap U\neq \emptyset$. In particular, $x\in Bd(A)$. Contradiction since $Bd(A)\cap A^c=\emptyset$. – Surb Feb 7 at 17:22

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, $$x\in A$$.
• I assumed that $x\notin A$ and then I proved that $x\in A$. – José Carlos Santos Feb 8 at 7:13
If $$\operatorname{Bd}(A) \subseteq A$$ then $$A$$ is closed:
For all $$A$$ we have: If $$x \in X$$ then either $$x \in \operatorname{Int}(A)$$ or $$x \in \operatorname{Bd}(A)$$ or $$x \notin \overline{A}$$ and these are mutually exclusive. This is a classic fact and follows straight from the definitions.
So if $$x \in \overline{A}$$ one of the first two must hold, as the last has been ruled out, and in both cases we have that $$x \in A$$ as always $$\operatorname{Int}(A) \subseteq A$$ and by assumption $$\operatorname{Bd}(A) \subseteq A$$. So $$A$$ is closed.