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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?

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  • $\begingroup$ Show that $A^c$ is open. $\endgroup$ – Surb Feb 7 at 17:17
  • $\begingroup$ 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? $\endgroup$ – Unknown x Feb 7 at 17:20
  • $\begingroup$ 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 $Bd(A)\cap A^c=\emptyset$. $\endgroup$ – Surb Feb 7 at 17:22
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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, $x\in A$.

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  • $\begingroup$ Sir, You are proving by contradiction. Where do we get a contradiction in our assumption? $\endgroup$ – Unknown x Feb 8 at 1:05
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    $\begingroup$ I assumed that $x\notin A$ and then I proved that $x\in A$. $\endgroup$ – José Carlos Santos Feb 8 at 7:13
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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.

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