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I'm trying to show that a closure of a set is equal to the union of the set and its boundary.

Let $A$ be a subset of a metric space $(X, d)$. Then show that $\overline A = A \cup \partial A$

Where $\overline A$ is the closure of $A$ and $\partial A$ is the boundary of $A$ and $A^o$ is the interior of $A$.

My attempt:

Let $a \in \overline A$. Then either $a \in \overline A$ \ $A^o$ or $a \in A^o$.

$a \in A^o$ part is trivial so I omit this part.

Suppose $a \in \overline A$ \ $A^o$

Since $\overline A$ \ $A^o$ is the smallest closed set containing $A$ and all the interior points of $A$ removed, only the boundary points of $A$ are left. So $a \in \overline A$ \ $A^o$ = $\partial A$ $\subset A \cup \partial A$.

Hence $\overline A \subset A \cup \partial A$

Now suppose $a \in A \cup \partial A$. Again $a \in A$ part is trivial so I omit this part. So we consider $a \in \partial A$. Then $a \in \partial A = \overline A$ \ $A^o$. So $a \in A \cup (\overline A$ \ $A^o)$ = $\overline A$.

So $A \cup \partial A \subset \overline A$

Therefore, $\overline A = A \cup \partial A$

Does this proof make sense?

Any comment / correction is appreciated

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  • $\begingroup$ I would like to comment, but please tell me first: which definition of boundary are you using? $\endgroup$ – José Carlos Santos Nov 21 '18 at 11:58
  • $\begingroup$ @JoséCarlosSantos Def : A point $a \in X$ is a boundary point of $A$ if $\forall r > 0$, $B_r (a) \cap A \neq \emptyset$ and $B_r (a) \cap A^c \neq \emptyset$ $\endgroup$ – TUC Nov 21 '18 at 12:02
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Let $x\in \overline{A}$. Let $x\notin A$. Thus $x\in X\setminus A\implies x\in \overline{X\setminus A}$. Thus $x\in \overline{A}\cap \overline{X\setminus A}=\partial A$.

Again by definition $A\subset \overline{A}$ and $\partial A\subset \overline{A}$. Hence $A\cup \partial A\subset \overline{A}$.

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Your proof is correct, except that you asserted that $\overline A\setminus\mathring A$ is the smallest closed set containing $A$. No; that would be $\overline A$.

Besides, since you use twice the fact that $\partial A=\overline A\setminus\mathring A$, I suggest that you prove this as a lemma first.

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