Vrftsjtryk

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

Let $xin overline{A}$. Let $xnotin A$. Thus $xin Xsetminus Aimplies xin overline{Xsetminus A}$. Thus $xin overline{A}cap overline{Xsetminus A}=partial A$.

Again by definition $Asubset overline{A}$ and $partial Asubset overline{A}$. Hence $Acup partial Asubset overline{A}$.

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

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