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
Best Answer
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 $A\cup \partial A\subset \overline{A}$.