# Show that Closure of a set is equal to the union of the set and its boundary

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

• I would like to comment, but please tell me first: which definition of boundary are you using? – José Carlos Santos Nov 21 '18 at 11:58
• @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 $B_r (a) \cap A^c \neq \emptyset$ – TUC Nov 21 '18 at 12:02

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