# Proof that $\overline{A}=A\cup \partial A$

Let $(X,d)$ be a metric space and $A\subset X$. Prove that $\overline{A}=A\cup \partial A$ (that is, the closure of $A$ is the union of $A$ with its boundary).

Let $x\in \overline{A}$, then $x\in A$ or $x\not\in A$. Suppose $x\not\in A$, then $\forall\varepsilon >0, B_\varepsilon(x)\cap X\setminus A\ne \emptyset$. Since $\overline{A}$ is closed, there is a sequence $(x_k)_k\subset A$ converging to $x$. Since $(x_k)_k\subset A$, for every $x_k$ there exists $\varepsilon_{x_k}$ such that $B_{\varepsilon_{x_k}}(x_k)\subset A$. Thus $\forall \varepsilon >0$, $B_\varepsilon(x)\cap A\ne \emptyset$. Hence, $x\in \partial A$. Thus $\overline{A}\subset A\cup \partial A$.

For the other direction, suppose $x\in A\cup \partial A$, then $x\in A$ or $x\in \partial A$. If $x\in A$ then $x\in\overline{A}$. If $x\in \partial A$ then $\forall\varepsilon > 0$, $B_\varepsilon(x)\cap A\ne \emptyset \ne B_\varepsilon(x)\cap X\setminus A$. Since $B_\varepsilon(x)\cap A\ne \emptyset$, there are infinitely many points in $B_\varepsilon(x)$ of some sequence contained in $A$. Let $\varepsilon'<\varepsilon$, then there are infinitely many points in $B_\varepsilon'(x)$ of some sequence contained in $A$. Proceeding by induction and taking $\varepsilon$ smaller and smaller, we see that for every $\varepsilon > 0$ there are infinitely many terms in $B_\varepsilon(x)$ of some sequence contained in $A$. Inductively, this means that there is a sequence contained in $A$ which converges to $x$. Thus, $x\in\overline{A}$.

Please let me know if you find my proof correct. Maybe it's a bit over-complicated, but it would be interesting to know if it's OK.

• What is the definition of $\partial$ that you are using? – Oiler Nov 2 '17 at 21:43
• $\partial A$ is the boundary of $A$. – sequence Nov 2 '17 at 22:12
• WLOG, A is open. – Jacob Wakem Aug 19 '18 at 18:07

Note that $\partial A = \overline{A} \setminus A^\circ = \overline{A} \cap (A^\circ)^c$. Then $$A \cup \partial A = A \cup \left(\overline{A} \cap (A^\circ)^c\right) = (A \cup\overline{A}) \cap (A \cup (A^\circ)^c).$$ Since $A \subseteq \overline{A}$, $A \cup \overline{A} = \overline{A}$. Further, $\overline{A} \subseteq A \cup (A^\circ)^c$ so the result follows.
Let $A^c$ be the complement of A. In any topological space we have $\partial A\subset \overline A$. Therefore $$\overline A= (\bar A\cap A)\cup (\bar A\cap A^c)=$$ $$=A\cup (\bar A\cap A^c)\subset$$ $$\subset A\cup (\bar A\cap \overline {A^c})=$$ $$=A\cup \delta A\subset$$ $$\subset A\cup \bar A=\overline A.$$