Open and dense set in metric space My professor asked me the next question:
If $A$ is an open subset of a metric space $X$ and $D=A\cup(A^c)^\circ$, then $D$ is dense.
Can you check my proof? I never used that $A$ is open.
Lemma 1. If $U,V$ are subsets of a metric space, then $\overline{U\cup V}=\overline{U}\cup\overline{V}$.
Lemma 2. If $U$ is a subset of a metric space, then $$\overline{U^c}=(U^\circ)^c\mbox{ and }(U^c)^\circ=(\overline{U})^c$$
My proof (of the excercise): Since $D=A\cup (A^c)^\circ$, by part 2 of Lemma 2 we have $D=A\cup(\overline{A})^c$. Then, by Lemma 1, $\overline{D}=\overline{A}\cup\overline{(\overline{A})^c}$. Now, from part 1 of Lemma 2, we have $D=\overline{A}\cup((\overline{A})^\circ)^c...(1)$
Now, since $(\overline{A})^\circ\subseteq\overline{A}$, then $(\overline{A})^c\subseteq((\overline{A})^\circ)^c$, so, taking union with $\overline{A}$ on both sides we have $X=\overline{A}\cup(\overline{A})^c\subseteq\overline{A}\cup((\overline{A})^\circ)^c$. But the right hand side of the last inclusion is, by (1), $\overline{D}$, so $X\subseteq\overline{D}$ and done.
Do you think that my proof is ok?
 A: Your proof is correct. Openness of $A$ is not necessary. However, there is a simpler proof: Let $U$ be any non-empty open set. If $U \cap A=\emptyset$ then $U \subseteq A^{c}$ which implies $U \subseteq (A^{c})^{0}$. This shows that $U$ always intersects $D$. So $D$ is dense.
A: 1... If $B,C$ are subsets of a space $X$ then $\overline {B\cup C}=\overline  B \cup \overline C.$
2... $\partial A=^{def}\,\overline A\cap \overline {A^c}.$ If $A$ is any subset of any space $X$ then  $$A^{\circ}\cup (A^c)^{\circ}\cup \partial A=X.$$
Proof: Suppose $p\in X$ but $p\not\in A^{\circ}\cup (A^c)^{\circ}.$ Every nbhd of $p$ contains a member of $A^c$ ...[ because $p\not\in A^{\circ}$]...so $p\in \overline {A^c}.$ Every nbhd of $p$ contains a member of $A$ ...[because $p\not\in (A^c)^{\circ}$]... so $p\in \overline A.$ Therefore $p\in\overline A\cap \overline {A^c}=\partial A.$
3...Therefore, for any space $X$ and any $A\subseteq X,$ we have
$\overline {A\cup (A^c)^{\circ}}=\overline A \cup \overline {(A^c)^{\circ}}\supseteq$
$\supseteq \overline A \cup  (A^c)^{\circ}=$
$= \overline {A\cup A^{\circ}}  \cup  (A^c)^{\circ}=$
$=\overline  A \cup \overline { A^{\circ}}\cup (A^c)^{\circ}\supseteq$
$\supseteq \overline  A \cup  A^{\circ}\cup (A^c)^{\circ}=$
$\supseteq (\overline  A \cap \overline {A^c})\cup  A^{\circ}\cup (A^c)^{\circ}=$
$=(\partial A)\cup  A^{\circ}\cup(A^c)^{\circ}=X.$
