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?

Solution 1:

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.