Show that closure of a separable space is separable.

Let $A$ be a separable subset of a metric space $(M,d)$.

Show that $\overline {A}$ is separable.

Since $A$ is separable ,$A$ has a countable dense subset say $D$.But $\overline D=A$.

I can't make $\overline D=\overline A$. What should I do?

• The definition of dense subset that you’re using is correct, but it’s often not the most useful one. It’s equivalent to the following definition, which is frequently more useful. A set $D\subseteq A$ is dense in $A$ if and only if every open set in $M$ that hits $A$ contains a point of $D$. Once you prove that equivalence, you can observe that if an open set $U$ hits $\operatorname{cl}A$, then $U\cap A\ne\varnothing$ (by the definition of closure), so $U\cap D\ne\varnothing$. Thus, $D$ is dense in $\operatorname{cl}A$. – Brian M. Scott Nov 17 '16 at 20:07

For the topology induced by $(M,d)$ on $A$, you have $\overline{D}=A$.

But for the topology $(M,d)$, you have $A \subset \overline{D}$.

So, $A \subset \overline{D}$. And $\overline{D}$ is closed. So $\overline{A} \subset \overline{D}$ because $\overline{A}$ is the smallest closed set containing $A$.

And $D \subset \overline{A}$, so $D$ is dense in $\overline{A}$.

There is a theorem of general topology that says that, for $Z\subseteq Y\subseteq X$, it holds $\overline Z^Y=Y\cap \overline Z^X$. So, you know that $A=\overline D^A=A\cap \overline D^M$

So $\overline D^M=\overline{\color{blue}{\overline D^M}}^M\color{blue}\supseteq \overline{\color{blue}A}^M$. On the other hand, $D\subseteq A\implies\overline D^M\subseteq \overline A^M$.

Since $A$ is separable so $A$ has a countable dense subset say $D$.

Let $U(\neq \emptyset)$ be a open set in $\bar A$.Let $x\in U$.

Since $x\in \bar A$ so any open set containing $x$ must intersect $A$.So $U\cap A\neq \emptyset$.

Now $U\cap A$ is open in $A$ and $D$ being dense in $A$ so $U\cap D=U\cap (A\cap D)=(U\cap A)\cap D\neq \emptyset$ .

So $\bar A$ has a countable dense subset $D$.