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?
Please help me.
Best Answer
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 $D$.