0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $\operatorname{cl}A$. $\endgroup$ – Brian M. Scott Nov 17 '16 at 20:07
1
$\begingroup$

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 $\overline{A}$.

$\endgroup$
0
$\begingroup$

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, $D\subseteq A\implies\overline D^M\subseteq \overline A^M$.

$\endgroup$
0
$\begingroup$

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 $D$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.