Separability of $A \subseteq X$ Let $(X, \tau)$ be a separable topological space. Let $A \subseteq X$. There exists a subset $D \subseteq A$ such that $D$ is dense over $A$ and $D$ is countable?
$X$ is separable if it contains a countable dense subset.
I supposed $D$ dense, and I would like to prove $D$ is not countable.
Thank you.
 A: If I understand correctly, you’re asking whether a subset of a separable space is necessarily separable; the answer is no. Here’s one example.
Let $Y$ be the Sorgenfrey line, also known as the reals with the lower limit topology: the underlying set is $\Bbb R$, and the half-open intervals of the form $[a,b)$ are a base for the topology. $Y$ is separable, because $\Bbb Q$ is a countable dense subset of $Y$. Let $X=Y\times Y$, the Sorgenfrey plane, with the product topology. $X$ is also separable, because $\Bbb Q\times\Bbb Q$ is a countable dense subset of $X$. Now let $A=\{\langle x,-x\rangle:x\in\Bbb R\}$; $A$ is clearly uncountable, and it is also a discrete subset of $X$, since
for each $x\in\Bbb R$, the set $[x,x+1)\times[-x,-x+1)$ is an open nbhd of $\langle x,-x\rangle$ that contains no other point of $A$. (You can easily check that it is also a closed subset.) Clearly $A$ cannot be separable.
Another example can be constructed from $\Bbb R$ in a different way. We construct the new topology as follows. Points of $\Bbb Q$ are isolated. If $x\in\Bbb R\setminus\Bbb Q$, the sets of the form $\{x\}\cup\big((a,b)\cap\Bbb Q\big)$, where $a<x<b$, form an open nbhd base at $x$. In other words, each rational is an isolated point, and basic nbhds of an irrational consist of that irrational and all of the rationals in an open interval around it. Call this space $X$. The rationals are clearly dense in $X$, so $X$ is separable. However, the irrationals are an uncountable closed discrete subset of $X$, so the set of irrationals is not separable in $X$.
A space in which every subspace is separable is said to be hereditarily separable; all second countable spaces, and hence all separable metric spaces, are hereditarily separable.
A: Yes, since X is separable, there exist a countable set, Y, that is dense in X.
What can you say about $A\cap Y$?
