TITLE: Can a dense set contain isolated points?
QUESTION [6 upvotes]: I was interested in this question, can a dense set contain an isolated point, because I was reading into the lexicographic order topology on the unit square. 
I read in here that:

$S$ is not separable, since the set of all points of the form $(x,\frac{1}{2})$ is discrete but is uncountable. 

I do understand why $T=\{ (x,\frac{1}{2} ) \,|\,\,\, 0\leq x \leq 1 \}$ is discrete, but I don't see how $T$ being uncountable assists us in showing $S$ is not separable? I thought maybe this is because a dense set cannot contain any isolated points? But I'm not sure on that. If anyone can clarify here, it'll be great!

REPLY [4 votes]: Even if $X$ has no isolated points, it is possible that a dense set $S$ has isolated points. For example if $X$ is indiscrete, then any one-point set is dense and this point is isolated. Or if $X=A\times B$ where $A$ is discrete and $B$ is indiscrete (but $|B|>1$), the the graph of any function $A\to B$  is dense in $X$ and is a discrete space.