Radially open set topology is separable? Consider the topology of radially open sets on $\mathbb{R^2}$ : a set $S$ is radially open iff, for every $x\in S$ and every line $L\subset\mathbb{R^2}$ that contains $x$, $S$ contains an open segment of $L$ centered at $x$. I'm trying to study the countability properties of this space. I think I can see it is not first countable(I'm still working at it) and it's easy to see that $\mathbb{R^2}$ with this topology isn't Lindelöf, as the circles are closed subspaces not Lindelöf (they are infinite and have discrete topology). As it is not Lindelöf, cannot be second countable.
However, about separability, I don't know how to proceed. If the world was fair, $\mathbb{Q}^2$ would still be a dense subset in thiss topology, but I haven't been able to prove it. An approach I tried was proving that every radially open set contains a usual open set, although I have not been able to prove it or refute it. On the other hand, trying to prove separability directly I find the problem, of course, with points with two irrational components, I cannot see which of the lines that pass through them will contain points of $\mathbb{Q}^2$ sufficiently near the point. Should I try looking for another dense subset ? It doesn't seem to be one better. Can you help me ?
EDIT: In fact, I don't know either how to prove it is not first countable
 A: Hint for separability: If $U$ is an open neighborhood of $x$, then $U$ not only needs to contain an open segment $S$ of any line through $x$, but also must contain an open segment of any line through any point of $S$.  Can you use that to show $U$ contains a point of $\mathbb{Q}^2$?
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 By taking the horizontal line through $x$, we see that $U$ must contain a point $y$ whose first coordinate is in $\mathbb{Q}$.  Taking the vertical line through $y$, we see that $U$ must contain a point of $\mathbb{Q}^2$.

Hint for first-countability: Since $\mathbb{Q}^2$ is dense, it suffices to show that not every point is a limit of a sequence in $\mathbb{Q}^2$.  Can you show that the limit of a sequence in $\mathbb{Q}^2$ must be on a line between two points of $\mathbb{Q}^2$?
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 Since there are only countably many lines between two points of $\mathbb{Q}^2$, there is some point $x\in\mathbb{R}^2$ that is not on any of those lines.  Suppose $(x_n)$ is a sequence in $\mathbb{Q}^2$ converging to $x$.  If infinitely many of the $x_n$ were on some line $L$, then $x$ would have to also be on $L$ (since $L$ is radially closed), which is impossible by assumption.  Thus, every line contains only finitely many of the $x_n$.  But now this implies that $\mathbb{R}^2\setminus\{x_n:n\in\mathbb{N}\}$ is radially open, so it is an open neighborhood of $x$ which contains no $x_n$, contradicting the assumption that $x_n$ converges to $x$.

