TITLE: a dense set in plane
QUESTION [6 upvotes]: Is there a dense set in $\Bbb{R^2}$ that every vertical line or horizontal line intersect in finite points. I think that we can consider $\Bbb{Q} ×\Bbb{Q}$ but every vertical line or horizontal line don't intersect in finite points.

REPLY [5 votes]: Yes, we can find an example by shifting each point in $\mathbb{Q}^2$ a little bit. 
In particular, enumerate the points of $\mathbb{Q}^2$ as $(x_1,y_1)$, $(x_2,y_2)$, ..., employing the fact that $\mathbb{Q}^2$ is countable.
Now I claim that the set
$$
S = \left\{\left(x_n + \frac{\pi}{n}, y_n + \frac{\pi}{n}\right) : n \in \mathbb{N} \right\}
$$
works. 
Each horizontal or vertical line intersects at most one point of $S$. This follows from the fact that $\pi$ is irrational.
Furthermore, $S$ lies dense in $\mathbb{R}^2$ because $\mathbb{Q}^2$ lies dense in $\mathbb{R}$: any open ball in $ \mathbb{R}^2$ contains infinitely many points from $\mathbb{Q}^2$ and therefore also points from $S$ (for large enough $n$).
In fact, we can even find a dense set in the plane that contains no three collinear points. Such a set can be constructed by adding the points one by one; see Timothy Gowers's blog for a nice explanation.