Example of discrete subspace on a plane which has a closure of continuum?

Question

I thought that I could take all points with rational coordinates, but this space is not discrete

Answer 1

For each $n\in\mathbb{N}$, let $$D_n=\{(k/2^n, 1/n)\in \mathbb{R}^2: k\in \mathbb{Z}\}.$$ Let $D=\cup_{n=1}^\infty D_n$. Then $D$ is discrete. To see this, note that if $x\in D_n$, then no other point of $D$ is closer than $\min\{2^{-n}, \frac{1}{n}-\frac{1}{n+1}\}$ to $x$.

But the closure of $D$ contains the entire $x$-axis.

Answer 2

In this note a classic fact is shown that in a metric space the boundary of an open set is the closure of a discrete set. So e.g. the circle in the plane, being the boundary of the open unit disk, is such a set.