Existence of a special set in $\mathbb{R}^2$

Question

Let $X$ be a non–empty subset of $\mathbb{R}^2$ satisfying the following properties:

1. Each point of $X$ is a limit point of $X$. ( Under the distance topology in $ \mathbb{R}^2$)

2. No rational point belongs to $X$. (We call a rational point to be a point whose two coordinates in $\mathbb{R}^2$ are both rational numbers)

3. $X$ is closed.

Does such a set $X$ exist?

If such a set $X$ exists, then any rational point in $\mathbb{R}^2$ is not a limit point of $X$. But I don't know how to go on. Any hint would be of great welcome!

Edit: Thanks for everybody's help! Consider another question: What if we change $\mathbb{R}^2$ in the problem into $\mathbb{R}$?

Answer 1

Try the graph $xy = \sqrt{2}$. Clearly it can contain no rational points, since $\sqrt{2}$ is irrational. The neighbourhood of any point contains another point, so each point is a limit point and it is clearly closed (as the inverse image of $\{\sqrt{2}\}$ under the continuous map $(x,y) \mapsto xy$.

For a simpler example, the line $x+y = \sqrt{2}$ for $x,y \geq 0$ also works for the same reasons.

Answer 2

Maybe you could try this:

$$X=\mathbb{R}\times\{\pi\}$$