Is the Cartesian square of the set of irrational numbers path connected?

Question

Let $X=\mathbb{R}\setminus \mathbb{Q}$. Is $X\times X$ path-connected?

I don't know where to start I think we need some number theory knowledge.

Answer 1

This space is not the same as $\mathbb R^2$ \ $\mathbb Q^2$.

$\mathbb R^2$ \ $\mathbb Q^2$ is path connected and if you search this site you will find a couple of nice proofs of that fact.

In this space, we remove the rational numbers before taking the product. Consider the partition of $X\times X$ into two half planes: $\{(x,y): x\lt 0\}$ and $\{(x,y): x\gt 0\}$. Each of these sets is open and since they partition the space, they are each the complement of an open set so they are also closed.

Therefore, $X\times X$ is disconnected and so it cannot be path connected.