TITLE: Is $\mathbb{Q} \times \mathbb{Q} \cup \mathbb{I} \times \mathbb{I}$ disconnected?
QUESTION [10 upvotes]: Let $\mathbb{I}$ be the set of all irrational real numbers and $\mathbb{Q}$ be the set of all rational numbers as usual. As a subspace of the Euclidean plane $\mathbb{R}^2$, is the set  $\mathbb{Q} \times \mathbb{Q} \cup \mathbb{I} \times \mathbb{I}$ disconnected?

The story is the following.
I was making a typical undergraduate level topology problem which concerns connectedness. The question which I concerned is the following.
As a subspace of the Euclidean plane $\mathbb{R}^2$, is the set $\mathbb{Q} \times \mathbb{I} \cup \mathbb{I} \times \mathbb{Q}$ connected? What about the set  $\mathbb{Q} \times \mathbb{Q} \cup \mathbb{I} \times \mathbb{I}$ ?
The former is relatively well known. For second, the solution which I had in mind turned out to be wrong. I guess lines of the form $y=qx$ where $q$ is some positive rationals will connect the set which is apparently wrong. Now I have no solution.
Please give some help to resolve this problem.

REPLY [2 votes]: Let $S = \mathbb{Q}^2 \cup \mathbb{I}^2$. For each nonzero rationals $q$, let $L_q$ be the line through the origin with slope $q$. Then $L_q \subseteq S$. Let $T=\bigcup_{q \in \mathbb{Q}\setminus \left\{0 \right\}} L_q$. Then $T$ is connected in $S$. Moreover $\overline{T}$ relative to $S$ itself is equal to $S$. Therefore $S$ is connected.