Prove that the complement of $\mathbb{Q} \times \mathbb{Q}$. in the plane $\mathbb{R}^2$ is connected. Prove that the complement of $\mathbb{Q} \times \mathbb{Q}$. in the plane $\mathbb{R}^2$ is connected. 

I have no idea how can I do that. If $\mathbb{R} \setminus\mathbb{Q}$ is connected then the proof is easy which is not true. somebody help me please.thanks for your time.
 A: You will have to use the fact that $\mathbb{Q} \times \mathbb{Q}$ is countable. And $\mathbb{R}^{2}\setminus A$ where $A$ is countable is always path connected$\implies$ connected.
A: You can easily show that $S=\Bbb R\times\Bbb R\setminus\Bbb Q\times\Bbb Q$ is pathwise connected, which is stronger than connectedness. Fix an irrational number $\alpha$; it suffices one connect any point of $S$ by a path in $S$ to $(\alpha,\alpha)$. If $(x,y)\in S$ with $x$ irrational, make a straight line path from $(x,y)$ to $(x,\alpha)$ and then from there straight to $(\alpha,\alpha)$. If $x$ is rational then $y$ must be irrational, and a similar path from $(x,y)$ to $(\alpha,y)$ and then to $(\alpha,\alpha)$ works.
A: This actually seems to be true for any subset $S\subsetneq \mathbb{R}$.
Hint: Can you connect any two points $(x_1,y_1)\in\mathbb{R}^2\setminus (S\times S)$ and  $(x_2,y_2)\in\mathbb{R}^2\setminus (S\times S)$ by a path?
A: Let $x,y \in \mathbb{R}^2 \backslash \mathbb{Q}^2$. There are uncountably many disjoint paths between $x$ and $y$ in $\mathbb{R}^2$ (you can exhibit such a family). Therefore, you find path connectedness by cardinality.
A: For any $(x,y) \in X = \mathbb{R}^2 \setminus \mathbb{Q}^2$, at least one of $x, y \notin \mathbb{Q}$. We can always connect it to $(\sqrt{2},\sqrt{2}) \in X$ by a polygon path in $X$:
$$\begin{cases}
(x,y) \to (x, \sqrt{2} ) \to (\sqrt{2}, \sqrt{2}), &\text{ for }x \notin \mathbb{Q}\\
(x,y) \to (\sqrt{2}, y ) \to (\sqrt{2}, \sqrt{2}), &\text{ for }x \in \mathbb{Q}, y \notin \mathbb{Q} 
\end{cases}$$
Any two points in $X$ can be connected by a polygonal path by joining their paths to $(\sqrt{2},\sqrt{2})$ and hence $X$ is path connected.
Update
In another answer, there is an interesting statement:

And $\mathbb{R}^2\setminus A$ where $A$ is countable is always path connected.

Let me give a proof of this statement and hence an alternative proof for this question. 
For any $\vec{x}_1 \in \mathbb{R}^2 \setminus A$, consider following collection of unit vectors:
$$X_1 = \left\{ \pm \frac{\vec{y} - \vec{x}_1}{|\vec{y} - \vec{x}_1|} : \vec{y} \in A\right\} \subset S^{1}$$
Since $A$ is countable, so does $X_1$. This means $S^{1} \setminus X_1 \ne \emptyset$. In fact, $S^{1} \setminus X_1$ is uncountable. Pick any unit vector $\vec{n}_1$ from $S^{1} \setminus X_1$ and construct a line $l_1$ passing through $\vec{x}_1$ in the direction of $\vec{n}_1$:
$$l_1 = \left\{ \vec{x}_1 + t \vec{n}_1 : t \in \mathbb{R} \right\}$$
By construction, it is clear $\vec{x}_1 \in l_1 \subset \mathbb{R}^2 \setminus A$. 
For another $\vec{x}_2 \in \mathbb{R}^2 \setminus A$, a similar argument allow us to construct another set of unit vectors $X_2$, unit vectors $\vec{n}_2$ and line $l_2$ such that $\vec{x}_2 \in l_2 \subset \mathbb{R}^2 \setminus A$. 
Furthermore, since $S^1 \setminus ( X_1 \cup X_2 )$ is infinite, we can choose a $\vec{n}_2$ not in the direction of $\pm \vec{n}_1$. 
Under this restriction, $l_1 \not\parallel l_2$ and intersect at some point $\vec{y} \in \mathbb{R}^2 \setminus A$. The polygon path $\vec{x}_1 \to \vec{y} \to \vec{x}_2$ lies completely outside $A$ and hence $\mathbb{R}^2 \setminus A$ is path connected.
Update 2
It turns out this question has been asked and answered before. 
A much simpler argument for the statement can be found at JDH's answer there. The basic idea is what TonyK given in the comment below and what is in Seirios' answer.
