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.
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.
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.
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.
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?
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.
