Show that $S=\mathbb R^2\setminus\{(a,b):a,b\in\mathbb Q\}$ is path connected.
By definition of path connected, there should exist continuous mapping $f:[0,1]\rightarrow \mathbb R^2$ s.t. $f(0)=a,f(1)=b$ for all $a,b\in\mathbb R^2$. I have utterly no idea how to find such map. Please help.
This result can be generalized to an arbitrary countable set $A \subset \mathbb{R}^2$. Since $A = \mathbb{Q}^2$ is countable, and there exist uncountably many lines in $\mathbb{R}^2$ through the point $x$, there must exist a line $\ell_1$ through $x$ that does not intersect the set $A$. Analogously, there exists a line $\ell_2$ through $y$ such that $\ell_2$ does not intersect $A$. Let $\ell_1$ and $\ell_2$ intersect at $p \in \mathbb{R}^2 - A$ and choose the path $x\to p \to y$.
Hint: Connect every point $(a,b)$ in your set to $(\pi, \pi)$ by first going horizontally and then vertically (or vertically and then horizontally).
It suffices to show that there is a path from $(e, e)$ to $(a, b)$ for all pairs $(a, b)\in \mathbb R^2-\mathbb Q^2$. WLOG, assume $a$ is irrational. Then we can draw a path from $(e, e)$ to $(a, e)$, and then draw a path from $(a, e)$ to $(a, b)$. Thus, all points in the set can be connected to $(e, e)$, and thus to each other.
