Let $E$ be the set of all points in $\mathbb{R}^2$ having both coordinates rational. Prove that the space $\mathbb{R}^2\setminus E$ is path-connected.
Path-connected definition: A topological space $(X,\tau)$ is said to be path-connected if given $a,b\in X$, there exists a continuous function $f:[0,1]\to X$ such that $f(0)=a$ and $f(1)=b$.
I have read a similar thread on mathstackexchange but I am failing to build the function that proves that any two points of $\mathbb{R}^2\setminus E$ are path-connected.
If we consider $(x_1,y_1),(x_2,y_2)\in\mathbb{R}^2\setminus E$ so that $x_1,y_2$ are irrational as proposed in the answer of another question.
I can build two functions $f:(x_1,y_1)\to(x_1,y_2)\\(x_1,y_1)\to(x_1,y_1+c)$
so that $c\in\mathbb{R}$
$g:(x_1,y_2)\to(x_2,y_2)\\(x_1,y_1)\to(x_1+d,y_2)$ so that $d\in\mathbb{R}$
So $f \circ g:(x_1,y_1)\to(x_2,y_2)$.
However this is not a generalization for all the points in $\mathbb{R}^2\setminus E$ and I cannot relate the function to the interval $[0,1]$.
Question:
How should I solve the exercise?
Thanks in advance!
Best Answer
Let $C=\{c_1,c_2,\dots\}\subset \mathbb R^2$ be any countable set. Then $\mathbb R^2\setminus C$ is path connected.
Proof: Suppose $p,q$ are distinct points in $\mathbb R^2\setminus C.$ Consider the set of rays emanating from $p$ that contain a point of $C;$ the set of such rays is countable. Same thing for for $q.$ Thus if we let $L$ be the perpendicular bisector of $[p,q],$ the set of intersections of these rays with $L$ is countable. Hence there exists $r\in L$ such that both $[p,r],[r,q]$ are disjoint from $C.$ We have therefore found an "isosceles" path from $p$ to $q$ within $R^2\setminus C.$