Prove that $\mathbb{R}^2\setminus E$ is path-connected

Question

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!

Answer 1

Let $(a,b), (a',b') \in \mathbb Q^2$ be two points of $\mathbb R^2$ with both rational coordinates. Consider two sequences $(a_n)_{n\in \mathbb Z},(b_n)_{n\in \mathbb Z}$ of irrational numbers such that $$ \lim_{-\infty}a_n = a\quad \lim_{+\infty} a_n=a', \quad \lim_{-\infty} b_n=b,\quad \lim_{+\infty} b_n=b',$$.

Now consider the $\mathbb Z-$sequence of points $$\left\{\begin{matrix}x_{2n+1}&=& (a_n+1,b_n) \\x_{2n}&=&(a_n,b_{n}) \end{matrix}\right.$$ and notice that $$\forall n\in \mathbb Z, \forall t\in [0,1],~~ (1-t)x_n+tx_{n+1} \notin E$$ you then construct a piecewise linear path $\gamma: [-1,1]\rightarrow \mathbb R^2$ such that $\forall n\in \mathbb Z, \gamma(\tanh(n))=x_n$. By definition, $\lim_{t\rightarrow -1}\gamma(t)=(a,b)$ and $\lim_{t\rightarrow 1} \gamma(t)=(a',b')$

Answer 2

The idea is good, since you are only one coordinate at a time. Instead of just adding a number, however, add something along the lines of $tb$ for $t\in[0,1]$.

In other words, for fixed $(x_1,y_1),(x_2,y_2)\in\mathbb{R}^2\setminus E$, first note that we may assume that all the coordinates are irrational. This is because at least one of $x_1$ and $y_1$ is irrational, so pick an irrational $z$ and use the path which is constant in the irrational coordinate, and $f_z(x,t)=(1-t)x+tz$ in the other coordinate.

Now, since all the above coordinates may be assumed irrational, use the two paths $f(t)=(f_{x_2}(x_1,t),y_1)$ and $g(t)=(x_2,f_{y_2}(y_1,t))$.

Answer 3

You write about a function $f : (x_1,y_1) \to (x_1,y_2)$, but $P = (x_1,y_1)$ and $Q = (x_1,y_2)$ are just single points in $\mathbb R^2$, and so this is a rather un-useful function whose domain is one point and whose range is another point.

If you are familiar with formulas for parameterizing line segments in $\mathbb R^2$, then you will know that the line segment from the point $P = (x_1,y_1)$ to the point $Q = (x_1,y_2)$ can be parameterized by the continuous function $f_1 : [0,1] \to \mathbb R^2$ given by $$f_1(t) = (1-t)P + t Q = ((1-t)x_1+tx_1,(1-t)y_1 + t y_2) = (x_1,(1-t)y_1 + t y_2) $$

Similarly, the line segment from the point $Q$ to the point $R = (x_2,y_2)$ can be parameterized by the continuous function $f_2 : [0,1] \to \mathbb R^2$ similarly given by $f_2(t)=(1-t)Q + t R$.

Finally, if you are familiar with concatenation of paths, you obtain a path from $P$ to $R$ using the concatenation $$f_1 * f_2(t) = \begin{cases} f_1(2t) & \text{if $0 \le t \le 1/2$} \\ f_2(2t-1) & \text{if $1/2 \le t \le 1$} \end{cases} $$ Now all you have to do is to convince yourself, in turn, that $f_1$ and $f_2$ and $f_1 * f_2$ are paths in $\mathbb R^2 \setminus E$.

Answer 4

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 $R^2\setminus C.$