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

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?

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 $$R^2\setminus C.$$

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 $$\lim_{t\rightarrow 1} \gamma(t)=(a',b')$$

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 $$g(t)=(x_2,f_{y_2}(y_1,t))$$.

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 $$\mathbb R^2 \setminus E$$.