$\mathbb{Q}\times(\mathbb{R}-\mathbb{Q})\cup(\mathbb{R}-\mathbb{Q})\times\mathbb{Q}$ in $\mathbb{R}^2$

Question

For $(a, b) \in \mathbb{R}^2,$ say $(a, b) \in A \subset \mathbb{R}^2$ if $a$ and $b$ are both rational numbers, or $a$ and $b$ are both irrational numbers; and $(a, b) \in B=\mathbb{R}^2\setminus A \subset \mathbb{R}^2$ otherwise.

Given any two distinct $x, y \in \mathbb{R}^2,$ can we draw a continuous curve from $x$ to $y$ in $\mathbb{R}^2$ such that it does not pass any element in $B?$

Context of this question: I am trying to prove that $B$ is connected in $\mathbb{R}^2$ with respect to the Euclidean metric.

Answer 1

It is an interesting question in general whether $A$ being dense in $X$ implies that $A\times A$ is dense in $X\times X$ for a general topological space $X$. Maybe someone can answer this general question.

However, in the case that $A=\Bbb Q$ and $X=\Bbb R$, it is easy to show directly. The pair $(\mathbb R,d)$ is a metric space, where $d(x,y)=\sqrt{(x-y)^2}$. Since we already know that $\Bbb Q$ is dense in $\Bbb R$, we know that for all $\epsilon>0$, that $$\bigcup_{q\in\Bbb Q}\mathbb B^1(q,\epsilon)=\mathbb R$$ Where $\mathbb B^1(x,r)$ is the one dimensional ball $\mathbb B^1(x,r)=\{s\in\mathbb R:d(x,s)<r\}$. This automatically means that $$\bigcup_{\boldsymbol q\in\mathbb Q^2}\mathbb B^2(\boldsymbol q,\epsilon)=\mathbb R^2$$ For all $\epsilon>0$. Because $\mathbb R^2$ is trivially dense in $\mathbb R^2$, we know that all of the sets $$S_n=\bigcup_{\boldsymbol q\in\mathbb Q^2}\mathbb B^2(\boldsymbol q,1/n)$$ are dense in $\mathbb R^2$ for all $n\in\mathbb N$. THEREFORE, $$\bigcap _{n\in\mathbb N}S_n = \mathbb Q^2$$ Is also dense in $\mathbb R^2$. QED.

Answer 2

First note that $A$ is arc-connected. This follows from the fact that a line segment which is a part of the graph of $ax + b$ where both $a,b$ are rational and $a$ is not $0$ contains only points from $A$. With this in mind, given two points, say $(x_1,y_1), (x_2,y_2) \in A$ one can constructs an arc connecting them and contained in $A$. The process is as follows. If both points are rational this is easy ( just take line segment connecting them). If one of them, say $(x_2,y_2)$ has both coordinates irrational then take a point with distance, say $< 1/2$ from it which has both coordinates rational and connect it to $(x_1,y_1)$ by an arc contained in $A$. Then, in the next step one finds a point, say with distance $<1/4$ from $(x_2,y_2)$ and extend the first arc by a line segment to its end to the new point. Then you continue by induction. You have a family of arcs which converges (uniformly) to and an arc joining $(x_1,y_1)$ and $(x_2,y_2)$. Similar construction works if both $(x_1,y_1), (x_2,y_2)$ are irrational. So, what matters here is not so much that $A$ is arc connected but the construction itself. That is, that we can build "circle like" (that closed) curves which are contained in $A$. Then to show that $B$ is not connected you take interior points and exterior points of such curve. Both sets are non-empty and open and their sum is $B$. So, $B$ is not connected hence it can not be arc-connected.