Connected component of $A=\mathbb{R}^2\setminus(\mathbb{Q}\times \mathbb{Q}^c)$

Question

I showed in a previous post that $$A=\mathbb{R}^2\setminus(\mathbb{Q}\times \mathbb{Q}^c)$$ looks something like $$\{(a,b)\;|\;a\in \mathbb{Q}^c, b\in \mathbb{R}\}\cup \{(a,b)\;|\;a\in \mathbb{R}, b\in \mathbb{Q}\}$$

I want to find the connected components of $A$?

My Efforts

I have two ways to approach in mind. Please have a look and point out the errors.

(1)

Take $a\in \mathbb{Q}^c$ then we are free to chose the second component so connected component of any point whose first coordinate is irrational will be a vertical line $x=a$

and if $a\in \mathbb{Q}$, the only choice for $b$ is that $b$ is also a rational number. So connected component of a point whose both coordinate are rational is just a singleton set.

Method 2

Take $b$ in $\mathbb{Q}^c$ then $a$ also belong to $\mathbb{Q}^c$, so connected component of such points is just the point itself, that is a sigleton set.

And if $b$ is in $\mathbb{Q}$ then we are free to chose $a$ to be any real. Connected component in this case is horizontal lines.

Why I am getting different connected components? I am confused.

Answer 1

Actually, $A$ is (path) connected: For $(x,y)$ in the set, either $\{x\}\times \Bbb R$ or $\Bbb R\times\{y\}$ is completely contained in $A$ and intersetcs the main diagonal $\{\,(t,t)\mid t\in\Bbb R\,\}$ that also belongs to $A$. Along this horizontal or vertical line plus the diagonal, we readily find a path in $A$ from $(x,y)$ to $(0,0)$.