Is $\Bbb R^2 -\Bbb Q^2$ connected? I need to prove whether or not $X=\Bbb R^2 -\Bbb Q^2$ is connected. The definition of connectedness I am using is a space X is connected if it is not the union of two disjoint nonempty open sets. This is a very new concept for me so I am not exactly sure how to set it up. I believe we will prove it by contradiction, but I do not know where to start.
Take two points $x,y \in X$. Since $\Bbb{Q}^2$ is countable and the number of lines going through $x$ is uncountable (in bijection with $[0,\pi)$), there are uncountably many lines going through $x$ and contained in $X$. The same applies to $y$. Therefore, you can find lines going through $x$ and $y$ respectivly that are not parralel, and thus intersect each other. This proves that $X$ is connected.
Fix two points $(x,y), (u,v) \in A:= \mathbb{R}^2\setminus \mathbb{Q}^2$. 

Suppose $x, v\notin \mathbb{Q}$ then the lines joining $(x,y)\to (x,v)$ and $(x,y) \to (u,v)$ are both in $A$.

Something similar happens if $y,u \notin \mathbb{Q}$.

If $x,u\notin \mathbb{Q}$, then assume WLOG that $v&lt;y, u&lt;x$ and choose $z \in (v,y)\setminus \mathbb{Q}$ and $w\in (u,x)\setminus \mathbb{Q}$. So the lines $(x,y) \to (z,w), (z,w) \to (u,w)$ and $(u,w) \to (u,v)$ are all in $A$.

Similarly if $y,v\notin \mathbb{Q}$. Hence $A$ is path-connected, and so connected.