Hi, I'm having trouble with this:
Prove that the set R2\ Q2 is connected
Recall that $\displaystyle \left( {a,b} \right) \notin \mathbb{R}^2 \backslash \mathbb{Q}^2 \Rightarrow \quad a \notin \mathbb{Q}^2 \vee b \notin \mathbb{Q}^2 $.
So suppose that $\displaystyle \left\{ {\left( {a,b} \right),\left( {c,d} \right)} \right\} \notin \mathbb{R}^2 \backslash \mathbb{Q}^2 $.
Construct a polygonal path from $\displaystyle (a,b) \mbox{ to } (c,d)$ which is a subset of $\displaystyle
\mathbb{R}^2 \backslash \mathbb{Q}^2 $.
There are several cases. I will help with one.
If $\displaystyle a \notin \mathbb{Q} \wedge d \notin \mathbb{Q}$ there is a linear path from $\displaystyle (a,b) \mbox{ to } (a,d)$: $\displaystyle \alpha _1 = \left\{ {\left( {a,t(d - b) + b} \right)} \right\};\;0 \leqslant t \leqslant 1$.
And there is a linear path from $\displaystyle (a,d) \mbox{ to } (c,d)$: $\displaystyle \alpha _2 = \left\{ {\left( {t(c-a)+a,d} \right)} \right\};\;0 \leqslant t \leqslant 1$.
Putting the two together we have a polygonal path from $\displaystyle (a,b) \mbox{ to } (c,d)$ which is a subset of $\displaystyle \mathbb{R}^2 \backslash \mathbb{Q}^2 $.
There are other cases. But in each you can show pathwise connectivity.