connected sets

**LAINHELL** · September 26th 2008, 07:44 AM

Hi, I'm having trouble with this:
Prove that the set R2\ Q2 is connected

**Plato** · September 26th 2008, 09:26 AM

Recall that $\left( {a,b} \right) \notin \mathbb{R}^2 \backslash \mathbb{Q}^2 \Rightarrow \quad a \notin \mathbb{Q}^2 \vee b \notin \mathbb{Q}^2$ .
So suppose that $\left\{ {\left( {a,b} \right),\left( {c,d} \right)} \right\} \notin \mathbb{R}^2 \backslash \mathbb{Q}^2$ .
Construct a polygonal path from $(a,b) \mbox{ to } (c,d)$ which is a subset of $<br /> \mathbb{R}^2 \backslash \mathbb{Q}^2$ .
There are several cases. I will help with one.
If $a \notin \mathbb{Q} \wedge d \notin \mathbb{Q}$ there is a linear path from $(a,b) \mbox{ to } (a,d)$ : $\alpha _1 = \left\{ {\left( {a,t(d - b) + b} \right)} \right\};\;0 \leqslant t \leqslant 1$ .
And there is a linear path from $(a,d) \mbox{ to } (c,d)$ : $\alpha _2 = \left\{ {\left( {t(c-a)+a,d} \right)} \right\};\;0 \leqslant t \leqslant 1$ .
Putting the two together we have a polygonal path from $(a,b) \mbox{ to } (c,d)$ which is a subset of $\mathbb{R}^2 \backslash \mathbb{Q}^2$ .
There are other cases. But in each you can show pathwise connectivity.

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