Does $\Bbb R-\Bbb Q$ have a well ordered subset of type $\omega\cdot\omega$ Does $\Bbb R- \Bbb Q$ have a well ordered subset of type $\omega\cdot\omega$?
I thought of taking the subset to be A={$n\cdot \sqrt{m}:n\in\Bbb N,m\in P$} where P is the set of all prime numbers, with the well ordering -
 $\sqrt{2}<\sqrt{3}<\sqrt{5}<\sqrt{7}<...<2\cdot\sqrt{2}<2\cdot\sqrt{3}<2\cdot\sqrt{5}<...<3\cdot\sqrt{2}<3\cdot\sqrt{3}<...<m\cdot\sqrt{2}<m\cdot\sqrt{3}<m\cdot\sqrt{5}<...$
A is indeed a subset of  $\Bbb R- \Bbb Q$ and the well ordering is of type $\omega\cdot\omega$.
Am I correct?
And if I have to use the regular order of numbers, does there still exist a subset with such an ordering type?

There's an incredibly easy answer to this.

Theorem. If $(A,<)$ is a dense linearly ordered set, then countable linear order can be embedded into $A$.

This is really a theorem about $\Bbb Q$ itself, and then a consequence from the fact that every dense linear order has a subset isomorphic to $\Bbb Q$.
But to see this more concretely with $\omega\cdot\omega$, just pick an $\omega$ sequence of irrational numbers, e.g. $x_n=\pi+n$, then for each $x_n$, pick a sequence of order type $\omega$ in $(x_{n-1},x_n)$. For example, in this case, $y_{n+1,k}=\pi+n+\frac k{k+1}$.
Now show that $\{x_n\mid n<\omega\}\cup\{y_{n,k}\mid n,k<\omega\}$ is the wanted sequence.
