Ordering of rationals

by taupi   Last Updated September 12, 2019 04:20 AM

Let $(\mathbb{Q},<)$ be the usual ordering of rationals. Show that there is a family $\mathcal F$ of subsets of $\mathbb{Q}$ such that $|\mathcal F|=2^\omega$ and for every $A, B \in \mathcal F, (A,<)\ncong (B,<).$

I know that the question is asking to find a family $\mathcal F$ such that it is an uncountable set as $2^\omega \geq \omega_1.$

Also in order to show that $(A,<)\ncong (B,<)$, i need to show that there is no bijection $f: A \to B$ such that for every $x,y\in A$, $x<y \iff f(x)<f(y)$.

I am unsure of how to do this construction.



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