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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.

taupi
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    $2^{\omega}\geq \omega_1$. You don’t know it is equal. (If you did, you would get the Continuum Hypothesis as a consequence). – Arturo Magidin Sep 12 '19 at 04:03
  • Thanks for pointing that out. – taupi Sep 12 '19 at 04:11
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    If you know that every countable (linearly) ordered set is isomorphic to a subset of $\mathbb Q$, they your problem boils down to exhibiting a family of $2^\omega$ non-isomorphic countable ordered sets. Can you do that? – bof Sep 12 '19 at 04:15

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For a given set $X\subseteq\mathbb N$ let $$A_X=\left(\mathbb Q\setminus\bigcup_{n\in\mathbb N}\left(n-\frac13,n+\frac13\right)\right)\cup X.$$ There are $2^\omega$ different subsets $X\subseteq\mathbb N$, and $X\ne Y\implies A_X\not\cong A_Y$.

bof
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