Subsets of the unit interval

Question

I have been working on a problem and I need to answer the following question:

Is there a family $\{F_\alpha: \alpha \in \omega_1\}$ of subsets of the interval $]0,1[$ such that:

(a) $F_\alpha=\{x_1^\alpha, x_2^\alpha\}$ with $x_1^\alpha< x_2^\alpha$;

(b) If $\alpha, \beta \in \omega_1$ with $\alpha \ne \beta$, then $x_1^\alpha<x_1^\beta<x_2^\alpha<x_2^\beta$ or $x_1^\beta<x_1^\alpha<x_2^\beta<x_2^\alpha$?

I have tried to build such a family without success. Does anybody see the answer?

Answer 1

As suggested YCor take $F_x=\{x,x+\frac{1}{2}\}$. Then use the axiom of choice to set a well-order $\prec$ of $(0;\frac{1}{2})$. Then you can get such a family $G_\alpha$ to be the $F_x$'s enumereted with respect to $\prec$.

Answer 2

Note that the condition is about embedding into the open unit interval the order $2 \times P$ for two total orders of cardinalities two and $\omega_1$. (Unless I got it backwards and really mean $P \times 2$.) This means not just two disjoint copies of $P$, but also (by condition (b)) every element of one copy is below every element of the other copy, and the family $F$ defines an order isomorphism between the copies. Thus both copies possess (or both lack) extremal elements. This means one cannot cover the open interval with such an embedding. If one does not need a cover, the other answer and comment provide a solution.

Gerhard "Is 1/2 Out Of Order?" Paseman, 2019.03.21