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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?

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migrated from mathoverflow.net Mar 23 at 1:56

This question came from our site for professional mathematicians.

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    $\begingroup$ What about the family of all $F_x=\{x,x+1/2\}$ for $0<x<1/2$? $\endgroup$ – YCor Mar 21 at 14:14
  • $\begingroup$ Which I think would naturally occur to almost everyone who posts in this forum. I suspect a cover of (0,1) is wanted but not stated, and thus also some set theoretical assumptions. Gerhard "And Where To Put 1/2?" Paseman, 2019.03.21. $\endgroup$ – Gerhard Paseman Mar 21 at 14:31
  • $\begingroup$ @YCor, thanks a lot! $\endgroup$ – Claudia Correa Mar 21 at 16:43
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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

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  • $\begingroup$ Thanks a lot! You really helped. $\endgroup$ – Claudia Correa Mar 21 at 15:08
  • $\begingroup$ @Claudia, you're welcome. Is this for a functional analysis course? Gerhard "Is Wondering A Little Bit" Paseman, 2019.03.21. $\endgroup$ – Gerhard Paseman Mar 21 at 15:45
  • $\begingroup$ If this was the intended question, you should edit the question and accept this answer. $\endgroup$ – YCor Mar 21 at 16:10
  • $\begingroup$ @YCor, I wanted exactly what I asked for. Thanks! $\endgroup$ – Claudia Correa Mar 21 at 16:14
  • $\begingroup$ @Gerhard, it is not for a course. I am trying to solve a problem and I simply couldnt see the answer ;-) $\endgroup$ – Claudia Correa Mar 21 at 16:16
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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 $\prec$.

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  • $\begingroup$ No, you cannot just steal complete ideas and get credit for them as answers. $\endgroup$ – mbsq Mar 21 at 22:19

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