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I was reading a question here the other day and find the next claim.

There are uncountabily many subset $\{E_\alpha\}_{\alpha\in I}$ of $\mathbb{Q}$ sucha that $E_\alpha \cap E_\beta$ is finite for $\alpha\neq \beta$.

I tried going with the asumption that they are numerable, an hence they can be order $E_1,E_2,\dots $ my idea was to take one element of to creat a new set, but the more I though about it the less it work or at least with the ideas I came up with either had problems with the fact that my set was new or that the intersection were finite.

Any help would be apreciated.

I'm not sure what are the corresponding tags for this question either.

juaaan
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  • This is a great question, but it has been asked before; see here, specifically Asaf Karagila's answer. – Noah Schweber Apr 08 '21 at 18:32
  • @NoahSchweber Thank you for point it out. – juaaan Apr 08 '21 at 18:34
  • It can be done by constructing the sets recursively, but you need transfinite recursion to do it; I second Noah’s advice. – Brian M. Scott Apr 08 '21 at 18:34
  • @BrianM.Scott True, but that's much less clean than the non-recursive construction (but I've deleted my comment re: recursive constructions since in retrospect it was overly strong). Moreover, I don't immediately see how to recursively get an AD family of size continuum, just of size $\omega_1$. – Noah Schweber Apr 08 '21 at 18:35
  • @Noah: I mentioned it only because the OP was apparently thinking along those lines; otherwise I’d have given the same advice that you gave. (I also like the version that uses $\Bbb Q^2$ and an infinite strip one unit wide centred at the origin.) I’m inclined to doubt that a straightforward recursive construction can produce an AD family of size $2^\omega$, but the OP asked only for an uncountable family, and I did want to acknowledge that a recursive approach wasn’t out of the question for the problem as stated. – Brian M. Scott Apr 08 '21 at 18:44
  • @BrianM.Scott Oh indeed, that wasn't meant as criticism, just elaboration for the OP. (Also, I haven't seen the strip version before - do you mean $\mathbb{Z}^2$ and counting lattice points, or something else?) – Noah Schweber Apr 08 '21 at 18:46
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    @Noah: Argh! Yes, I most certainly did mean $\Bbb Z^2$. That one and the Cantor tree are my favorites, I think. – Brian M. Scott Apr 08 '21 at 18:56

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