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Lets consider the set $$ A = \{R\subset\mathbb{N}^2:R \text{ is a well-order of } \mathbb{N}\} $$ Now, it's clear that $\aleph_1\leq|A|\leq2^{\aleph_0}$(Since all the well-orders of $\mathbb{N}$ are isomorphic to some $\alpha<\aleph_1$ which gives $\aleph_1\leq|A|$ and every well-order of $\mathbb{N}$ is a subset of $\mathbb{N}^2$, of which there are $2^{\aleph_0}$, giving $|A|\leq2^{\aleph_0}$)

If we were to assume the Continuum hypothesis then it clearly follows $|A|=\aleph_1=2^{\aleph_0}$(By Cantor-Berstein theorem)

However in ZFC, I'm not sure how to procced, I think that $|A|=2^{\aleph_0}$, but I do not know how to prove this or if this is right. Any idea on what to do?


marked as duplicate by Asaf Karagila cardinals May 9 at 15:45

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  • $\begingroup$ Btw, I just realised it's been answered before: <math.stackexchange.com/questions/165010/…>. and here mathoverflow.net/questions/112651/… Please tell me if you want me to delete this post $\endgroup$ – miraunpajaro May 9 at 13:45
  • $\begingroup$ No, don't delete it. We don't delete duplicate questions, we close them as duplicates, with a pointer to the previously answered question. In fact you can close it yourself with the "close" button at the bottom of your question. $\endgroup$ – bof May 9 at 13:51

Fix some well-order on $\Bbb N^2$.

Take a countably infinite sequence of natural numbers $a = (a_1, a_2, \ldots)$, and define a well-order on $\Bbb N^2$ as follows:

  • $(i,a_i)$ comes before $(j, a_j)$ iff $i$ is less than $j$ (under the standard ordering on $\Bbb N$)
  • Any point of the form $(i, a_i)$ is before any point not of that form
  • For any two points not on the form $(i, a_i)$, compare using the fixed well-order

(don't forget to prove that this is actually a well-order). This gives an injection (show this too, of course) from the set of infinite sequences of natural numbers (which has cardinality $2^{\aleph_0}$) to the set of well-orderings of $\Bbb N^2$.

  • $\begingroup$ Thank you so much, I'll try that! $\endgroup$ – miraunpajaro May 9 at 13:44
  • $\begingroup$ OK, but why not just show that $\mathbb N$ admits $2^{\aleph_0}$ orderings which are isomorphic to the standard ordering, i.e., orderings of type $\omega$? There is one of those for each bijection $f:\mathbb N\to\mathbb N$. $\endgroup$ – bof May 9 at 13:47
  • $\begingroup$ @bof I could've done that. Ordering each copy of $\{n\}\times \Bbb N$ according to some $\omega$-type ordering. But I don't see that as much easier, really. $\endgroup$ – Arthur May 9 at 13:57
  • $\begingroup$ Who said anything about ordering copies og $\{n\}\times\mathbb N$? I mean, for each bijection $f:\mathbb N\to\mathbb N$, define an ordering $\lt_f$ of $\mathbb N$ by $$m\lt_f n\iff f(m)\lt f(n).$$ Zap, $2^{\aleph_0}$ well-orderings of $\mathbb N$. $\endgroup$ – bof May 9 at 14:06
  • $\begingroup$ @bof Men, I just feel so stupid now $\endgroup$ – miraunpajaro May 9 at 14:06

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