# Cardinal of all well-orders of $\mathbb{N}$ [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 9 at 15:45

• 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 – miraunpajaro May 9 at 13:45
• 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. – 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 $$\Bbb N^2$$.

• Thank you so much, I'll try that! – miraunpajaro May 9 at 13:44
• 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 $f:\mathbb N\to\mathbb N$. – bof May 9 at 13:47
• @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. – Arthur May 9 at 13:57
• 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 $\mathbb N$. – bof May 9 at 14:06
• @bof Men, I just feel so stupid now – miraunpajaro May 9 at 14:06