Cardinal of all well-orders of $\mathbb{N}$

Question

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?

Answer 1

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 $\Bbb N^2$.