Is the cardinality of two countably infinite, order-isomorphic sets equal?

Question

I am trying to get a better sense for the cardinality between infinite sets. If two sets are countably infinite, well-ordered, and share an order isomorphism, is it said that these two sets must be equal in cardinality, due to a bijection that maps each element of one set to the other, i.e. through the "back-and-forth" method? If yes, is there any other rationale for this equality in cardinality?

Answer 1

Two sets that are both countably infinite have the same cardinality, no matter whether they are also ordered, well-ordered, or have some other structure or no structure.

Since they are countably infinte, there are bijections $f:\mathbb N\to A$ and $g:\mathbb N\to B$. Now $g\circ f^{-1}$ is a bihection $A\to B$.

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At the same time, two ordered sets that are order-isomorphic have the same cardinality, no matter whether you know that they are countably infinite (or finite or uncountable or whatever). An order isomorphism, like every other kind of isomorphism, is a bijection, and therefore in and of itself means that the two sets are equinumerous.