# Ordinals with the same cardinality

1. I was wondering how a bijection between two ordinal numbers with the same cardinality is constructed generally? if there is no general solution, some examples please?
2. Since any well-ordered set is order-isomorphic to a unique ordinal, is it correct that any bijection between two ordinals with the same cardinality cannot be an order-isomorphism?

Thanks and regards!

• Does ordinal belong to number theory? – Tim May 27 '11 at 5:15
• Not usually (except, of course, for the finite ones...) – Arturo Magidin May 27 '11 at 5:22

## 1 Answer

For (2), you need to exclude the case where the two ordinals are equal; but if $\alpha\neq \beta$ are ordinals, then no bijection between them can be an order isomorphism.

For (1), there's lots of ways. For instance, there are uncountably many bijections of $\omega+\omega$ with $\omega$: for each infinite subset $A$ of $\omega$ with infinite complement (there are $2^{\aleph_0}$ of them), both $A$ and $\omega\setminus A$ are well-ordered by the induced order, each of them isomorphic to $\omega$. So map $\omega+\omega$ to $\omega$ by mapping the first copy to $A$ and the second copy to $\omega\setminus A$. You can biject $\omega\times\omega$ with $\omega$ by using the zig-zag mapping that one often uses to show that there is a surjection from $\mathbb{Q}$ to $\mathbb{N}$ (or that one uses to show that a countable union of copies of $\mathbb{N}$ is countable). Etc.

• @Arturo: the countable union of countable sets needs not be countable. :-) In fact there is a model that the real numbers are a countable union of countable sets, and another (much more complicated) by Gitik in which all limit ordinals are of cofinality $\omega$ (this assumes some very strong axioms in addition though). – Asaf Karagila May 27 '11 at 4:51
• @Asaf: I was trying to get around that by saying "well-ordered" (as opposed to just "countable") so that you don't need to pick a bijection, but I really should have said "copies of $\omega$"... Unless I'm mistaken, to prove that a countable union of copies of $\omega$ is countable, you don't need Choice. Right? – Arturo Magidin May 27 '11 at 5:04
• @Arturo: I'm not sure anymore... (I've been working in one of the aforementioned models recently and all my intuition is off. You just start to assume that you can't really do anything ;-)) I'll get back to you with an answer later today. – Asaf Karagila May 27 '11 at 5:15
• @Asaf: Fair enough: but, $\mathbb{N}\times\mathbb{N}$ is countable. If you have a countable set of copies of $\mathbb{N}$, say $\{\mathbb{N}\times\{i\}\}_{i\in I}$, then pick an injection from $I$ to $\mathbb{N}$ (no choice needed, since the set of such injections is nonempty), and then embed the disjoint union to $\mathbb{N}\times\mathbb{N}$ by mapping $x$ from the $i$th copy to $(x,f(i))$. You can embed the union to the disjoint union also using $f$. – Arturo Magidin May 27 '11 at 5:21
• One way I've seen of putting this is that you can prove without choice that a countable union of counted sets is countable, where a counted set is a set equipped with an injection into $\mathbb{N}$. All you need choice for is to show you can turn your countably many countable sets into countably many counted sets, by simultaneously choosing an injection for each one. – Chris Eagle May 27 '11 at 9:04