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  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!

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    Not usually (except, of course, for the finite ones...)2011-05-27

1 Answers 1

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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.

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    @Arturo: This is slightly stronger though. You require that there exists some sequence of bijections with $\omega$, which might not be extracted from the mere fact that each set has a bijection (not without the axiom of countable choice, anyway).2011-05-27