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I am trying to understand this proof of the existence of an uncountable ordinal. I don't see why $\mathcal{P}(\omega \times \omega)$ contains a copy of every countable ordinal as it is said.

For example, what element of $\mathcal{P}(\omega \times \omega)$ would correspond to $\omega\cdot 2$ ?

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M.G
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1 Answers1

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Let $\alpha$ be a countable ordinal, and let $f\colon\omega\to\alpha$ be a bijection. Then the relation $\{\langle m,n\rangle\mid f(m)\in f(n)\}$ is a well-ordering of $\omega$ with order type $\alpha$ (with $f$ being the isomorphism), and it is an element of $\mathcal P(\omega\times\omega)$.

If you want a particular relation which is isomorphic to $\omega\cdot2$, take the following: $$\{\langle m,n\rangle\mid (m<n\text{ and }m\equiv n\pmod 2)\lor m\text{ is odd, and }n\text{ is even}\}.$$

Asaf Karagila
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  • If I got it right, a relation on $\omega$ can be coded by a collection of ordered pairs of elements natural numbers. In particular, any well-order on subsets of $\omega$ is a subset of $\mathcal{P}(\omega \times \omega)$. – M.G May 20 '13 at 23:23
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    @M.G: No, a relation on a set $A$ is a subset of $A\times A$. So a well-order of $\omega$ is a set of pairs of natural numbers, that is to say, an element of $\mathcal P(\omega\times\omega)$. – Asaf Karagila May 20 '13 at 23:26