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Define (a,b) < (a',b') if \max(a,b) < \max(a',b') or \max(a,b) = \max(a',b') and b < b' or \max(a,b) = \max(a',b') and b = b' and a < a'.

Now I want to prove that the order-type of $(\{(b,c) : \max(b,c) = a\}, <)$ is equal to $a + a + 1$, does someone have a hint how to do this? I can't find the bijection. All elements are ordinals.

Thanks.

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    @JDH: Oops; quite right. I was just $f$ocusing on the issue o$f$ maxima. Thanks, Joel.2010-12-13

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Hint: The set naturally breaks up into three disjoint parts: $\{(a,a)\}$, the set $\{(b,a)\mid b\lt a\}$; and the set $\{(a,b)\mid b\lt a\}$. Every element of the latter is less than each element of the middle one, and each element of the middle one is stricly smaller than $(a,a)$.

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    Oh, haha. I staired at this for some while but I was thinking about a $\leq$ in the definition instead of $=$. Thanks!2010-12-13