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By Cantor's normal form theorem, any ordinal $r$ can be expressed as: $r=\omega^{k_1}a_1 + \omega^{k_2}a_2 + \ldots$. ($k_1>k_2>\ldots$)

I want to know whether the class of all $a_i$'s is countable or not.

If it is not countable how do i prove a problem such as: $\omega^{k_1}a_1 + \omega^{k_2}a_2 + \omega^{k_3} + \ldots < \omega^{k_1}a_1 + \omega^{k_1}a_2 +\omega^{k_1}a_3 + \ldots$?

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You should look at a definition of Cantor normal form. –  Chris Eagle May 29 '12 at 15:39
    
I just read it. Now i know it's true that such class is countable but i dont know how to prove that.. –  Katlus May 29 '12 at 15:48
    
Oh.. I proved it thank you –  Katlus May 29 '12 at 15:56

2 Answers 2

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If you require that $k_i$ and $a_i$ are all strictly less than $\omega$ then this simply corresponds to finite sequences of natural numbers, which is a countable collection.

Now remember that every ordinal has a unique Cantor normal form, so every ordinal can be expressed like this. If, however, you require this to be non-degenerate everywhere (e.g. $\omega_1=\omega^{\omega_1}$ is the Cantor normal form of $\omega_1$) then you indeed have only countably many ordinals.

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Each $a_i$ in the Cantor normal form $\omega^{k_1}\cdot a_1 + \cdots + \omega^{k_n}\cdot a_n$ is a positive natural number.

So the "class of all $a_i$'s" is the set of all positive natural numbers, which is countable.

If you meant "the class of all tuples $(a_1,\ldots, a_n)$", then this is the set of all finite tuples of positive natural numbers, again a countable set.

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