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The Cantor Normal Form Theorem states that every ordinal $\alpha > 0$ can be uniquely expressed in the form $$\omega^{\beta_1}k_1 + \omega^{\beta_2}k_2 + \dots + \omega^{\beta_n}k_n$$ for some $n \ge 1$, positive integers $k_1,k_2,\dots,k_n$ and ordinals $\alpha \ge \beta_1 > \beta_2 > \dots > \beta_n$.

If I understand this correctly, then every countable ordinal's Cantor Normal Form is a finite-degree polynomial in $\omega$. My reasoning is that if $\beta$ is infinite, then $\omega^{\beta}$ is uncountable. This must be incorrect because it would imply that there are only countably many countable ordinals. Since $\aleph_1$ is the set of all countable ordinals and is uncountable, this is a contradiction.

My main question is where did I go wrong? If every countable set can't be written as a polynomial in $\omega$, what is a counter-example?

Thank you for your time.

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    $\begingroup$ No, exponentiation here is in the ordinal sense, it is not cardinal exponentiation. This is defined so $\alpha^\beta=\sup\{\alpha^\gamma\mid\gamma<\beta\}$ when $\beta$ is limit. But this easily gives (by an inductive argument) that $\alpha^\beta$ is countable if $\alpha$ and $\beta$ are countable. $\endgroup$ Apr 5 '12 at 3:03
  • $\begingroup$ Well, $\omega^\omega = \sup \{\omega^n | n \in \omega \}$ but it is not countable. What am I missing? $\endgroup$ Apr 5 '12 at 3:18
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    $\begingroup$ $\omega^\omega$ is countable, as is any countable limit or countable ordinals. $\endgroup$ Apr 5 '12 at 3:24
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    $\begingroup$ @Emily: ...it's a countable union of countable sets $\endgroup$ Apr 5 '12 at 3:30
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    $\begingroup$ $\omega^\omega$ here does not mean the set of functions from $\omega$ to $\omega$, rather here it means the set of polynomials in $\omega$ of finite degree with integer coefficients. $\endgroup$ Apr 5 '12 at 3:33
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As per the many comments, the issue is confusing ordinal and cardinal arithmetic. The number of such expressions that yield countable ordinals is uncountable, as it should be. This is because the $\beta$ can range over all countable ordinals (of which there $\aleph_1$ many). Thank you to all commenters for replying so quickly.

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