most recent 30 from http://mathoverflow.net 2013-06-19T14:21:34Z http://mathoverflow.net/feeds/user/22663 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93183/cantors-normal-form-and-aleph-1/93186#93186 Emily Sergel 2012-04-05T03:48:54Z 2012-04-05T03:48:54Z

<p>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.</p>

http://mathoverflow.net/questions/93183/cantors-normal-form-and-aleph-1 Emily Sergel 2012-04-05T03:00:56Z 2012-04-05T03:48:54Z

<p>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 $\alpha \ge \beta_1 > \beta_2 > \dots > \beta_n$.</p> <p>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.</p> <p>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?</p> <p>Thank you for your time.</p>

http://mathoverflow.net/questions/93183/cantors-normal-form-and-aleph-1 Emily Sergel 2012-04-05T03:43:49Z 2012-04-05T03:43:49Z

Oh, I see that Andres Caicedo was correct and I am confusing cardinal and ordinal arithmetic. I knew I must have been missing something. Thank you all!

http://mathoverflow.net/questions/93183/cantors-normal-form-and-aleph-1 Emily Sergel 2012-04-05T03:18:56Z 2012-04-05T03:18:56Z

Well, $\omega^\omega = \sup \{\omega^n | n \in \omega \}$ but it is not countable. What am I missing?