most recent 30 from http://mathoverflow.net 2013-05-22T15:57:54Z http://mathoverflow.net/feeds/question/32308 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32308/can-one-really-construct-an-ordinal-table ashpool 2010-07-17T19:51:27Z 2010-07-18T10:02:47Z

<p>Many books describe how one can construct "by hand" a table of ordinals $1,\ 2,\ \ldots,\ \omega,\ \omega +1,\ \omega +2,\ \ldots,\ \omega\cdot 2,\ \omega\cdot 2 +1,\ \ldots,\ \omega^{2},\ \ldots,\ \omega^{3},\ \ldots\ \omega^{\omega},\ \ldots,\ \omega^{\omega^{\omega}},\ \ldots, \epsilon_{0},\ \ldots$. But does this span the entire ordinal class? For some reason I can't seem to prove it. Is there an easy way to see that? Thanks!</p>

http://mathoverflow.net/questions/32308/can-one-really-construct-an-ordinal-table/32330#32330 Asaf Karagila 2010-07-18T00:14:37Z 2010-07-18T00:14:37Z

<p>Since ordinal numbers have a unique division, logarithm and subtraction properties, when given an ordinal $\alpha$ you can write any other ordinal as a finite polynomial in $\alpha$, when $\alpha = \omega$ you get what's known as "Cantor normal form of $\gamma$ for the base $\omega$".</p> <p>I.e., any ordinal $\gamma$ can be written as a finite sum: $$\gamma = \sum_{i=0}^n \omega^{\alpha_i}\cdot\beta_i$$ Where $\alpha_i$ is a decreasing chain of ordinals, and $\beta_i$ is finite.</p> <p>(More generally, you can take some base $\zeta$ and then $\beta_i &lt; \zeta$)</p> <p>Thing is that we only have a finite number of symbols, so at most we can represent (uniquely) a countable number of numbers, since we have a proper class of ordinals, which is a mind boggling concept of infinitude, you obviously can't write them all. But still, any given ordinal can be presented as a finite polynomial in $\omega$, thus spanning the table discussed.</p>