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It seems that one can construct ordinals from bottom up by successively introducing a new symbol each time a limit is taken: $$1,\ 2,\ \ldots,\ \omega,\ \omega +1,\ \omega +2,\ \ldots,\ \omega\cdot 2,\ \omega\cdot 2 +1,\ \ldots,\ \omega^{2},\ \ldots,\ \omega^{3},\ \ldots\ \omega^{\omega},\ \ldots,\ \omega^{\omega^{\omega}},\ \ldots, \epsilon_{0},\ \ldots$$ Can this be taken as a (mechanical) definition of ordinals? More abstract definitions like "an ordinal is a transitive well-ordered set satisfying certain properties" are much more appealing to me. Is this mechanical definition sufficient to prove things like "each well-ordered set is order isomorphic to exactly one ordinal?"

hmakholm left over Monica
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ashpool
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    What do you do when you reach an uncountable ordinal - do you have an uncountable number of symbols to choose from? What if we just use each ordinal as a symbol for itself? – Carl Mummert Oct 11 '11 at 01:37
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    Worse yet, the first uncountable ordinal $\omega_1$ cannot be reached as the limit of a countable sequence of smaller ordinals. So your process will give you at most the countable ordninals. – hmakholm left over Monica Oct 11 '11 at 01:44
  • @Henning: this is an argument in favor of taking each ordinal as a symbol for itself. – Carl Mummert Oct 11 '11 at 01:50
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    @Carl, they are awfully hard to write down on paper (except by use of other symbols, and even then we don't get most of them), which strikes me as a rather basic requirement for symbols. – hmakholm left over Monica Oct 11 '11 at 01:53
  • It is perhaps off-topic here, but maybe this is why induction to $\varepsilon_0$ is (by some people) considered "almost finitistic". (I've seen this mentioned in connection with Hilbert's program.) Ordinals up to $\varepsilon_0$ are basically ordinals we can "imagine to write down". – Martin Sleziak Oct 11 '11 at 02:24
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    In fact, the intent of the OP's method won't even get you to a nonrecursive ordinal, but it will get you to things that make $\varepsilon_{0}$ pale into insignificance (e.g. $\Gamma_{0}$, $\Gamma_{\varepsilon_{0}}$, the Bachmann-Howard ordinal, etc.). See http://en.wikipedia.org/wiki/Recursive_ordinal and http://en.wikipedia.org/wiki/Large_countable_ordinal – Dave L. Renfro Oct 11 '11 at 14:52

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As remarked in the comments, this is far from sufficient to cover even the countable ordinals. Personally, I see the problem with the three dots at the end which imply both an undefined idea of continuing this sequence, as well something that will terminate after at most $\omega_1$ many steps.

I imagine you might get to some large countable ordinals, perhaps $\epsilon_{\epsilon_0}$ or even higher. However this will terminate long before $\omega_1$.

Why is that a problem? Well, of course that we know about well-ordered sets whose order type is uncountable. But think of this reason: $\mu_0=\{\text{all those ordinals you wrote above}\}$, ordered by $\in$ this would be a transitive and well-ordered set. However it is not isomorphic to any of its members.

More generally, if you have a process which can only be used to generate set many ordinals then this set itself would be a well-ordered set not isomorphic to any of them. This would essentially break the theorem that every well-ordered set is isomorphic to exactly one ordinal.

Asaf Karagila
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