Ordinal collapsing function: Difference between revisions

Revision as of 11:04, 23 February 2013

Revision as of 20:11, 9 May 2013

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* If <math>\psi(\alpha) = \psi(\beta)</math> with <math>\beta<\alpha</math> then necessarily <math>C(\alpha) = C(\beta)</math>. Indeed, no ordinal <math>\beta'</math> with <math>\beta\leq\beta'<\alpha</math> can belong to <math>C(\alpha)</math> (otherwise its image by <math>\psi</math>, which is <math>\psi(\alpha)</math> would belong to <math>C(\alpha)</math> &mdash; impossible); so <math>C(\beta)</math> is closed by everything under which <math>C(\alpha)</math> is the closure, so they are equal.

* Any value <math>\gamma=\psi(\alpha)</math> taken by <math>\psi</math> is an <math>\varepsilon</math>-number (i.e., a fixed point of <math>\beta\mapsto\omega^\beta</math>). Indeed, if it were not, then by writing it in [[Ordinal arithmetic#Cantor normal form|Cantor normal form]], it could be expressed using sums, products and exponentiation from elements less than it, hence in <math>C(\alpha)</math>, so it would be in <math>C(\alpha)</math>, a contradiction.

* Lemma: Assume <math>\delta</math> is an <math>\varepsilon</math>-number and <math>\alpha</math> an ordinal such that <math>\psi(\beta)<\delta</math> for all <math>\beta<\alpha</math>: then the <math>\Omega</math>-pieces (defined [[#A note about base representations|above]]) of any element of <math>C(\alpha)</math> are less than <math>\delta</math>. Indeed, let <math>C'</math> be the set of ordinals all of whose <math>\Omega</math>-pieces are less than <math>\delta</math>. Then <math>C'</math> is closed under addition, multiplication and exponentiation (because <math>\delta</math> is an <math>\varepsilon</math>-number, so ordinals less than it are closed under addition, multiplication and exponentitionexponentiation). And <math>C'</math> also contains every <math>\psi(\beta)</math> for <math>\beta<\alpha</math> by assumption, and it contains <math>0</math>, <math>1</math>, <math>\omega</math>, <math>\Omega</math>. So <math>C'\supseteq C(\alpha)</math>, which was to be shown.

* Under the hypothesis of the previous lemma, <math>\psi(\alpha) \leq \delta</math> (indeed, the lemma shows that <math>\delta \not\in C(\alpha)</math>).

* Any <math>\varepsilon</math>-number less than some element in the range of <math>\psi</math> is itself in the range of <math>\psi</math> (that is, <math>\psi</math> omits no <math>\varepsilon</math>-number). Indeed: if <math>\delta</math> is an <math>\varepsilon</math>-number not greater than the range of <math>\psi</math>, let <math>\alpha</math> be the least upper bound of the <math>\beta</math> such that <math>\psi(\beta)<\delta</math>: then by the above we have <math>\psi(\alpha)\leq\delta</math>, but <math>\psi(\alpha)<\delta</math> would contradict the fact that <math>\alpha</math> is the ''least'' upper bound &mdash; so <math>\psi(\alpha)=\delta</math>.