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Motivated by a problematic exercise in an analysis textbook, I decided to search for an example of an ordered field which satisfies the nested interval property yet fails to be complete.

I first considered the ultraproduct construction as a natural starting point. Let $U$ be a non-principal ultrafilter on a set $X$, and let $F = \Bbb{R}^{X} / U$ be the ultraproduct of the copies of $\Bbb{R}$ indexed by $X$. Clearly $F$ is an ordered field strictly containing $\Bbb{R}$. In particular, $F$ is not complete.

But still it seemed to me that the assumptions are too weak to prove that $F$ has the nested interval property. This naturally led me to consider the $\sigma$-completeness of $U$:

  • For any $A_1, A_2, \cdots \in U$, we have $A_1 \cap A_2 \cap \cdots \in U$.

This is a quite strong condition, and with this $\sigma$-completeness I could easily prove that $F$ satisfies the nested interval property. But still some questions remained:

  1. I heard that the existence of a $\sigma$-complete non-principal ultrafilter cannot be either proven or disproved in ZFC. In this case, I'm curious if adding this claim as a new axiom into ZFC will result in an acceptable axiomatic system.

  2. Otherwise, is there a way to construct an example by circumventing the need of $\sigma$-completeness? (I heard of a surreal 'Field' and its restriction to $\omega_1$, but I'm not sure if the latter can be achieved within our familiar axiomatic system or not.)

I googled and searched on this site, but the results were too technical for me to understand, if not irrelevant. Indeed, my set-theoretic knowledge is limited largely to a very naive and elementary level. So I would be glad if someone helps me with this topic.

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Answering your 1: The existence of such an ultrafilter implies the existence of a measurable cardinal. This is a large cardinal axiom. This axiom is not provable from ZFC and furthermore its consistency relative to ZFC is not provable (i.e. you cannot prove that you cannot prove the negation of the axiom). Studying this axiom and even stronger ones has been a very fruitful endeavour in set theory. Just keep in mind that what you prove using this is not necessarily provable from ZFC. –  Apostolos Mar 31 '13 at 14:58
    
To add on @Apostolos' comment, measurable cardinals do appear naturally in topology from time to time. So do not be deterred from using something like a measurable cardinal. –  Asaf Karagila Mar 31 '13 at 15:21
    
Unfortunately, if $U$ is $\sigma$-complete, then the ultrapower of $\mathbb R$ with respect to $U$ will be isomorphic to $\mathbb R$. –  Andreas Blass Mar 31 '13 at 19:21
    
Thanks @Apostolos and Asaf Karagila. Your comments just relieve my mind. :) –  sos440 Mar 31 '13 at 19:36
    
@AndreasBlass, Do you mean that such an event possible even if $U$ is non-principal? –  sos440 Mar 31 '13 at 19:41

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