ordered fields with the bounded value property, without choice

2011-08-19T04:11:58Z

In his answer to my question http://mathoverflow.net/questions/71432/ordered-fields-with-the-bounded-value-property, Ali Enayat showed that if one assumes the countable axiom of choice, then there exists a non-Archimedean ordered field $F$ with the bounded value property, by which I mean: for all $a < b$ in $F$ and for every continuous function $f$ from $[a,b]_F := ${$x \in F: a \leq x \leq b$} to $F$, there exists $B$ in $F$ such that $-B \leq f(x) \leq B$ for all $x$ in $[a,b]_F$. (Here we say $f$ is continuous if it satisfies the usual $\epsilon,\delta$ definition of continuity, where all quantification is over $F$.)

In the absence of $AC_\omega$, what can one prove? E.g., could we perhaps prove the assertion using an explicit subfield of the Field of surreal numbers, such as the set of surreal numbers created prior to day $\omega_1$? (I'm not a logician, so it's possible that such notions as "the set of surreal numbers created prior to day $\omega_1$" intrinsically depend on $AC_\omega$ in ways I'm not seeing.)

Answer by Ali Enayat for ordered fields with the bounded value property, without choice

2011-08-19T22:58:37Z

In my answer to the related question, $AC_{\omega}$ was only used to ensure that one can get hold of a regular uncountable cardinal (i.e., $\omega_1$). And of course Gitik's remarkable theorem assures us that, assuming the consistency of a proper class of strongly compact cardinals, there is a model of $ZF$ with no uncountable regular cardinals.

But despite the limitations imposed by Gitik's theorem, we can construct, in $ZF$ alone, a proper class field $F$ such that the statement "$F$ has the bounded value property" is provable in $NBG$ (i.e., von-Neumann-Bernays-Gödel theory of classes).

Explanation: $NBG$ is a "conservative" extension of $ZF$, designed to handle "large objects" such as the class V of sets, the class Ord of ordinals, and the field No of surreal numbers. $NBG$ can prove that the class of ordinals Ord is regular in the sense that every function from Ord to Ord with bounded range is constant on an unbounded subclass of Ord. On the other hand, Schmerl's proof of Sikorski's theorem, when implemented in $NBG$ shows that for any regular uncountable cardinal $\kappa \leq$ Ord there is an ordered field $F$ of cardinality and cofinality $\kappa$ that satisfies $BW(\kappa)$. Moreover, The analysis of Schmerl's proof reveals that $F$ can be arranged to be well-orderable, which, coupled with THE LEMMA established in my answer, shows that $NBG$ proves that $F$ has the bounded value property.

I will close this note by relating the discussion to surreals No.

In the presence of $CH$ (the continuum hypothesis), and $AC$, every ordered field of cardinality at most $\aleph_1$ is isomorphic to a subfield of No($<\omega_1$), where No($<\omega_1$) is the collection of surreals "born" before $\omega_1$. Coupled with my answer to the other question, this shows that in the presence of $AC+CH$, we have:

(1) There is a subfield of No($<\omega_1$) with the bounded value property.

Moreover, using the "resplendence" property of saturated models, one can show:

(2) $ZFC+CH$ proves that No($<\omega_1$) does NOT have the bounded value property.

(3) $NBG$ plus global choice proves that No does NOT have the bounded value property.

My proofs of (2) and (3) using resplendence are nonconstructive, but there might be explicit failures of the bounded value property for No and No($<\omega_1$) already in $ZF$.

ordered fields with the bounded value property, without choice - MathOverflow

ordered fields with the bounded value property, without choice

Answer by Ali Enayat for ordered fields with the bounded value property, without choice