Say that an ordered field $F$ satisfies the bounded value property if, 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 $-B < f(x) < B$ for all $x$ in $[a,b]_F$. (Here we say $f$ is continuous if it satisfies the usual $\epsilon$, $\delta$ definition, where all quantification is over $F$.)

Does there exist a non-Archimedean ordered field with the bounded value property?

I show in http://jamespropp.org/reverse.pdf (see the second paragraph on page 9) that every Archimedean ordered field satisfying the bounded value property is isomorphic to the reals, but my proof that the bounded value property implies the Archimedean property (see the first paragraph on page 9) is incorrect (thanks to Ricky Demer for pointing out my mistake).

In attempting to fix my proof, I am starting to wonder if in fact the implication fails. For instance, does the surreal number system have the bounded value property? I don't see how to prove that it doesn't.

All I can show is that if $F$ satisfies the bounded value property and contains a cofinal set $S$ whose cardinality is less than or equal to that of the continuum, then $F$ is Archimedean. (Proof: Let $g:[0,1] \rightarrow F$ be a function that takes on all values in $S$, and for all $x$ in $[a,b]_R$ with standard part $\overline{x}$ let $f(x) = g(\overline{x})$. If $F$ is non-Archimedean, $f$ is continuous on $[a,b]_F$ and unbounded.) But, even leaving aside constructivist qualms about how one constructs $g$ from $S$, clearly this approach won't work for the surreal numbers or for sufficiently large fields within the Field of surreal numbers.

(1)that was not specified anywhere in your question, and(2)there is very little to say about uncountable model theory and algebra in the absence of choice. – Ali Enayat 0 secs ago $\endgroup$ – Ali Enayat Jul 29 '11 at 21:44