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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.

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related to your paper but not this question: For the cut property, wouldn't it be nicer to drop the condition that $A\cup B = R$ and use the second conclusion? –  Ricky Demer Jul 29 '11 at 2:15
    
@Ricky: That was Tarski's feeling too. His version of the cut property is more flexible but I don't think students find it as intuitive, since the range of possibilities for the sets $A$ and $B$ becomes vastly greater. I personally find the narrower statement easier to visualize and hence more compelling, but that's purely an esthetic preference. If one were to develop real analysis from my version of the cut principle, your version should probably be Theorem 1! –  James Propp Jul 29 '11 at 15:41
    
@James, this is a reply to your latest comment on my answer, but the comments there were piling up. I cannot see how to eliminate AC from the proof given in my answer, if one uses the Schmerl method; and I suspect the same holds for Sikorski's construction. I am also a bit puzzled about requiring no use of $AC$ in the counterexample since (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 –  Ali Enayat Jul 29 '11 at 21:44
    
@James: despite the pessimism expressed in my previous comment about eliminating the axiom of choice $AC$ , a "miracle has happened", and I can now see how to take advantage of a key feature of Schmerl's construction to eliminate $AC$; by tomorrow I will put a PS on my answer in which I will outline how this is done. –  Ali Enayat Jul 31 '11 at 1:57
    
I don't see why that cardinal must be regular. For that matter, I don't see why it can't have countable cofinality. –  Ricky Demer Aug 1 '11 at 7:42
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1 Answer

up vote 6 down vote accepted

EDIT NOTE: A postscript has been added to indicate why the answer does not change if one is forced to work in $ZF+AC_\omega$ (prompted by a query of James Propp). Thanks to James Propp, Ricky Demmer, and Emil Jeřábek for catching infelicities of the past versions.

There are nonarchimedean fields with the bounded value property.

Let's begin with a key definition: an ordered field $F$ satisfies the $\kappa$-Bolzano-Weiestrass property, abbreviated $BW(\kappa)$, if every bounded sequence $x_\alpha$ of length $\kappa$ in $F$ has a convergent subsequence of length $\kappa$.

So the Bolzano-Weirestrass theorem says that $\Bbb{R}$ satisfies $BW (\aleph_{0})$.

Sikorski (1948) proved that for every uncountable regular cardinal $\kappa$ there is an ordered field of cardinality and cofinality $\kappa$ that satisfies $BW(\kappa)$. Since every archimedean ordered field has countable cofinality, the following Lemma, when coupled with Sikorski's theorem above (with $\kappa$ chosen as $\aleph_1$) shows that nonarchimedean fields with the bounded value property exist.

Note that the proof is of the Lemma is an adaptation of the usual real-analysis proof of the boundedness of continuous functions on closed bounded intervals, using $BW (\aleph_{0})$.

Lemma. Let $\kappa$ be a regular cardinal. If $F$ is an ordered field of cofinality $\kappa$ such that $F$ satisfies $BW(\kappa)$, then $F$ has the bounded value property.

Proof: Choose an increasing unbounded sequence $x_\alpha$ of elements of $F$, where $\alpha \in \kappa$. If $f[a,b]$ has no upper bound for a continuous function $f$, then for each $\alpha < \kappa$ there is some $t_{\alpha}$ $\in [a,b]$ with $f(t_{\alpha}) > x_{\alpha}$.

By $BW(\kappa)$ there is some unbounded subset $U$ of $\kappa$ such that the subsequence $S$ := {$ t_{\alpha} : x \in U $} converges to some $c\in [a,b]$. Therefore by continuity of $f$, the sequence $f(S)$ converges to $f(c)$.

But a convergent sequence of length $\kappa$ must be bounded (the regularity of $\kappa$, and the assumption that $F$ has cofinality $\kappa$ comes to the rescue here), and yet $f(S)$ is clearly unbounded by construction. This contradiction shows that $f[a,b]$ is bounded above; a similar reasoning shows that $f[a,b]$ is bounded below (or just replace $f$ by its absolute value). QED

Some references: Sikorski's Theorem appears in:

Roman Sikorski, On an ordered algebraic field. Soc. Sci. Lett. Varsovie. C. R. Cl. III. Sci. Math. Phys. 41 (1948), 69–96 (1950).

A proof of Sikorski's theorem can also be found in the following paper (Cor. 2.7), as a corollary of a vast generalization of Sikorski's theorem; the paper is an impressive showcase for the interaction between deep methods of models of arithmetic and higher set theory with field theory.

James Schmerl, Models of Peano arithmetic and a question of Sikorski on ordered fields. Israel J. Math. 50 (1985), no. 1-2, 145–159.

PS. One can show, using some machinery from the model theory of arithmetic, that working only in $ZF+AC_\omega$, Schmerl's proof can produce a well-orderable field $F$ of cardinality and cofinality $\aleph_1$ that satisfies $BW(\aleph_1)$. This allows one to one obtain a non-archimedean field with the bounded value property, entirely within $ZF+AC_\omega$.

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Thanks for your reply, Ali. The first place where I don't follow you is in your restatement of the Bolzano-Weierstrass Theorem. Did you really mean to say that the ordered field has cardinality $\kappa$? Note that the Bolzano-Weiestrass Theorem concerns sequences of length $\aleph_0$ chosen from a set of cardinality $\aleph_1$. So something seems amiss here. –  James Propp Jul 27 '11 at 22:47
    
@James, you are right, I should have just said: "the usual Bolzano-Weierstrass Theorem", I will fix that. –  Ali Enayat Jul 27 '11 at 23:13
    
@James: I ended-up removing any cardinality restrictions from $BW(\kappa)$ which does not effect my answer; but in case you look at Schmerl's paper note that he stipulates the cardinality of $F$ being greater than $\kappa$ as part of the definition of $BW(\kappa)$. –  Ali Enayat Jul 27 '11 at 23:28
    
How do you get the third sentence of your proof? –  Ricky Demer Jul 28 '11 at 1:25
    
I also appear to be missing something: doesn't Cantor's original (en.wikipedia.org/wiki/…) proof show that any countable ordered field does not satisfy $BW(\aleph_0)$ ? –  Ricky Demer Jul 28 '11 at 1:52
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