ordered fields with the bounded value property - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:02:53Z http://mathoverflow.net/feeds/question/71432 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71432/ordered-fields-with-the-bounded-value-property ordered fields with the bounded value property James Propp 2011-07-27T20:09:13Z 2011-08-03T15:29:32Z <p>Say that an ordered field $F$ satisfies the bounded value property if, for all $a &lt; b$ 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 $-B &lt; f(x) &lt; B$ 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 $F$.)</p> <p>Does there exist a non-Archimedean ordered field with the bounded value property?</p> <p>I show in <a href="http://jamespropp.org/reverse.pdf" rel="nofollow">http://jamespropp.org/reverse.pdf</a> (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).</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/71432/ordered-fields-with-the-bounded-value-property/71443#71443 Answer by Ali Enayat for ordered fields with the bounded value property Ali Enayat 2011-07-27T22:25:27Z 2011-08-03T15:29:32Z <p><strong>EDIT NOTE:</strong> 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.</p> <blockquote> <p>There are nonarchimedean fields with the bounded value property.</p> </blockquote> <p>Let's begin with a key definition: an ordered field $F$ satisfies the <strong>$\kappa$-Bolzano-Weiestrass property</strong>, abbreviated $BW(\kappa)$, if every bounded sequence $x_\alpha$ of length $\kappa$ in $F$ has a convergent subsequence of length $\kappa$.</p> <p>So the Bolzano-Weirestrass theorem says that $\Bbb{R}$ satisfies $BW (\aleph_{0})$.</p> <blockquote> <p>Sikorski (1948) proved that for every <strong>uncountable</strong> 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. </p> </blockquote> <p>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 $BW (\aleph_{0})$.</p> <p><strong>Lemma.</strong> <em>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.</em></p> <p><strong>Proof:</strong> 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 $\alpha &lt; \kappa$ there is some $t_{\alpha}$ $\in [a,b]$ with $f(t_{\alpha}) > x_{\alpha}$. </p> <p>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 $f(c)$. </p> <p>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). <strong>QED</strong></p> <p><strong>Some references</strong>: Sikorski's Theorem appears in:</p> <p>Roman Sikorski, <em>On an ordered algebraic field</em>. Soc. Sci. Lett. Varsovie. C. R. Cl. III. Sci. Math. Phys. 41 (1948), 69–96 (1950). </p> <blockquote> <p>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. </p> </blockquote> <p>James Schmerl, <em>Models of Peano arithmetic and a question of Sikorski on ordered fields</em>. Israel J. Math. 50 (1985), no. 1-2, 145–159.</p> <blockquote> <p><strong>PS</strong>. 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 <em>well-orderable</em> 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 $ZF+AC_\omega$. </p> </blockquote>