EDIT NOTE: In the proof of the Lemma, A postscript has been added to indicate why the answer does not change if one is forced to work in $ZF+AC_\omega$. 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)$. Since every archimedean ordered field has countable cofinality(in$ZF+AC_\omega$), , the following Lemma, when coupled with Sikorski's theorem above (with$\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$. 6 deleted 244 characters in body EDIT NOTE1: "uncountable" was missing from In the statement proof of Sikorski's Theorem in the previous version. Lemma, A postscript has been added to indicate why the answer does not change if one is forced to work in$ZF+AC_\omega$. Thanks to James Propp, Ricky Demmer, and Emil Jeřábek for catching infelicities of the past versions. EDIT NOTE 2: A PS has been added at the end to indicate why the answer does not change if one is barred from using the the axiom of choice. Sikorski (1948) proved that for every uncountable regular cardinal$\kappa$there is an ordered field of cardinality and cofinality$\kappa$that satisfies$BW(\kappa)$. Since every archimedean ordered field has cardinality at most countable cofinality (in$2^{\aleph_0}$, ZF+AC_\omega$), the following Lemma, when coupled with Sikorski's theorem above (with $\kappa$ chosen as any regular cardinal greater than $2^{\aleph_0}$) \aleph_1$) shows that nonarchimedean fields with the bounded value property exist. PS. One can show, using some machinery from the model theory of arithmetic, that without the use of working only in$AC$, ZF+AC_\omega$, Schmerl's proof can produce a well-orderable field $F$ of cardinality and cofinality $\kappa$ for any uncountable cardinal $\kappa$. Therefore by using $\kappa$ as the cardinal known as "Hartogs" of $\Bbb{R}$" (which is a cardinal \aleph_1$that cannot be injected into satisfies$\Bbb{R}$) BW(\aleph_1)$. This allows one obtains to one obtain a non-archimedean field with the bounded value property, entirely within $ZF$. ZF+AC_\omega$. 5 added 631 characters in body EDIT NOTE 1: "uncountable" was missing from the statement of Sikorski's Theorem in the previous version. Thanks to James Propp, Ricky Demmer, and Emil Jeřábek for catching infelicities of the past versions. EDIT NOTE 2: A PS has been added at the end to indicate why the answer does not change if one is barred from using the the axiom of choice. There are nonarchimedean fields with the bounded value property. Let's begin with a key definition: an ordered field$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 cardinality at most$2^{\aleph_0}$, the following Lemma, when coupled with Sikorski's theorem above (with$\kappa$chosen as any regular cardinal greater than$2^{\aleph_0}$) 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 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$by its absolute value). QED Some references: Sikorski's Theorem appears in: Roman Sikorski, On an ordered algebraic field. Soc. Sci. Lett. Varsovie. C. R. Cl. III. Sci. Math. Phys. 41 (1948), 69–96 (1950). A proof of Sikorski's theorem can also be found in the following paper (Cor. 2.7), as a corollary of a vast generalization of Sikorski's theorem; the paper is an impressive showcase for the interaction between deep methods of models of arithmetic and higher set theory with field theory. James Schmerl, Models of Peano arithmetic and a question of Sikorski on ordered fields. Israel J. Math. 50 (1985), no. 1-2, 145–159. PS. One can show, using some machinery from the model theory of arithmetic, that without the use of$AC$, Schmerl's proof can produce a well-orderable field$F$of cardinality and cofinality$\kappa$for any uncountable cardinal$\kappa$. Therefore by using$\kappa$as the cardinal known as "Hartogs" of$\Bbb{R}$" (which is a cardinal that cannot be injected into$\Bbb{R}$) one obtains a non-archimedean field with the bounded value property, entirely within$ZF\$.