Lobachevskii Journal of Mathematics Vol. 11, 2002, 13-18 \begin{matrix}c \\ ◯ \end{matrix}M. Moniri, J. S. Eivazloo

M. Moniri, J. S. Eivazloo

PATHOLOGICAL FUNCTIONS ON PUISEUX SERIES ORDERED FIELDS AND OTHERS

(submitted by M. M. Arslanov)

Abstract. We show that over closed bounded intervals in certain Archimedean ordered fields as well as in all non-Archimedean ones of countable cofinality, there are uniformly continuous 1-1 functions not mapping interior to interior. For the latter kind of fields, there are also uniformly continuous 1-1 functions mapping all interior points to interior points of the image which are, nevertheless, not open. In particular the ordered Laurent and Puiseux series fields with coefficients in any ordered field accommodate both kinds of such strange functions.

1. Introduction and Preliminaries

A subset C of an ordered field F is a cut in there if C < F \ C. A nontrivial cut is a gap whenever it fails to have a least upper bound in the field. A gap G in F is regular when ∀ϵ ∈ F^{>0}, G + ϵ ⊈ G. Ordered fields with no regular gaps are called Scott complete. An ordered field is Archimedean (has no infinitesimals) just in case it is (isomorphic to) a subfield of the real ordered field. The latter is, up to isomorphism, the unique Dedekind Complete ordered field (i.e. has no gaps). It is easy to see that any Dedekind incomplete ordered field has gaps in all its non-degenerated intervals.

Monotone complete ordered fields, introduced in [1], are those with no bounded strictly increasing divergent functions. There exist monotone complete ordered fields of any uncountable regular cardinality, see [5] (Cor. 2.7), so there are many monotone complete ordered fields not isomorphic to ℝ. No such ordered field can be of countable cofinality. Any non-degenerated interval in a monotone incomplete ordered field, contains the image of a strictly increasing divergent function.

2000 Mathematical Subject Classification. 12J15, 26A03.

Key words and phrases. Ordered Fields, gaps, Scott completeness, monotone completeness, cofinality, open maps, Puiseux Series.

The authors thank Institute for Studies in Theoretical Physics and Mathematics for partial support.

For an ordered field F and ordered Abelian group G, recall that the ordered field of generalized power series with exponents in G and coefficients in F is:

[[F^{G}]] = { ∑_{g∈G} a_{g}t^{g} | the support {g|a_{g} ≠ 0} of ∑_{g∈G} a_{g}t^{g} is well ordered }.

It is equipped with the natural addition, multiplication, and order defined by ∑_{g∈G} a_{g}t^{g} > 0 if a_{g_{0}} > 0, where g_{0} is the minimum of Supp(∑_{g∈G} a_{g}t^{g}).

A well known necessary and sufficient condition for [[F^{G}]] to be real closed is that G be divisible and F be real closed, see [4] (6.10). Also [[F^{G}]] is always Scott complete, see [2] (Thm. 3.2 and the Note Added in Proof therein).

The field of Puiseux series with ascending exponents over an ordered field F is defined to be P_{F}^{a} = ∪_{n≥1}[[F^{( \frac{1}{n} ℤ)}]]. They, like other non-Archimedean ordered fields of cofinality ω, are not monotone complete as, e.g., the sequence of natural numbers in them shows. Moreover P_{F}^{a} is real closed if and only if F is so.

The goal in this paper is to show that an important feature of the real ordered field fails in certain real closed ordered fields, namely in all P_{F}^{a}’s with F real closed. An arbitrary ordered field F was proved in [2] (Thm. 1.2) to be Dedekind complete if (and obviously only if) all continuous 1-1 functions defined on some (equivalently all) non-degenerated closed bounded interval(s) of F map interior points [of the interval(s)] to interior points [of their range(s)]. These were further observed in op.cit. (Cor. 1.3) to be equivalent to all continuous 1-1 functions on some (equivalently all) non-degenerated closed bounded interval(s) being open (note that all open maps on an interval in any ordered field map interior points to points in the interior of the image). On the other hand, we proved in [2] (Thm. 2.1) that, over all closed bounded intervals in arbitrary monotone incomplete ordered fields, there are continuous non uniformly continuous unbounded functions whose ranges are not closed and continuous 1-1 functions mapping all interior points to interior of the image which, nevertheless, are not open. Here we show that, in the Archimedean case and provided that their unique continuous extensions to ℝ are 1-1, such rather strange functions can not be uniformly continuous. Also, on closed bounded intervals of certain Archimedean as well as all non-Archimedean ordered fields of cofinality ω, there are uniformly continuous 1-1 functions which do not map interior to interior. In the mentioned non-Archimedean case, there are also uniformly continuous 1-1 functions which do map interior to interior but still are not open. We will finish by presenting some classes of ordered fields, among which the Puiseux fields, over them these phenomena occur.

2. Partial Results for Some Archimedean Ordered Fields

In this section, the matter described above is treated for proper subfields of ℝ.

Lemma 2.1. Let F be an Archimedean ordered field. If f : [0, 1]_{F} → [0, 1]_{F} is a uniformly continuous function, then f can be extended to a unique (uniformly) continuous function \overline{f} on [0, 1]_{ℝ}.

Proof Given x ∈ [0, 1]_{ℝ}, there exists a sequence (r_{n})_{n≥1} in [0, 1]_{ℚ} such that lim_{ℝ} r_{n} = x. As f is uniformly continuous on [0, 1]_{F} which contains [0, 1]_{ℚ} and the sequence (r_{n})_{n≥1} is Cauchy, (f(r_{n}))_{n≥1} is Cauchy in [0, 1]_{F} and so has a limit in [0, 1]_{ℝ}. Let \overline{f}(x) = lim_{ℝ} f(r_{n}). This is well defined, since if (r_{n})_{n≥1} and (s_{n})_{n≥1} are two Cauchy sequences in [0, 1]_{ℚ} such that lim_{ℝ} r_{n} = lim_{ℝ} s_{n}, then by uniform continuity of f, we have lim_{ℝ} f(r_{n}) = lim_{ℝ} f(s_{n}). Notice that \overline{f} is indeed the unique such extension of f. It is also continuous on [0, 1]_{ℝ}: Let x_{0} ∈ [0, 1]_{ℝ} and ϵ ∈ ℝ^{>0}. By uniform continuity of f, there exists δ > 0 such that ∀x, y ∈ [0, 1]_{F} (|x − y| < δ → |f(x) − f(y)| < \frac{ϵ}{2}) (∗). We claim that for this δ, (∀x ∈ [0, 1]_{ℝ})(|x − x_{0}| < δ → |\overline{f}(x) − \overline{f}(x_{0}| < ϵ). Let x ∈ [0, 1]_{ℝ} be such that |x − x_{0}| < δ. There exist sequences (r_{n})_{n≥1} and (s_{n})_{n≥1} in [0, 1]_{ℚ} such that lim_{ℝ} r_{n} = x_{0} and lim_{ℝ} s_{n} = x respectively. Now let N ∈ ℕ be such that (∀n ≥ N)(|r_{n} − s_{n}| < δ), so from (∗) we have (∀n ≥ N)(|f(r_{n}) − f(s_{n})| < \frac{ϵ}{2}). Thus |\overline{f}(x) − \overline{f}(x_{0})| = lim_{ℝ} |f(r_{n}) − f(s_{n})| ≤ \frac{ϵ}{2} < ϵ. □

Proposition 2.2. Let F be an Archimedean ordered field. If f : [0, 1]_{F} → [0, 1]_{F} is a uniformly continuous function whose unique extension (as above) to ℝ is one-to-one, then it maps every open subset of [0, 1]_{F} onto an open subset of f([0, 1]_{F}).

Proof Let U be an open subset of [0, 1]_{F}. There is an open subset V of ℝ such that U = V ∩ [0, 1]_{F}. We have f(U) = \overline{f}(U) = \overline{f}(V ∩ [0, 1]_{F}) = \overline{f}(V ∩ [0, 1]_{ℝ}) ∩ \overline{f}([0, 1]_{F}) = \overline{f}(V ∩ [0, 1]_{ℝ}) ∩ f([0, 1]_{F}). But \overline{f} is open, hence \overline{f}(V ∩ [0, 1]_{ℝ}) is a relatively open subset of \overline{f}([0, 1]_{ℝ}) and so there is an open subset W of ℝ, such that \overline{f}(V ∩ [0, 1]_{ℝ}) ∩ f([0, 1]_{F}) = (W ∩ \overline{f}([0, 1]_{ℝ})) ∩ f([0, 1]_{F}) = W ∩ f([0, 1]_{F}). Therefore f(U) is a relatively open subset of f([0, 1]_{F}). □

It would be nice to know whether there are proper subfields of ℝ for which the unique extension of any uniformly continuous 1-1 function on its unit interval to the unit interval in the reals is still 1-1. We are able to provide examples in the other direction:

Proposition 2.3. There is a uniformly continuous 1-1 function on [0, 1]_{ℚ} whose range has empty interior.

Proof Consider the function f(x) = |\frac{1}{4}x^{2} +x− \frac{1}{2}|. The proof that it is 1-1 is based, after the change of variable x = 2(u − 1), on the fact that the equation r^{2} + s^{2} = 3 is not solvable in ℚ. A further argument shows that complement of the range of f with respect to [0, 1]_{ℚ} is dense in [0, 1]_{ℚ} and so the range, as a subspace of [0, 1]_{ℚ}, has empty interior. □

Similar arguments will work for purely transcendental field extensions of ℚ inside ℝ.

3. The Non-Archimedean Case of Cofinality ω

Theorem 3.1. Let F be a non-Archimedean ordered field with cf(F) = ω. Then for all a < b in F, there exist 1-1 uniformly continuous functions f : [a, b]_{F} → [a, b]_{F} whose ranges are closed with either of the two further (nonoverlapping) properties:

(i) f maps some interior point of [a, b]_{F} to a boundary point of its image. (ii) f does satisfy f((a, b)_{F}) ⊆ (f([a, b]_{F}))^{o}, but is still non-open.

Proof (i) Let [a, b] be a non-degenerated interval in F and c = \frac{a+b}{2}. Fix a strictly increasing sequence (a_{k})_{k∈ω} in [a, c) such that a_{0} = a, (∀k ≥ 1)|a_{k+1} − a_{k}| ≤ \frac{1}{4}|a_{k} − a_{k−1}| and lim_{k} a_{k} = c. Put b_{0} = b, and for all k ≥ 1, b_{k} = b_{0} − (a_{k} − a_{0}). Downward closure of the monad of 0 is an irregular gap G in [0, 1]_{F}. Fix (a ℚ-infinitesimal) γ ∈ G^{>0}. Note that (∀x ∈ G)(∀y ∈ F \ G)(y − x > γ). For each k ∈ ω, let U_{k} be the image of G ∩ [0, 1] under the linear increasing function from [0, 1] onto [a_{k}, a_{k+1}] and V_{k} the image of G ∩ [0, 1] under the linear increasing function from [0, 1] onto [b_{k+1}, b_{k}]. Let S_{0} = U_{0}, T_{0} = [a, b] \ (downward closure of V_{0}), and for all k ≥ 1, S_{k} = U_{k}\U_{k−1}, T_{k} = V_{k−1}\V_{k}. For all k ≥ 1, we have (∀x ∈ S_{k})(∀y ∈ S_{k+1})(y − x > γ(a_{k+1} − a_{k})) (∗) and, from a similar observation for the T_{k}’s, I_{k} = [b_{k} − γ(b_{k} − b_{k+1}), b_{k} + γ(b_{k} − b_{k+1})] ⊆ T_{k}. For k ∈ ω \{0}, let h_{k} be the linear increasing function which maps [a_{k−1}, a_{k+1}] onto I_{2k}. For k = 0, first pick out some d_{0} ∈ T_{0}\{b} such that b−d_{0} < a_{1}−a and put I_{0} = [d_{0}, b]. Then let h_{0} : [a, a_{1}] → I_{0} be the onto linear increasing function. Let f_{k} be the restriction of h_{k} to S_{k}; so f_{k} is a function which maps S_{k} linearly and increasingly into I_{2k} ⊆ T_{2k}. Similarly, for all k ∈ ω, we can get linear increasing functions g_{k} : T_{k} → T_{2k+1}. Let f = (⋃_{k∈ω} f_{k}) ∪ (⋃_{k∈ω} g_{k}) ∪ {(c, c)}. Then f is a one-to-one function from [a, b] into [c, b] with a closed range which maps c ∈ (a, b) to the boundary point c ∈ f([a, b]) ⊆ [c, b].

To prove that f is uniformly continuous, we proceed as follows. Given ϵ ∈ F^{>0} which we may assume without loss of generality to be less than a_{2} − a_{1}, let δ = \frac{1}{4}γϵ. Take x, y ∈ [a, b] with |x − y| < δ. If either x or y (but to avoid trivialities not both) equals c or x < c < y, then one easily checks that |f(x) − f(y)| < |x − y|. So assume both x and y are strictly less or greater than c, they will be either both in two S’s or both in two T ’s. The arguments will be similar, we only deal with the S’s. Suppose x ∈ S_{k} and y ∈ S_{l}. There are the following exclusive cases of how l is compared to k.

(A) l = k. Here f = f_{k} = h_{k} ↾ S_{k}. If k = 0, then by the condition on d_{0}, we have |f(x) − f(y)| = |f_{0}(x) − f_{0}(y)| < |x − y|. Assume k ≥ 1. Then h_{k} : [a_{k−1}, a_{k+1}] → I_{2k} is linear and |I_{2k}| < (a_{k+1} − a_{k−1}), so |f(x) − f(y)| = |f_{k}(x) − f_{k}(y)| < |x − y| < δ which is infinitely smaller than ϵ.

(B) l = k + 1. By the choice of δ, we must have k ≥ 2. The reason is that we will have δ < γ(a_{2} − a_{1}) (and in particular δ < γ(a_{1} − a_{0})). We have |f(x) − f(y)| = |f_{k}(x) − f_{k+1}(y)| < |b_{2(k+1)+1} − b_{2k−1}| ≤ |b_{k+1} − b_{k}| = |a_{k+1} − a_{k}| < [by (∗)] \frac{1}{γ}|x − y| < \frac{1}{γ}δ = \frac{1}{4}ϵ < ϵ.

(C) k ≥ 2 and l = k + 2. We will have |f(x) − f(y)| ≤ |f(x) − f(a_{k+1})| + |f(a_{k+1}) − f(y)|. Both these terms are less than \frac{1}{4}ϵ by case (B).

(D) k ≥ 2 and l ≥ k + 3. We will have |f(x) − f(y)| < |b_{2l+1} − b_{2k−1}|. Using |a_{i+1} − a_{i}| ≤ \frac{1}{4}|a_{i} − a_{i−1}|, one gets |b_{2l+1} − b_{2k−1}| < |a_{k+1} − a_{l−1}|. The latter value is less than |x − y|.

(E) k = 0 and l ≥ 2. We will have |f(x) − f(y)| ≤ |f(x) − f(a_{1})| + |f(a_{1}) − f(a_{2})| + |f(a_{2}) − f(y)|. The first two terms on the right side are, by case (B), less than \frac{1}{4}ϵ. The last one is also less than \frac{1}{4}ϵ: If l = 2, resp. l = 3, resp. l = 4, resp. l ≥ 5, then one uses the case (A), resp. (B), resp. (C), resp. (D).

(F) k = 1 and l ≥ 3. We will have |f(x) − f(y)| ≤ |f(x) − f(a_{2})| + |f(a_{2}) − f(y)|. The first term on the right is, by case (B), less than \frac{1}{4}ϵ. The second one is also less than \frac{1}{4}ϵ: If l = 3, resp. l = 4, resp. l ≥ 5, then one uses the case (B), resp. (C), resp. (D).

(ii) Let G, γ, a_{n}, b_{n}, U_{n}, V_{n}, S_{n} and T_{n} be as above. Consider the function f : [a, b] → [a, b] mapping S_{k} to T_{2k}, for all k ∈ ω, and T_{k} to T_{2k+1}, for k > 0, linearly and increasingly and c to c as in the proof of part (i), while mapping T_{0} onto [a, c) in the following fashion. Let (d_{i})_{i≥0}, with d_{0} = b, be a strictly decreasing ω-sequence which traverses T_{0} backwards (this exists since F is monotone incomplete and of cofinality ω). Now let f map (d_{i+1}, d_{i}] linearly and decreasingly onto [a_{i}, a_{i+1}).

By construction, the resulting function f is one-to-one, has a closed range, maps every interior point to an interior one but, nevertheless, is not open [for example c is a boundary point of f((a_{1}, b_{1}))].

To see that f is uniformly continuous, we argue as follows. Given ϵ ∈ F^{>0} which once again is assumed less than a_{2} − a_{1}, let δ = min\{δ_{1}, \frac{1}{4}γϵ\}, where δ_{1} = min\{\frac{a_{i+1}−a_{i}}{d_{i}−d_{i+1}} , d_{i} − d_{i+1} : i < N\} in which N is chosen so that for all n ≥ N, |a_{n} − c| < \frac{1}{3}ϵ. Take x, y ∈ [a, b] with |x − y| < δ. If neither x nor y belongs to T_{0}, then by part (i), |f(x) − f(y)| < ϵ. Note that it can not be the case that exactly one of these two points belongs to T_{0}, since δ < γ(a_{1} − a_{0}). The case when both belong to T_{0} is further treated in three subcases. If x, y ≤ d_{N}, then |f(x) − f(y)| < c − a_{N} < \frac{1}{3}ϵ < ϵ. If x, y ≥ d_{N}, then there will exist i < N such that either x, y ∈ (d_{i+1}, d_{i}] (in which case |f(x) − f(y)| < \frac{1}{3}ϵ) or x ∈ (d_{i+1}, d_{i}] and y ∈ (d_{i}, d_{i−1}] (in which case we will have |f(x) − f(y)| < |f(x) − a_{i}| + |a_{i} − f(y)| < \frac{1}{3}ϵ + \frac{1}{3}ϵ < ϵ). If x ≤ d_{N} ≤ y, then |f(x) − f(y)| < |f(x) − a_{N}| + |a_{N} − f(y)| < \frac{1}{3}ϵ + \frac{2}{3}ϵ = ϵ. □

We finish by presenting some non-Archimedean ordered fields allowing both kinds of the mentioned strange functions.

Example 3.2. Consider the ordered field (ℚ(x), +, ·, <) with x > ℚ. It is non-Archimedean and of cofinality ω, hence monotone incomplete. So there are uniformly continuous one-to-one functions f, g : [0, 1]_{ℚ(x)} → [0, 1]_{ℚ(x)} such that f maps some interior point to a boundary point of its image, while g does not do that but is still non-open.

Example 3.3. Let F be an ordered field and G an ordered group of cofinality ω. Then [[F^{G}]] is non-Archimedean of cofinalty ω, and so monotone incomplete. E.g., all ordered Laurent series fields (where exponents come from the integers and coefficients from an arbitrary ordered field) are such. Once again, there will exist both of the mentioned kinds of functions over closed bounded intervals of these. As mentioned before, these are all Scott complete (see [3]), moreover if G is divisible and F is real closed, then [[F^{G}]] is also real closed.

Example 3.4. For any real closed field F, the Puiseux field P_{F}^{a} is real closed, Scott incomplete, non-Archimedean of cofinality ω. Therefore these fields allow such wild functions.

18 M. MONIRI, J. S. EIVAZLOO

References

[1] H. J. Keisler, Monotone Complete Fields, in: Victoria Symposium on Nonstanard Analysis (A. Hurd and P. Loeb, eds.), Lecture Notes in Mathematics 369, pp. 113-115, Springer-Verlag, Berlin, 1974.

[2] M. Moniri and J.S. Eivazloo, Using Nets in Dedekind, Monotone, or Scott Incomplete Ordered Fields and Definability Issues, Proceedings of the Ninth Prague Topological Symposium (P. Simon, ed.) pp. 195-203, Topology Atlas (available electronically at http://at.yorku.ca), Toronto, 2002.

[3] M. Moniri and J.S. Eivazloo, Some Monotone or Scott Complete Ordered Fields without Integer Parts, Abstract of a talk at Algebra Conference, Venezia 2002.

[4] P. Ribenboim, Fields: Algebraically Closed and Others, Manuscripta Math. 75 (1992) 115-150.

[5] J. H. Schmerl, Models of Peano Arithmetic and a Question of Sikorski on Ordered Fields, Israel J. Math. 50 (1985) 145-159.

Institute for Studies in Theoretical Physics and Mathematics (IPM),Tehran, Iran E-mail address: mojmon@ipm.ir, eivazloo@ipm.ir Department of Mathematics, Tarbiat Modarres University, Tehran, Iran E-mail address: moniri m@modares.ac.ir, eivazl j@modares.ac.ir

Received May 14, 2002