Abstract
An ultraproduct can be a helpful organizing principle in presenting solutions of problems at many levels, as argued by Terence Tao. We apply it here to the solution of a calculus problem: every infinite sequence has a monotone infinite subsequence, and give other applications.
1 Introduction
Solutions to even elementary calculus problems can be tricky but in many cases, enriching the foundational framework available enables one to streamline arguments, yielding proofs that are more natural than the traditionally presented ones.
We explore various proofs of the elementary fact that every infinite sequence has a monotone infinite subsequence, including some that proceed without choosing a convergent one first.
An ultraproduct can be a helpful organizing principle in presenting solutions of problems at many levels, as argued by Terence Tao Tao in [1]. We apply it here to the solution of the problem mentioned above. A related but different problem of proving that every infinite totally ordered set contains a monotone sequence is treated by Hirshfeld in [2, Exercise 1.2, p. 222]. We first present the ultrapower construction in Section 2. Readers familiar with ultraproducts can skip ahead to the proof in Section 3.
2 Ultrapower construction
Let us outline a construction (called an ultrapower) of a hyperreal extension ℝ ↪ ∗ℝ exploited in our solution in Section 3. Let ℝℕ denote the ring of sequences of real numbers, with arithmetic operations defined termwise. Then we have a totally ordered field ∗ℝ = ℝℕ/MAX where “MAX” is a suitable maximal ideal. Elements of ∗ℝ are called hyperreal numbers. Note the formal analogy between the quotient ∗ℝ = ℝℕ/MAX and the construction of the real numbers as equivalence classes of Cauchy sequences of rational numbers. In both cases, the subfield is embedded in the superfield by means of constant sequences, and the ring of sequences is factored by a maximal ideal.
We now describe a construction of such a maximal ideal MAX ⊆ ℝℕ exploiting a suitable finitely additive measure ξ: 𝓟(ℕ) → {0, 1} (thus ξ takes only two values, 0 and 1) taking the value 1 on each cofinite set, [1] where 𝓟(ℕ) is the set of subsets of ℕ. The ideal MAX consists of all “negligible” sequences 〈un〉, i.e., sequences which vanish for a set of indices of full measure ξ, namely, ξ({n ∈ ℕ: un = 0}) = 1. The subset 𝓤 = 𝓤ξ ⊆ 𝓟 (ℕ) consisting of sets of full measure ξ is called a free ultrafilter (these can be shown to exist using Zorn’s lemma). A similar construction applied to ℚ produces the field ∗ℚ of hyperrational numbers. The construction can also be applied to a general ordered set F to obtain an ultrapower extension denoted ∗F = Fℕ/𝓤.
Definition 2.1
The order on the field∗Fis defined by setting
or equivalently {n ∈ ℕ: un < vn} ∈ 𝓤.
In particular, every element x ∈ F is canonically identified with the class [〈x〉] of the constant sequence 〈x〉 with general term x. Then x ∈ ∗F satisfies x < v if and only if {n ∈ ℕ: x < vn} ∈ 𝓤.
3 Solution
Let F be an ordered field. We are mainly interested in the cases F = ℚ and F = ℝ though the arguments go through in greater generality for an arbitrary totally ordered set.
Theorem 3.1
A sequence 〈un〉 of elements ofFnecessarily contains a subsequence 〈unk〉 such that eitherunk ≥ unℓwheneverk > ℓ, orunk ≤ unℓwheneverk > ℓ.
This is an immediate consequence of the following more detailed result.
Theorem 3.2
Letu ∈ ∗F = Fℕ/𝓤 be the element obtained as the equivalence class of the sequence 〈un〉. Consider the partition ℕ = A ⊔ B ⊔ Cwhere A = {n ∈ ℕ: un < u}, B = {n ∈ ℕ: un = u}, C = {n ∈ ℕ: un > u}. Then exactly one of the following three possibilities occurs:
B ∈ 𝓤 and then 〈un〉 contains an infinite constant subsequence;
A ∈ 𝓤 and then 〈un〉 contains an infinite strictly increasing subsequence;
C ∈ 𝓤 and then 〈un〉 contains an infinite strictly decreasing subsequence.
Proof
By the property of an ultrafilter, exactly one of the sets A, B, C is in 𝓤. If B ∈ 𝓤 then u is an element of the subfield F ⊆ ∗F (embedded via constant sequences). Since B ⊆ ℕ is necessarily infinite, enumerating it we obtain the desired subsequence.
Now assume A ∈ 𝓤. We choose any element un1 ∈ A to be the first term in the subsequence. We then inductively choose the index nk+1 > nk in A so that unk+1 is the earliest term greater than unk and therefore closer to u than the previous term unk If the subsequence were to terminate at, say, up, this would imply that {n ∈ ℕ: un ≤ up} ∈ 𝓤 and therefore u ≤ up, contradicting the definition of the set A. Therefore we necessarily obtain an infinite increasing subsequence.
The case C ∈ 𝓤 is similar and results in a decreasing sequence. □
Remark 3.3
The proof is essentially a two-step procedure: (1) we plug the sequence into the ultrapower construction, producing an elementu ∈ ∗F; (2) in each of the cases specified by the element u, we inductively find a monotone subsequence.
The approach exploiting ∗F has the advantage that the proof does not require constructing a completion of the field in the case F = ℚ. To work with the ultrapower, one needs neither advanced logic nor a crash course in NSA, since the ultrapower construction involves merely quotienting by a maximal ideal as is done in any serious undergraduate algebra course (see Section 2).
A monotone sequence can also be chosen by the following more traditional consideration. If the sequence is unbounded, one can choose a sequence that diverges to infinity. If the sequence is bounded, one applies the Bolzano-Weierstrass theorem (each bounded sequence has a convergent subsequence) to extract a convergent subsequence. Finally, a convergent sequence contains a monotone one by analyzing the terms lying on one side of the limit (whichever side has infinitely many terms).
The proof via an ultrapower allows one to bypass the issue of convergence. Once one produces a monotone subsequence, it will also be convergent in the bounded case but only when the field is complete. Furthermore, one avoids the use of the Bolzano–Weierstrass theorem.
Since in the case of F = ℚ the Bolzano–Weierstrass theorem is inapplicable, one would need first to complete ℚ to ℝ by an analytic procedure which is arguably at least as complex as the algebraic construction involved in the ultrapower of Section 2.
There is a clever proof of the same result, as follows (see e.g., problem 6 on page 4 in Newman [3]). Call a term in the sequence a peak if it is larger than everything which comes after it. If there are infinitely many peaks, they form an infinite decreasing subsequence. If there are finitely many peaks, start after the last one. From here on every term has a larger term after it, so one inductively forms an increasing subsequence (from this lemma one derives a simple proof of the Bolzano–Weierstrass theorem).
Remark 3.4
The proof in Newman consists of two steps: (1) introduce the idea of a peak; (2) consider separately the cases when the number of peaks is finite or infinite to produce the desired monotone subsequence. While the basic structure of the proof is similar to that using the ultrapower (see Remark 3.3), the basic difference is that step (1) in Newman is essentially ad-hoc, is tailor-made for this particular problem, and is not applicable to solving other problems. Mean while, the ultrapower construction is applicable in many other situations (see e.g., Section 4).
While the proof in Newman does not rely on an ultrapower, the idea of the ultrapower proof is more straightforward once one is familiar with the ultrapower construction, since it is natural to plug a sequence into it and examine the consequences.
We provide another illustration of how the element u = [〈un〉] can serve as an organizing principle that allows us to detect properties of monotone subsequences. To fix ideas let F = ℝ. An element u ∈ ∗ℝ is called finite if –r < u < r for a suitable r ∈ ℝ. Let 𝔥ℝ ⊆ ∗ℝ be the subring of finite elements of ∗ℝ. The standard part function st: 𝔥ℝ → ℝ rounds off each finite hyperreal u to its nearest real number u0 = st(u).
Proposition 3.5
Ifu ∈ 𝔥ℝ and u > u0then the sequence 〈un〉 possesses a strictly decreasing subsequence.
Proof
Since u > u0 we have {n ∈ ℕ: un > u0} ∈ 𝓤. We start with an arbitrary n1 ∈ {n ∈ ℕ: un > u0} and inductively choose nk+1 so that unk+1 is closer to u than unk. We argue as in the proof of Theorem 3.2 to show that the process cannot terminate and therefore produces an infinite subsequence. □
4 Compactness
A more advanced application is a proof of the nested decreasing sequence property for compact sets (Cantor’s intersection theorem) using the property of saturation. Such a proof exbibits compactness as closely related to the more general property of saturation, shedding new light on the classic property of compactness.
A typical proof of Cantor’s intersection theorem for a nested decreasing sequence of compact subsets An ⊆ ℝ would use the monotone sequence 〈un〉 where un is the minimum of each An. We will present a different and more conceptual proof.
Each set A ⊆ ℝ has a natural extension denoted ∗A ⊆∗ℝ. Similarly, the powerset P = 𝓟(ℝ) has a natural extension ∗ identified with a proper subset of 𝓟(∗ℝ). Each element of ∗𝓟 is naturally identified with a subset of ∗ℝ called an internal set.
The principle of saturation holds for arbitrary nested decreasing sequences of internal sets but we will present it in a following special case.
Theorem 4.1 (Saturation)
If 〈An: n ∈ ℕ〉 is a nested decreasing sequence of nonempty subsets of ℝ then the sequence 〈∗An: n ∈ ℕ〉 has a common point.
Proof
Let ℙ = 𝓟(ℝ) be the set of subsets of ℝ. We view the sequence 〈An ∈ ℙ: n ∈ ℕ〉 as a function f: ℕ → ℙ, n ↦ An. By the extension principle we have a function ∗f: ∗ℕ → ∗ℙ. Let Bn = ∗f(n). For each finite n we have Bn = ∗An ∈ ∗ℙ. For each infinite value of the index n = H the entity BH ∈ ∗ℙ is by definition internal but is not (necessarily) the natural extension of any subset of ℝ.
If 〈An〉 is a nested sequence in ℙ then by transfer 〈Bn: n ∈ ∗ℕ〉 is a nested sequence in ∗ℙ with each Bn nonempty. Let H be a fixed infinite index. Then for each finite n the set ∗An ⊆ ∗ℝ includes BH. Choose any element c ∈ BH. Then c is contained in ∗An for each finite n so that c ∈ ∩n ∈ ℕ∗An as required. □
Remark 4.2
An equivalent formulation of Theorem 4.1 is as follows. If the family of subsets {An}n ∈ ℕhas the finite intersection property then ∃c ∈ ∩n ∈ ℕ∗An.
Let X be a topological space. Let p ∈ X. The halo of p, denoted 𝔥(p) is the intersection of all ∗U where U runs over all neighborhoods of p in X (a neighborhood of p is an open set that contains p). A point y ∈ ∗X is called nearstandard in X if there is p ∈ X such that y ∈ 𝔥(p).
Theorem 4.3
A spaceXis compact if and only if every y ∈ ∗Xis nearstandard inX.
Proof
To prove the direction ⇒, assume X is compact, and let y ∈ ∗X. Let us show that y is nearstandard (this direction does not require saturation). Assume on the contrary that y is not nearstandard. This means that it is not in the halo of any point p ∈ X. This means that every p ∈ X has a neighborhood Up such that y ∉ ∗Up. The collection {Up}p ∈ X is an open cover of X. Since X is compact, the collection has a finite subcover Up1, …, Upn, so that X = Up1 ∪ … ∪ Upn. But for a finite union, the star of union is the union of stars. Thus ∗X is the union of ∗Up1, …, ∗Upn, and so the point y is in one of the sets ∗Up1, …, ∗Upn, a contradiction.
Next we prove the direction ⇐ (this direction exploits saturation). Assume every y ∈ ∗X is nearstandard, and let {Ua} be an open cover of X. We need to find a finite subcover.
Assume on the contrary that the union of any finite collection of Ua is not all of X. Then the complements of Ua are a collection of (closed) sets {Sa} with the finite intersection property. It follows that the collection {∗Sa} similarly has the finite intersection property. By saturation (see Remark 4.2), the intersection of all ∗Sa is non-empty. Let y be a point in this intersection. Let p ∈ X be such that y ∈ 𝔥(p). Now {Ua} is a cover of X so there is a Ub such that p ∈ Ub. But y is in ∗Sa for all a, in particular y ∈ ∗Sb, so it is not in ∗Ub, a contradiction to y ∈ 𝔥(p) □.
Theorem 4.4 (Cantor’s intersection theorem)
A nested decreasing sequence of nonempty compact sets has a common point.
Proof
Given a nested sequence of compact sets Kn, we consider the corresponding decreasing nested sequence of internal sets, 〈∗Kn: n ∈ ℕ〉. This sequence has a common point x by saturation. But for a compact set Kn, every point of ∗Kn is nearstandard (i.e., infinitely close to a point of Kn) by Theorem 4.3. In particular, st(x) ∈ Kn for all n, as required. □
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