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LECTURE NOTES ON NONSTANDARD ANALYSIS
UCLA SUMMER SCHOOL IN LOGIC
ISAAC GOLDBRING
Date: November 10, 2014.
Nonstandard analysis was invented by Abraham Robinson in the 1960s as a way to rescue the na¨ıve use of infinitesimal and infinite elements favored by mathematicians such as Leibniz and Euler before the advent of the rigorous methods introduced by Cauchy and Weierstrauss. Indeed, Robinson realized that the compactness theorem of first-order logic could be used to provide fields that “logically behaved” like the ordered real field while containing
- The hyperreals Contents
- 1.1. Basic facts about the ordered real field
- 1.2. The nonstandard extension
- 1.3. Arithmetic in the hyperreals
- 1.4. The structure of N∗
- 1.5. More practice with transfer
- 1.6. Problems
- Logical formalisms for nonstandard extensions
- 2.1. Approach 1: The compactness theorem
- 2.2. Approach 2: The ultrapower construction
- 2.3. Problems
- Sequences and series
- 3.1. First results about sequences
- 3.2. Cluster points
- 3.3. Series
- 3.4. Problems
- Continuity
- 4.1. First results about continuity
- 4.2. Uniform continuity
- 4.3. Sequences of functions
- 4.4. Problems
- Differentiation
- 5.1. The derivative
- 5.2. Continuous differentiability
- 5.3. Problems
- Riemann Integration
- 6.1. Hyperfinite Riemann sums and integrability
- 6.2. The Peano Existence Theorem
- 6.3. Problems
- Weekend Problem Set #1
- Many-sorted and Higher-Type Structures
- 8.1. Many-sorted structures
- 8.2. Higher-type sorts 2 ISAAC GOLDBRING
- 8.3. Saturation
- 8.4. Useful nonstandard principles
- 8.5. Recap: the nonstandard setting
- 8.6. Problems
- Metric Space Topology
- 9.1. Open and closed sets, compactness, completeness
- 9.2. More about continuity
- 9.3. Compact maps
- 9.4. Problems
- Banach Spaces
- 10.1. Normed spaces
- 10.2. Bounded linear maps
- 10.3. Finite-dimensional spaces and compact linear maps
- 10.4. Problems
- Hilbert Spaces
- 11.1. Inner product spaces
- 11.2. Orthonormal bases and `^2
- 11.3. Orthogonal projections
- 11.4. Hyperfinite-dimensional subspaces
- 11.5. Problems
- Weekend Problem Set #2
- The Spectral Theorem for compact hermitian operators
- 13.1. Problems
- The Bernstein-Robinson Theorem
- Measure Theory
- 15.1. General measure theory
- 15.2. Loeb measure
- 15.3. Product measure
- 15.4. Integration
- 15.5. Conditional expectation
- 15.6. Problems
- Szemer´edi Regularity Lemma
- 16.1. Problems
- References
4 ISAAC GOLDBRING
- if a ≤ c for all a ∈ A, then b ≤ c (b is the least upper bound for A).
Such b is easily seen to be unique and is called the least upper bound of A, or the supremum of A, and is denote sup(A).
Exercise 1.2. Show that if A is nonempty and bounded below, then A has a greatest lower bound. The greatest lower bound is also called the infimum of A and is denoted inf(A).
1.2. The nonstandard extension. In order to start “doing” nonstandard analysis as quickly as possible, we will postpone a formal construction of the nonstandard universe. Instead, we will pose some postulates that a nonstan- dard universe should possess, assume the existence of such a nonstandard universe, and then begin reasoning in this nonstandard universe. Of course, after we have seen the merits of some nonstandard reasoning, we will return and give a couple of rigorous constructions of nonstandard universes. We will work in a nonstandard universe R∗^ that has the following prop- erties:
(NS1) (R; +, ·, 0 , 1 , <) is an ordered subfield of (R∗; +, ·, 0 , 1 , <). (NS2) R∗^ has a positive infinitesimal element, that is, there is ∈ R∗^ such that > 0 but < r for every r ∈ R>^0. (NS3) For every n ∈ N and every function f : Rn^ → R, there is a “natural extension” f : (R∗)n^ → R∗. The natural extensions of the field operations +, · : R^2 → R coincide with the field operations in R∗. Similarly, for every A ⊆ Rn, there is a subset A∗^ ⊆ (R∗)n^ such that A∗^ ∩ Rn^ = A. (NS4) R∗, equipped with the above assignment of extensions of functions and subsets, “behaves logically” like R. The last item in the above list is, of course, extremely vague and imprecise. We will need to discuss some logic in order to carefully explain what we mean by this. Roughly speaking, any statement that is expressible in first-order logic and mentioning only standard numbers is true in R if and only if it is true in R∗. This is often referred to as the Transfer Principle, although, logically speaking, we are just requiring that R∗, in a suitable first-order language, be an elementary extension of R. We will explain this in more detail later in these notes. That being said, until we rigorously explain the logical formalism of non- standard analysis, we should caution the reader that typical transferrable statements involve quantifiers over numbers and not sets of numbers. For example, the completeness property for R says that “for all sets of num- bers A that are nonempty and bounded above, sup(A) exists.” This is an example of a statement that is not transferrable; see Exercise 1.8 below.
Definition 1.3. R∗^ is called the ordered field of hyperreals.
Remark. If f : A → R is a function, where A ⊆ Rn, we would like to also consider its nonstandard extension f : A∗^ → R∗. We will take care of this matter shortly.
LECTURE NOTES ON NONSTANDARD ANALYSIS 5
1.3. Arithmetic in the hyperreals. First, let’s discuss some immediate consequences of the above postulates. Since R∗^ is an ordered field, we can start performing the field operations to our positive infinitesimal . For example, has an additive inverse −, which is then a negative infinitesimal. Also, we can consider π · ; it is reasonably easy to see that π · is also a positive infinitesimal. (This will also follow from a more general principle that we will shortly see.) Since 6 = 0, it has a multiplicative inverse −^1. For a given r ∈ R>^0 , since < (^1) r , we see that −^1 > r. Since r was an arbitrary positive real number, we see that −^1 is a positive infinite element. And of course, −−^1 is a negative infinite element. But now we can continue playing, considering numbers like
2 · −^3 and so on... And besides algebraic manipulations, we also have transcendental matters to consider. Indeed, we have the nonstandard extension of the function sin : R∗^ → R∗; what is sin()? All in due time... First, let’s make precise some of the words we have been thus far freely tossing around.
Definition 1.4.
(1) The set of finite hyperreals is Rfin := {x ∈ R∗^ | |x| ≤ n for some n ∈ N}. (2) The set of infinite hyperreals is Rinf := R∗^ \ Rfin. (3) The set of infinitesimal hyperreals is
μ := {x ∈ R∗^ | |x| ≤
n
for all n ∈ N>^0 }.
The notation μ comes from the more general notion of monad, which we will encounter later in the notes. Observe that μ ⊆ Rfin, R ⊆ Rfin, and μ ∩ R = { 0 }. Also note that if δ ∈ μ \ { 0 }, then δ−^1 ∈/ Rfin.
Lemma 1.5.
(1) Rfin is a subring of R∗: for all x, y ∈ Rfin, x ± y, x · y ∈ Rfin. (2) μ is an ideal of Rfin: μ is a subring of R∗^ and for all x ∈ Rfin and y ∈ μ, we have xy ∈ μ.
Proof. (1) Fix x, y ∈ Rfin. Choose r, s ∈ R>^0 such that |x| ≤ r and |y| ≤ s. Then |x ± y| ≤ r + s and |xy| ≤ rs, whence x ± y, xy ∈ Rfin. (2) Suppose x, y ∈ μ. We need to show that x ± y ∈ μ. Fix r ∈ R>^0 ; we need |x ± y| ≤ r. Well, since x, y ∈ μ, we have that |x|, |y| ≤ r 2. Then, |x ± y| ≤ |x| + |y| ≤ r 2 + r 2 = r. Now suppose x ∈ Rfin and y ∈ μ. We need xy ∈ μ. Fix r ∈ R>^0 ; we need |xy| ≤ r. Choose s ∈ R>^0 such that |x| ≤ s. Since y ∈ μ, we have |y| ≤ rs. Thus, |xy| = |x||y| ≤ s · rs = r.
A natural question now arises: What is the quotient ring Rfin/μ? The answer will arrive shortly.
LECTURE NOTES ON NONSTANDARD ANALYSIS 7
(1) x ≈ y if and only if st(x) = st(y). (2) If x ≤ y, then st(x) ≤ st(y). The converse of this statement is false; give an example. (3) If x ∈ R, then st(x) = x.
Theorem 1.11. st : Rfin → R is a surjective ring homomorphism: for all x, y ∈ Rfin, st(x + y) = st(x) + st(y) and st(xy) = st(x) st(y).
Proof. This follows immediately from Exercises 1.7(2) and 1.10(3).
Corollary 1.12. Rfin/μ ∼= R.
Proof. The kernel of st is precisely μ; now use the First Isomorphism The- orem for rings.
Corollary 1.13. μ is a maximal ideal of Rfin.
Proof. This follows from the fact that Rfin/μ is a field.
Exercise 1.14. Give a direct proof of the last corollary, that is, show di- rectly that μ is a maximal ideal of Rfin.
1.4. The structure of N∗. In this subsection, let’s take a brief look at the picture of N∗. First, let’s establish that N∗^ \ N 6 = ∅. To see this, let y ∈ R∗ be positive infinite. Since the statement “for all x ∈ R, if x > 0, then there is n ∈ N such that x ≤ n” is true in R, the statement “for all x ∈ R∗, if x > 0, then there is n ∈ N∗^ such that x ≤ n” is true in R∗^ by the transfer principle. Thus, there is N ∈ N∗^ such that y ≤ N. However, if N ∈ N, then y is finite, a contradiction. Thus, N ∈ N∗^ \ N. Also note that the same argument implies that N is positive infinite. The last sentence of the previous paragraph holds for all N ∈ N∗^ \ N: if N ∈ N∗^ \ N, then N is positive infinite. Indeed, the statement “for all n ∈ N, n ≥ 0” is true in R, so the statement “for all n ∈ N∗, n ≥ 0” is true in R∗, whence N ≥ 0. Also, if N ∈ Rfin, then there is n ∈ N such that n ≤ N ≤ n + 1. However, the statement “for all m ∈ N, if n ≤ m ≤ n + 1, then m = n or m = n+1” is true in R; applying the transfer principle to this statement, we have N = n or N = n + 1, whence N ∈ N, a contradiction. Thus, N ∈ Rinf. Now that we know that all nonstandard natural numbers are positive infinite, let’s ask the question: “How many nonstandard natural numbers are there?” To examine this question, let’s first establish some notation and terminology. For N ∈ N∗, set γ(N ) := {N ± m | m ∈ N}, the galaxy or archimedean class of N. Clearly, N ∈ N if and only if γ(N ) = Z; we call this the finite galaxy, while all other galaxies will be referred to as infinite galaxies.
Lemma 1.15. If N ∈ N∗^ \ N, then γ(N ) ⊆ N∗.
Proof. By transfer, N +1 ∈ N∗; by induction, this shows that N +m ∈ N∗^ for all m ∈ N. We now show, also inductively, that N − m ∈ N∗^ for all m ∈ N.
8 ISAAC GOLDBRING
Suppose that the result is true for a given m. Notice that N − m 6 = 0, else N = m ∈ N. Applying transfer to the statement “for all n ∈ N, if n 6 = 0, then n − 1 ∈ N,” we see that (N − m) − 1 = N − (m + 1) ∈ N∗.
Since we know we have at least one nonstandard natural number, we now know that we have an entire galaxy of them. Notice that a galaxy looks just like a copy of Z and that γ(M ) = γ(N ) if and only if |M − N | ∈ N. Observe that if γ(M ) = γ(M ′) and γ(N ) = γ(N ′) and γ(M ) 6 = γ(N ), then M < N if and only if M ′^ < N ′. Consequently, we can define an ordering on galaxies: if γ(M ) 6 = γ(N ), then we say γ(M ) < γ(N ) if and only if M < N. When γ(M ) < γ(N ), we think of M as being infinitely less than N. What can be said about the ordering of the set of galaxies? In particular, are there more than just two galaxies?
Lemma 1.16. The set of infinite galaxies is densely ordered without end- points, meaning:
(1) there is no largest infinite galaxy, that is, for every M ∈ N∗^ \ N, there is N ∈ N∗^ \ N such that γ(M ) < γ(N ); (2) there is no smallest infinite galaxy, that is, for every M ∈ N∗^ \ N, there is N ∈ N∗^ \ N such that γ(N ) < γ(M ); (3) between any two infinite galaxies, there is a third (infinite) galaxy, that is, for every M 1 , M 2 ∈ N∗^ \ N such that γ(M 1 ) < γ(M 2 ), there is N ∈ N∗^ \ N such that γ(M 1 ) < γ(N ) < γ(M 2 ).
Proof. (1) Given M ∈ N∗^ \ N, we claim that γ(M ) < γ(2M ). Otherwise, 2 M = M + m for some m ∈ N, whence M = m, a contradiction. (2) Since γ(M ) = γ(M −1), we may as well suppose that M is even. Then γ( M 2 ) < γ(M ) from the proof of (1); it remains to note that M 2 ∈ N∗^ \ N. (3) Again, we may as well assume that M 1 and M 2 are both even. In this case, arguing as before, one can see that γ(M 1 ) < γ( M^1 + 2 M^2 ) < γ(M 2 ).
Under suitable richness assumptions on the nonstandard extension (to be discussed later), one can go even further: if (Nα)α<κ is a descending sequence of nonstandard natural numbers, then there is N ∈ N∗^ \ N such that N < Nα for all α < κ.
1.5. More practice with transfer. In order to get some practice with the Transfer Principle, we will prove the assertion made in Remark 1.3. More precisely:
Theorem 1.17. The statement “every finite element of R∗^ has a standard part” implies the Completeness Property of the ordered real field.
Proof. Suppose that A ⊆ R is nonempty and bounded above. We must show that sup(A) exists. Let b ∈ R be an upper bound for A. Let’s define a function f : R → R as follows: If r ∈ R \ N, set f (r) = 0. Otherwise, set f (n) = the least k ∈ Z such that kn is an upper bound for A; such a k exists
10 ISAAC GOLDBRING
(b) x · y ≈ x′^ · y′; (c) xy ≈ x
′ y′^ if^ y^6 ≈^ 0. Show that (c) can fail if y, y′^ ∈ μ \ { 0 }.
Problem 1.2. Suppose x, y ∈ R∗^ and x ≈ y. Show that if b ∈ Rfin, then bx ≈ by. Show that this can fail if b /∈ Rfin.
Problem 1.3.
(1) Show that R∗^ is not complete by finding a nonempty subset of R∗ which is bounded above that does not have a supremum. (2) Show that if A ⊆ R is unbounded, then A has no least upper bound when considered as a subset of R∗. (This may even be how you solved part (a).)
Problem 1.4. Let F be an ordered field. F is said to be archimedean if for any x, y ∈ F with x, y > 0, there is n ∈ N such that y < nx. Show that R∗^ is not archimedean. (It is a fact that archimedean ordered fields are complete, so this problem strengthens the result of the previous problem.)
Problem 1.5. Construct a sequence of subsets (An) of R such that
⋃^ ∞
n=
An)∗^6 =
⋃^ ∞
n=
A∗ n.
Problem 1.6. If F is a field and V is an F -vector space, let dimF (V ) denote the dimension of V as an F -vector space.
(1) Observe that R∗^ is an R∗-vector space. (More generally, any field F is naturally an F -vector space.) What is dimR∗ (R∗)? (2) Observe that R∗^ is also a vector space over R. Show that dimR(R∗) = ∞. (Hint: Let x ∈ R∗^ \ R. Show that { 1 , x, x^2 ,... , xn,.. .} is an R- linearly independent set.) (3) Show that Rfin is an R-subspace of R∗. What is dimR(Rfin)? (4) Show that μ is an R-subspace of R∗. What is dimR(μ)? (5) What is dimR(R∗/Rfin)? (6) What is dimR(Rfin/μ)?
Problem 1.7. Show that card(N∗) ≥ 2 ℵ^0 ; here card(A) denotes the cardi- nality of the set A. (Hint: First show that card(Q∗) ≥ 2 ℵ^0 ).
Problem 1.8. Give a direct proof that μ is a maximal ideal of Rfin, that is, show that if I is an ideal of Rfin such that μ ⊆ I, then I = μ or I = Rfin.
- Logical formalisms for nonstandard extensions
At this point, you might be wondering one of two things: (1) What else can I do with these wonderful postulates for nonstandard extensions? or (2) Does such a nonstandard extension exist or was everything done in Section 1 all magical nonsense?
LECTURE NOTES ON NONSTANDARD ANALYSIS 11
If you asked the former question, you can safely skip this section and discover the wonders of the nonstandard calculus to come in the follow- ing sections. (But please, at some point, return and read this section!) If you asked the latter question, we will ease your trepidations by offering not one, but two, different logical formalisms for nonstandard extensions. The first formalism will rely heavily on the Compactness Theorem from first-order logic, but, modulo that prerequisite, this route is the quickest way to obtain nonstandard extensions. The second formalism is the Ultra- product Approach, which is the most algebraic and “mainstream” way to explain nonstandard methods to “ordinary” mathematicians. Of course, at some point, some logic must be introduced in the form of Los’ (pronounced “Wash’s”) theorem, which will be discussed as well.
2.1. Approach 1: The compactness theorem. In this section, some familiarity with first-order logic is assumed. We let L denote the first-order language consisting of the following symbols:
- for every r ∈ R, we have a constant symbol cr;
- for every n > 0 and every f : Rn^ → R, we have a n-ary function symbol Ff ;
- for every n > 0 and every A ⊆ Rn, we have a n-ary relation symbol PA. We let R be the L-structure whose universe is R and whose symbols are interpreted in the natural way, namely (cr)R^ = r, (Ff )R^ = f , and (PA)R^ = A. Let Γ = Th(R) ∪ {c 0 < v < cr | r ∈ R>^0 }, a set of L-formulae. Then Γ is finitely satisfiable: for any r 1 ,... , rn ∈ R>^0 , choose r ∈ R>^0 such that r < ri for i = 1,... , n. Then R |= Th(R) ∪ {c 0 < v < cri | i = 1,... , n}JrK. By the Compactness Theorem, Γ is satisfiable, say by some L-structure A and a ∈ A. Intuitively, a will be our desired positive infinitesimal. But first, we need R to be a substructure of A; some “abstract nonsense” can help take care of this.
Exercise 2.1.
(1) The function h : R → A given by h(r) = (cr)A^ is an injective homomorphism of L-structures. (2) Use (1) to find some L-structure A′^ isomorphic to A such that R is a substructure of A′.
By the result of the previous exercise, we may suppose that R is a sub- structure of A. In this case, we denote A by R∗^ and denote the universe of R∗^ by R∗. For f : Rn^ → R, we let f : (R∗)n^ → R∗^ denote (Ff )R
∗
. For A ⊆ Rn, we set A∗^ := (PA)R
∗
. These are the extensions that axiom (NS3) postulates. Since being an ordered field is part of Th(R), we have that R∗ is an ordered field and, since R is a substructure of R∗, we have that R is an ordered subfield of R∗, verifying postulate (NS1). Let ∈ R∗^ be such
LECTURE NOTES ON NONSTANDARD ANALYSIS 13
rational numbers (qn), where (qn) and (q′ n) are equivalent if they “represent the same real number,” or, more formally, if limn→∞(qn − q n′) = 0. We run into the same issue here: many sequences of real numbers should represent the same hyperreal number. For instance, it should hopefully be clear that the sequence (1, 2 , 3 ,... , n, n + 1,.. .) and (π, e, − 72 , 4 , 5 , 6 ,... , n, n + 1,.. .) should represent the same (infinite) hyperreal number as they only differ in a finite number of coordinates. More generally, we would like to say that two sequences of real numbers represent the same hyperreal number if they agree on “most” coordinates. But what is a good notion of “most” coordinates? A first guess might be that “most” means all but finitely many; it turns out that this guess is insufficient for our purposes. Instead, we will need a slightly more general notion of when two sequences agree on a large number of coordinates; this brings in the notion of a filter.
Definition 2.7. A (proper) filter on N is a set F of subsets of N (that is, F ⊆ P(N)) such that:
- ∅ ∈ F/ , N ∈ F;
- if A, B ∈ F, then A ∩ B ∈ F ;
- if A ∈ F and A ⊆ B, then B ∈ F. We think of elements of F as “big” sets (because that’s what filters do, they catch the big objects). The first and third axioms are (hopefully) intuitive properties of big sets. Perhaps the second axiom is not as intuitive, but if one thinks of the complement of a big set as a “small” set, then the second axiom asserts that the union of two small sets is small (which is hopefully more intuitive).
Exercise 2.8. Set F := {A ⊆ N | N \ A is finite}. Prove that F is a filter on N, called the Frechet or cofinite filter on N.
Exercise 2.9. Suppose that D is a set of subsets of N with the finite intersec- tion property, namely, whenever D 1 ,... , Dn ∈ D, we have D 1 ∩ · · · ∩ Dn 6 = ∅. Set
〈D〉 := {E ⊆ N | D 1 ∩ · · · Dn ⊆ E for some D 1 ,... , Dn ∈ D}.
Show that 〈D〉 is the smallest filter on N containing D, called the filter generated by D.
If F is a filter on N, then a subset of N cannot be simultaneously big and small (that is, both it and it’s complement belong to F), but there is no requirement that it be one of the two. It will be desirable (for reasons that will become clear in a second) to add this as an additional property:
Definition 2.10. If F is a filter on N, then F is an ultrafilter if, for any A ⊆ N, either A ∈ F or N \ A ∈ F (but not both!).
Ultrafilters are usually denoted by U. Observe that the Frechet filter on N is not an ultrafilter since there are sets A ⊆ N such that A and N \ A
14 ISAAC GOLDBRING
are both infinite (e.g. the even numbers). So what is an example of an ultrafilter on N?
Definition 2.11. Given m ∈ N, set Fm := {A ⊆ N | m ∈ A}.
Exercise 2.12. For m ∈ N, prove that Fm is an ultrafilter on N, called the principal ultrafilter generated by m.
We say that an ultrafilter U on N is principal if U = Fm for some m ∈ N. Although principal ultrafilters settle the question of the existence of ultrafilters, they will turn out to be useless for our purposes, as we will see in a few moments.
Exercise 2.13. Prove that an ultrafilter U on N is principal if and only if there is a finite set A ⊆ N such that A ∈ U.
Exercise 2.14. Suppose that U is an ultrafilter on N and A 1 ,... , An are pairwise disjoint subsets of N such that A 1 ∪ · · · ∪ An ∈ U. Prove that there is a unique i ∈ { 1 ,... , n} such that Ai ∈ U.
We are now ready to explain the ultrapower construction. Fix an ul- trafilter U on N. If (an) and (bn) are infinite sequences of real numbers, we say that (an) and (bn) are equal modulo U, written (an) ∼U (bn), if {n ∈ N| | an = bn} ∈ U. (This is the precise meaning of when two sequences agree on “most” coordinates.)
Exercise 2.15. Show that ∼U is an equivalence relation on the set of infinite sequences of real numbers. (The “ultra” assumption is not used in this exercsie.)
For an infinite sequence (an), we write [(an)]U , or simply [(an)], for the equivalence class of (an) with respect to ∼U. We let RU^ denote the set of ∼U -equivalence classes. We want to turn RU^ into an ordered field. The natural guess for the field operations are:
- [(an)] + [(bn)] := [(an + bn)]
- [(an)] · [(bn)] := [(anbn)]. Of course, since we are working with equivalence classes, we need to make sure that the above operations are well-defined. For example, if (an) ∼U (a′ n) and (bn) ∼U (b′ n), we need to know that (an + bn) ∼U (a′ n + b′ n) (and similarly for multiplication). However, if we set A := {n ∈ N | an = a′ n} and B := {n ∈ N | bn = b′ n}, then for n ∈ A ∩ B, an + bn = a′ n + b′ n; since A ∩ B ∈ U and A ∩ B ⊆ {n ∈ N | an + bn = a′ n + b′ n}, this shows that (an + bn) ∼U (a′ n + b′ n). Next we define an order on RU^ by declaring [(an)] < [(bn)] if and only if {n ∈ N | an < bn} ∈ U. One needs to verify that this is in fact a linear order. For example, to check the “linearity” axioms, fix [(an)], [(bn)] ∈ RU^. Set A := {n ∈ N | an < bn}, B := {n ∈ N | an = bn}, and C = {n ∈ N| an > bn}. Then N = A ∪ B ∪ C and A, B, C are pairwise disjoint.
16 ISAAC GOLDBRING
N \ X is finite, whence U is principal by Exercise 2.13. Consequently X ∈ U and α is infinite. For axiom (NS3), we define nonstandard extensions of sets and functions as follows:
- For f : Rn^ → R, we define f : (RU^ )n^ → RU^ by
f ([(a^1 m)],... , [(anm)]) := [(f (a^1 m,... , anm))].
- For A ⊆ Rn, we define A∗^ ⊆ (RU^ )n^ by
([(a^1 m)],... , [(anm)]) ∈ A∗^ if and only if {m ∈ N | (a^1 m,... , anm) ∈ A} ∈ U.
As before, one must check that these operations are well-defined (we leave this to the reader) and, after identifying R with d(R), these functions and relations really do “extend” the original functions and relations (again rele- gated to the lucky reader to verify). Finally, what about (NS4)? It is here that logic must reenter the picture in some shape or form. (Up until this point, the nonlogician aiming to use nonstandard methods via the ultrapower approach has been content.) Let L be the first-order language described in the previous subsection. We make RU^ into an L-structure RU^ by interpreting Ff and PA as the extensions defined above. Then the precise formulation of (NS4) is the following:
Theorem 2.19 (Los’). Suppose that ϕ(v 1 ,... , vn) is an L-formula and [(a^1 m)],... , [(anm)] ∈ RU^. Then
RU^ |= ϕJ[(a^1 m)],... , [(anm)]K if and only if {m ∈ N | R |= ϕJa^1 m,... , anmK} ∈ U.
Proof. A useful exercise in logic; proceed by induction on the complexity of ϕ.
Observe that, as a corollary of Los’ theorem, that RU^ is an ordered field (as these axioms are first-order). The analyst trying to refrain from logic surely avoids Los’ theorem, but in practice, ends up repeatedly verifying its conclusion on a case-by-case basis. In summary, we obtain:
Theorem 2.20. RU^ is a nonstandard universe.
2.3. Problems.
Problem 2.1. Discuss how to make define the nonstandard extension of functions f : A → B with A ⊆ Rm^ and B ⊆ Rn.
Problem 2.2. Let A ⊆ Rm^ and B ⊆ Rn.
(1) Suppose f : A → B is 1-1. Show that if a ∈ A∗^ \ A, then f (a) ∈ B∗^ \ B. (2) Show that A is finite if and only if A∗^ = A.
LECTURE NOTES ON NONSTANDARD ANALYSIS 17
- Sequences and series
3.1. First results about sequences. OK, so let’s start doing some calcu- lus nonstandardly. We start by studying sequences and series. A sequence is a function s : N → R. We often write (sn | n ∈ N) or just (sn) to de- note a sequence, where sn := s(n). By Exercise 2.3, we have a nonstandard extension s : N∗^ → R∗, which we also often denote by (sn | n ∈ N∗).
Notation: We write N > N to indicate N ∈ N∗^ \ N.
Definition 3.1. (sn) converges to L, written (sn) → L or limn→∞ sn = L, if: for all ∈ R>^0 , there is m ∈ N such that, for all n ∈ N, if n ≥ m, then |sn − L| < .
We now give our first nonstandard characterization of a standard concept.
Theorem 3.2. sn → L if and only if sN ≈ L for all N > N.
This theorem lends solid ground to the heuristic expression: sn → L if and only if, for really large N , sN is really close to L.
Proof. (⇒) Suppose sn → L. Fix N > N. We want sN ≈ L. Fix ∈ R>^0. We want |sN − L| < . By assumption, there is m ∈ N such that
R |= ∀n ∈ N(n ≥ m → |sn − L| < ).
(Here, we are mixing formal logic with informal notation. If we were being polite, we would write
R |= ∀n((PNn ∧ P≥nm) → P<F|∗−∗|FsncLc).
After seeing the formal version, hopefully you will forgive our rudeness and allow us to write in hybrid statements as above!) Thus, by the Transfer Principle, R∗^ |= ∀n ∈ N∗(n ≥ m → |sn − L| < ). Since m ∈ N and N > N, we have N ≥ m. Thus, |sN − L| < , as desired. (⇐) We now suppose sN ≈ L for N > N. Fix ∈ R>^0. We need m ∈ N such that, n ∈ N and n ≥ m implies |sn − L| < . But how are we to find such m? Well, R∗^ knows of such an m (satisfying the ∗- version of the desired condition). Indeed, if m > N, then n ∈ N∗^ and n ≥ m implies n > N, whence sn ≈ L and, in particular, |sn − L| < . So, R∗^ |= ∃m ∈ N∗∀n ∈ N∗(n ≥ m → |sn − L| < ). Thus, by the Transfer Principle, R |= ∃m ∈ N∀n ∈ N(n ≥ m → |sn − L| < ), as desired.
We used transfer in each direction of the previous proof. The first appli- cation is often called “Upward Transfer” as a fact from below (in the “real world”) was transferred up to the nonstandard world. Similarly, the second application is often called “Downward Transfer” for a similar reason.
Theorem 3.3 (Monotone Convergence). Let (sn) be a sequence.
(1) Suppose (sn) is bounded above and nondecreasing. Then (sn) con- verges to sup{sn | n ∈ N}.
LECTURE NOTES ON NONSTANDARD ANALYSIS 19
Exercise 3.9. (sn) is Cauchy if and only if, for all M, N > N, sM ≈ sN.
Proposition 3.10. (sn) converges if and only if (sn) is Cauchy.
Proof. The (⇒) direction is an easy exercise, so we only prove the (⇐) direction. Suppose (sn) is Cauchy; then (sn) is bounded. Fix M > N; then sM ∈ Rfin. Set L := st(sM ). If N > N is another infinite natural number, then by the previous exercise, sN ≈ sM , so sN ≈ L. Thus, (sn) converges to L.
3.2. Cluster points. If (sn) is a sequence and L ∈ R, then we say that L is a cluster point of (sn) if, for each ∈ R>^0 , the interval (L − , L + ) contains infinitely many sn’s. It will be useful for us to write this in another way: L is a cluster point of (sn) if and only if, for every ∈ R>^0 , for every m ∈ N, there is n ∈ N such that n ≥ m and |sn − L| < . We can also recast this notion in terms of subsequences. A subsequence of (sn) is a sequence (tk) such that there is an increasing function α : N → N satisfying tk = sα(k). We often write (snk ) for a subsequence of (sn), where nk := α(k).
Exercise 3.11. L is a cluster point of (sn) if and only if there is a subse- quence (snk ) of (sn) that converges to L.
Recall that (sn) converges to L if sN ≈ L for all N > N. Changing the quantifier “for all” to “there exists” gives us the notion of cluster point:
Proposition 3.12. L is a cluster point of (sn) if and only if there is N > N such that sN ≈ L.
Proof. (⇒): Apply the transfer principle to the definition of cluster point. Fix ∈ μ>^0 and M > N. Then there is N ∈ N∗, N ≥ M , such that |sN − L| < . This is the desired N since N > N and is infinitesimal. (⇐): Fix N > N such that sN ≈ L. Fix ∈ R>^0 , m ∈ N. Then
R∗^ |= (∃n ∈ N∗)(n ≥ m ∧ |sn − L| < );
indeed, N witnesses the truth of this quantifier. It remains to apply transfer to this statement.
We immediately get the famous:
Corollary 3.13 (Bolzano-Weierstraß). Suppose that (sn) is bounded. Then (sn) has a cluster point.
Proof. Fix N > N. Then since (sn) is bounded, sN ∈ Rfin. Let L = st(sN ); then L is a cluster point of (sn) by the last proposition.
Suppose s = (sn) is a bounded sequence. Let C(s) denote the set of cluster points of s. Then, by the previous proposition, we have
C(s) = {L ∈ R | sN ≈ L for some N > N}.
Exercise 3.14. C(s) is a bounded set.
20 ISAAC GOLDBRING
We may thus make the following definitions:
Definition 3.15.
(1) lim sup sn := sup C(s); (2) lim inf sn := inf C(s).
It turns out that lim sup sn and lim inf sn are themselves cluster points of s, that is, these suprema and infima are actually max and min:
Proposition 3.16. lim sup sn, lim inf sn ∈ C(s).
Proof. We only prove this for lim sup sn; the proof for lim inf sn is similar. For simplicity, set r := lim sup sn. Fix > 0 and m ∈ N. We need to find n ∈ N such that |r−sn| < . Since r := sup C(s), there is L ∈ C(s) such that r − < L ≤ r. Take N > N such that L = st(sN ). Then r − < sN < r + . Now apply transfer.
Theorem 3.17. (sn) → L if and only if lim sup sn = lim inf sn = L.
Proof. (sn) → L if and only if st(sN ) = L for all N > N if and only if C(s) = {L} if and only if lim sup sn = lim inf sn = L.
There is an alternate description of lim sup sn and lim inf sn in terms of tail sets which we now explain. For n ∈ N, set Tn := {sn, sn+1, sn+2,.. .}, the nth^ tailset of s. Set Sn := sup Tn. Since Tn+1 ⊆ Tn, we have Sn+1 ≤ Sn, whence (Sn) is a nonincreasing sequence. Since (sn) is bounded, so is (Sn). Thus, by the Monotone Convergence Theorem, (Sn) converges to inf Sn.
Theorem 3.18. lim sup sn = lim Sn = inf Sn.
Before we can prove this result, we need a preliminary result:
Lemma 3.19. Let ∈ R>^0. Then there is m ∈ N such that, for all n ∈ N, if n ≥ m, then sn < lim sup sn + .
Proof. Fix N > N. Then st(sN ) ≤ lim sup sn, so sN < lim sup sn + . Thus, any M > N witness that
R∗^ |= (∃M ∈ N∗)(∀N ∈ N∗)(N ≥ M ⇒ sN < lim sup sn + ).
Now apply the transfer principle.
Proof. (of Theorem 3.18) Fix m ∈ N. Then by definition, if n ∈ N and n ≥ m, then sn ≤ Sm. By transfer, this holds for all n ∈ N∗^ with n ≥ m. Take N > N such that lim sup sn = st(sN ). Then sN ≤ Sm, so lim sup sn = st(sN ) ≤ Sm. Since m ∈ N is arbitrary, we see that lim sup sn ≤ inf Sn. Now suppose, towards a contradiction, that lim sup sn < inf Sn. Choose ∈ R>^0 such that lim sup sn+ < inf Sn. By the previous lemma, there is m ∈ N such that, for all n ≥ m, sn < lim sup sn + . Thus, Sm ≤ lim sup sn + < inf Sn, a contradiction.
Exercise 3.20. State and prove an analog of Theorem 3.18 for lim inf sn.
