Perspectives in Constructive Reverse
Mathematics
Josef Berger
Habilitationsschrift
Fakultät für Mathematik, Informatik und Statistik der
Ludwig-Maximilians-Universität München
July 1, 2014
1
Introduction
Reverse mathematics was founded by Friedman [14]. It is based on a subsystem of second-order arithmetic as a basic formal system. Using classical logic,
he investigates systematically which additional set existence assumptions are
needed to prove certain theorems. Moreover and crucially, those extra axioms should be as weak as possible; over the base system, the respective
proposition should also imply the additional set existence assumptions.
Constructive mathematics is distinguished from its classical counterpart by
avoiding the law of excluded middle (for every statement A, either A is true,
or its negation ¬A is true) as a proof tool. Consequently, the phrase ‘there
exists’ is strictly interpreted as ‘we can construct’. It is not sufficient to derive a contradiction out of the assumption that no object with the desired
properties exists. One occurrence of this discipline is constructive mathematics in the style of Bishop [6, 7, 8, 9]. In this setting, you work as intuitively
as a normal mathematician, you only avoid the law of excluded middle and
indirect proofs. Another way to conduct constructive mathematics is to work
with an intuitionistic formal systems like Heyting arithmetic [26]. Here, you
have a concrete set of objects, axioms, methods to construct new objects,
and methods to prove new statements. Schwichtenberg [25, Part 3] has elaborated a rigorous formal approach to computability at higher types, based
on information systems and computable functionals.
The axiom of choice is a function existence axiom:
∀x ∈ X ∃y ∈ Y P (x, y) ⇒ ∃f : X → Y ∀x ∈ X P (x, f (x))
In classical reverse mathematics, it serves as a tool for classifying propositions. In constructive mathematics it can be admissible or not depending on
its version and on the type of constructivism within one is working. In its
general form, it is not permitted, since it implies the law of excluded middle
[12]. Special cases, for example when the set X is countable, are freely used
in Bishop-style constructivism [8, p. 12] but not in Heyting arithmetic [26].
The essence of constructive reverse mathematics is the following question.
Which extra axioms are necessary and sufficient to find a constructive proof for a nonconstructive theorem?
Bishop himself tended to classify propositions into two types: those which are
constructively valid and those which are not. The first type can be identified
by giving a constructive proof. The second type can be stigmatised by finding a Brouwerian counterexample, that means showing that the proposition
implies a ‘non-constructive’ principle like LPO.
1
LPO
∀α ∈ {0, 1}N (∃n (αn = 1) ∨ ∀n (αn = 0)) .
This axiom is non-constructive in the following sense: a constructive proof
of it would either exclude the possibility of the existence of an n such that
αn = 1 or it would render a concrete n such that αn = 1. This would
give us a very short proof of Fermat’s last theorem. A typical result in
Bishop-style constructive reverse mathematics is the equivalence between the
statement ‘every uniformly continuous, real-valued function with compact
domain attains its minimum’ and the principle LLPO [15].
LLPO
∀α ∈ {0, 1}N (∀n, m (αn = αm = 1 ⇒ n = m) ⇒
∀n (α2n = 0) ∨ ∀n (α2n+1 = 0))
Another example is the equivalence of Brouwer’s fan theorem FAN and the
statement that ‘every positive-valued uniformly continuous function on the
unit interval has a positive infimum’ [18]. The term ‘constructive reverse
mathematics’ itself first appeared in [16], where Ishihara classified propositions in terms of logical principles (like LPO, LLPO) and function existence
principle (like choice axioms), based on a formal system. This could be labeled formal constructive reverse mathematics. One motivation for this approach is the possibility of building a bridge to classical reverse mathematics.
Starting around 2005, there has been a wave of papers in constructive reverse
mathematics, largely in the informal style of Bishop [1, 2, 3, 4, 13, 20, 24].
The title ‘perspectives in constructive reverse mathematics’ indicates that we
are applying a variety of methods to a variety of mathematical fields. Each
of the Sections 2, 3, and 4 are self-contained.
In Section 2, we prove that the weak König lemma constructively implies the
uniform continuity theorem for integer-valued functions on Cantor space: ‘every pointwise continuous function F : {0, 1}N → N is uniformly continuous.’
This is little surprising when working within Bishop-style constructivism.
But it is remarkable that the implication still holds in absence of any choice
axiom. The result is published in [5], where we used Heyting arithmetic as
a base system. For the sake of better readability, we do without rigorous
formalisation in this thesis.
Section 3 is about Dickson’s lemma,
DL For functions f, g : N → N, there exist i < j such that f (i) ≤ f (j) and
g(i) ≤ g(j).
2
and about Higman’s lemma,
HL For every function f : N → {0, 1}∗ , there exist i < j such that f (i) v
f (j).
Here, v is the embedding relation on the set {0, 1}∗ of binary words. Propositions of this type play a fundamental role in computer science, because they
are a tool for proving termination of algorithms, for example in the context
of term rewriting systems [23]. The first part of Section 3 contains a new
proof of Dickson’s lemma (for two functions); this is joint work with Helmut
Schwichtenberg. This proof is constructive, choice-free and requires induction for purely existential formulas only. Furthermore, the proof comes with
a bound k = T (f, g) such that the indices i, j can be chosen below k. In the
second part of Section 3, we show the equivalence of Dicksons’s lemma (for
arbitrarily many functions) and Higman’s lemma. Although both propositions are valid constructively [10, 27], this is of interest, since we establish
a correspondence in the following sense: start with a ‘Higman-sequence’,
transform it into a ‘Dickson-sequence’, apply Dicksons’s lemma and transfer the result back into the Higman-situation. And vice versa. Therefore, a
‘Higman-algorithm’ is the same as a ‘Dickson-algorithm’; this result alters
the perception of Dickson-style axioms considerably. It can be regarded a
special variant of constructive reverse mathematics, namely to establish the
equivalence of axioms which are in fact constructively valid. Section 3 is
written in a rather concise and formal style, the results are yet unpublished.
Section 4 constitutes an application of constructive reverse mathematics to
mathematical finance; this is joint work with Gregor Svindland, and it is
work in progress. A natural candidate for a proposition to be investigated
is the fundamental theorem of asset pricing, which says that a market model
is arbitrage-free if and only if there exists a martingale measure. The first
condition could be paraphrased as ‘there is no free lunch’, the second one
as ‘the world is possibly fair’. This theorem is one of the backbones of
mathematical finance; for example, the proof of the famous Black–Scholes
formula [11] is based on the notion of martingale measure. When we look at
proofs of this theorem in classical textbooks [22], we come across two related
propositions.
• Two disjoint convex subsets of a finite-dimensional Hilbert space can
be separated by a linear functional.
• The lemma on two alternatives, which states that for any matrix, either a linear combination (with positive coefficients) of the rows is
component-wise positive or else a linear combination (with positive coefficients) of the columns is component-wise negative.
3
We prove that all the aforementioned propositions are equivalent to LPO if
they are formulated in a disjunctive manner: either there exists an arbitrage
strategy or there exists a martingale measure; either the convex sets intersect or they can be separated. Moreover, we show that both the original
formulation of the fundamental theorem of asset pricing implies Markov’s
principle.
MP
∀α ∈ {0, 1}N (¬∀n (αn = 0) ⇒ ∃n (αn = 1))
In this section we work informally in the style of Bishop. The first reason
for this is that when working with the real numbers, or continuous objects
in general, it is rather a kerfuffle to keep an account of the usage of choice
axioms. Second, we have to go rather deeply into analysis and topology and
therefore we want to refer to textbooks which are written in the style of
Bishop.
Let N = {0, 1, 2, . . .} denote the set of natural numbers and let N+ =
{1, 2, 3, . . .} denote the set of positive natural numbers. A set X is finite
if there is a k ∈ N such that there exists a bijection between {0, 1, . . . , k} and
X. A set X is finitely enumerable if there is a k ∈ N such that there exists
a surjection from {0, 1, . . . , k} onto X.
This thesis is dedicated to Horst, Júlia, and Nicola.
2
The weak König lemma and the uniform
continuity theorem
We show that the weak König lemma constructively implies the uniform
continuity theorem for integer-valued functions on Cantor space. Hereby we
do not assume that the pointwise continuous functions have a modulus of
pointwise continuity. This result is based on two observations:
• Without applying any instance of the axiom of choice, we can show
that the weak König lemma implies that for every sequence of infinite
binary trees there is a sequence of infinite branches.
• Under the weak König lemma, we can use a quantifier-free formula for
the statement that a pointwise continuous, integer-valued function is
constant.
This result had been shown with classical logic [19].
4
The elements of N are denoted by the variables m, n, k, l, and N . If X and
Y are sets, then X → Y or Y X denotes the set of all functions from X into
Y . If F ∈ Y N , we write Fn for F (n). If F ∈ {0, 1}X and x ∈ X, we write
x ∈ F for F (x) = 0. The elements of {0, 1}N are denoted by the variables
α, β, and γ. Set 0 = (0, 0, 0, . . .). Let {0, 1}∗ denote the set of finite binary
sequences. The elements of {0, 1}∗ are denoted by the variables u and v.
The empty sequence is denoted by (). The length of u is denoted by |u|.
Therefore, |()| = 0. The concatenation of u and v is denoted by u ∗ v. That
means that for
u = (u0 , . . . , un−1 ) and v = (v0 , . . . , vm−1 )
we have
|u| = n and u ∗ v = (u0 , . . . , un−1 , v0 , . . . , vm−1 ) .
Furthermore, for every α set
u ∗ α = (u0 , . . . , un−1 , α0 , α1 , α2 , . . .) .
For every k set
αk = (α0 , . . . , αk−1 )
and, if k ≤ n, set
uk = (u0 , . . . , uk−1 ) .
Note that α0 = u0 = ().
A function B : {0, 1}∗ → {0, 1} is a bar if
∀α ∃n (αn ∈ B)
and a uniform bar if
∃N ∀α ∃n ≤ N (αn ∈ B) .
FAN Every bar is a uniform bar.
A function T : {0, 1}∗ → {0, 1} is an infinite tree if
∀n ∃u (|u| = n ∧ u ∈ T) and ∀u, v (u ∗ v ∈ T ⇒ u ∈ T) .
If the second property holds, we say that T is closed under restriction. A
sequence α is called an infinite branch of the infinite tree T if
∀n (αn ∈ T) .
5
WKL For every infinite tree there exists an infinite branch.1
A function T : N → ({0, 1}∗ → {0, 1}) is an infinite multitree if every Tn is
an infinite tree. A function A : N → {0, 1}N is an infinite multibranch of T
if every An is an infinite branch of Tn .
Proposition 1. WKL implies that every infinite tree has an infinite multibranch.
Proof. Let T : N → ({0, 1}∗ → {0, 1}) be an infinite multitree. We use the
pairing function J : N × N → N, defined by
(k, n) 7→ 12 (k + n) (k + n + 1) + n.
n
k
0
1
2
3
4
0
0
1
3
6
10
1
2
4
7
11
16
2
5
8
12
17
23
3
9
13
18
24
31
4
14
19
25
32
40
Note that
• J is bijective,
• k ≤ J(k, n), n ≤ J(k, n),
• k < k 0 ⇒ J(k, n) < J(k 0 , n),
• n < n0 ⇒ J(k, n) < J(k, n0 ).
Define a function τ : {0, 1}∗ → {0, 1} by
u ∈ τ ⇔ ∀k, n J(k, n) < |u| ⇒
uJ(0,n) , . . . , uJ(k,n) ∈ Tn .
We show that τ is an infinite tree. Clearly, τ is closed under restriction.
Now fix an l. Choose w0 , . . . , wl−1 ∈ {0, 1}l such that
∀n < l (wn ∈ Tn ) .
1
In the presence of countable choice, the axioms WKL and LLPO are equivalent [15].
6
We show that the u of length l which is uniquely characterised by
∀n, k J(k, n) < l ⇒ uJ(k,n) = (wn )k
belongs to τ . Therefore, fix k and n with J(k, n) < l. Then
uJ(0,n) , . . . , uJ(k,n) = wn (k + 1) ∈ Tn .
By WKL, there is an infinite branch α of τ . Define A : N → {0, 1}N by
(An )k = αJ(k,n) .
We show that A is an infinite multibranch of T . Fix n. We show that An is
an infinite branch of Tn . Fix therefore k with k ≥ 1. We have
α(J(k, n) ∈ τ,
which implies that
An k = ((An )0 , . . . , (An )k−1 ) = αJ(0,n) , . . . , αJ(k−1,n) ∈ Tn .
A function F : {0, 1}N → N is pointwise continuous if
∀α ∃n ∀β αn = βn ⇒ F (α) = F (β)
and uniformly continuous if
∃N ∀α, β αN = βN ⇒ F (α) = F (β) .
Note that we do not assume that the pointwise continuous functions have a
modulus of pointwise continuity.
UC Every pointwise continuous function F : {0, 1}N → N is uniformly continuous.
Both pointwise and uniform continuity can be expressed in terms of constancy. For F : {0, 1}N → N and u let
F u : {0, 1}N → N
be given by
α 7→ F (u ∗ α).
7
Lemma 1. Fix F : {0, 1}N → N.
• F is pointwise continuous ⇔ ∀α ∃n F αn is constant
• F is uniformly continuous ⇔ ∃N ∀u (|u| = N ⇒ F u is constant)
Proposition 2. Assume WKL and fix a function
F : N → {0, 1}N → N
such that every Fn is pointwise continuous. Then there exists a function
β : N → {0, 1}
such that
n ∈ β ⇔ Fn is constant
(n ∈ N) .
Proof. Define g : N → ({0, 1}∗ → {0, 1}) by
u ∈ gn ⇔ Fn (u ∗ 0) 6= Fn (0)
(u ∈ {0, 1}∗ , n ∈ N) .
Define T : N → ({0, 1}∗ → {0, 1}) by
u ∈ Tn ⇔
∀m ≤ |u| (∃v (|v| = m ∧ v ∈ gn ) ⇒ ∃k ≤ m (uk ∈ gn )) ∧
∀m ≤ |u| (um ∈ gn ⇒ ∀k ∈ {m, . . . , |u| − 1} (uk = 0))
(u ∈ {0, 1}∗ , n ∈ N).
We show that T is an infinite multitree. Fix n. We show that Tn is an
infinite tree. It follows directly from its definition that Tn is closed under
restriction. Now fix an l. We show that there exists an u ∈ Tn with |u| = l.
Let us distinguish two cases. In the first case, for every v with |v| ≤ l we
have v ∈
/ gn . Then every u with |u| = l belongs to Tn . In the second case,
there is m ≤ l such that there exists v ∈ gn with |v| = m. Then there exists
a minimal m with that property. Fix a corresponding v. Define
u = v ∗ ( 0, . . . , 0 ).
| {z }
l−m times
Then u ∈ Tn and we have shown that T is an infinite multitree.
Proposition 1 implies the existence of an infinite multibranch
A : N → {0, 1}N
8
of T . Fix n and set α = An . Note that
(u ∈ {0, 1}∗ ) .
u ∈ gn ⇒ ∃m ≤ |u| (αm ∈ gn )
(1)
Next we show that
∃m (αm ∈ gn ) ⇒ Fn (α) 6= Fn (0).
(2)
Suppose that there is an m with αm ∈ gn . By the pointwise continuity of
Fn , there is an l ≥ m such that
Fn (αl ∗ 0) = Fn (α).
Since αl ∈ Tn and αm ∈ gn , we find that αm ∗ 0 = αl ∗ 0 and therefore
Fn (α) = Fn (αl ∗ 0) = Fn (αm ∗ 0) 6= Fn (0).
This completes the proof of (2).
Now we show that
Fn (α) = Fn (0) ⇒ Fn is constant.
(3)
Assume that Fn (α) = Fn (0). Assume further that there exists a γ such that
Fn (γ) 6= Fn (0). By the pointwise continuity of Fn , there exists an l such
that Fn (γl ∗ 0) 6= Fn (0) and hence γl ∈ gn . By (1), there exists m ≤ l such
that αm ∈ gn . Therefore, (2) yields Fn (α) 6= Fn (0), a contradiction. This
completes the proof of (3).
Since n was chosen arbitrarily, we have shown
Fn (An ) = Fn (0) ⇔ Fn is constant
(n ∈ N) .
Finally, we can define a function β : N → {0, 1} by
n ∈ β ⇔ Fn (An ) = Fn (0)
This function has the desired property.
The following result is from [17].
Proposition 3. WKL implies FAN.
Proposition 4. WKL implies UC.
9
(n ∈ N) .
Proof. Fix a pointwise continuous function F : {0, 1}N → N. Define
G : {0, 1}∗ → {0, 1}N → N
by G(u) = F u . By Proposition 2, there exists a function B : {0, 1}∗ → {0, 1}
such that
u ∈ B ⇔ G(u) is constant (u ∈ {0, 1}∗ ) .
The pointwise continuity of F and Lemma 1 imply that B is a bar. By
FAN, which follows from WKL, we obtain that B is a uniform bar. Applying
Lemma 1 again yields the uniform continuity of F .
3
Dickson’s lemma and Higman’s lemma
3.1
A new proof of Dickson’s lemma for two functions
(joint work with Helmut Schwichtenberg)
Lemma 2. Fix k, n ∈ N+ and a1 , . . . , an ∈ N with
• ai < k
(i ∈ {1, . . . , n}) ,
• i 6= j ⇒ ai 6= aj
(i, j ∈ {1, . . . , n}) .
Then n ≤ k.
Lemma 3. Fix k, l, n ∈ N+ and suppose that a1 , . . . , an and b1 , . . . , bn are
natural numbers with
• a1 , . . . , an < k,
• b1 , . . . , bn < l,
• i 6= j ⇒ (ai , bi ) 6= (aj , bj )
(i, j ∈ {1, . . . , n}) .
Then n ≤ k · l.
Proof. Fix l and n. We conduct induction on k. For k = 1, we can apply
Lemma 2. Now assume that a1 , . . . , an and b1 , . . . , bn are natural numbers
with
• a1 , . . . , an < k + 1,
• b1 , . . . , bn < l,
10
• i 6= j in {1, . . . , n} implies (ai , bi ) 6= (aj , bj ).
If a1 , . . . , an < k, we are done. Otherwise, assume, without loss of generality,
that the ai are descending. Let m ∈ {1, . . . , n} be the biggest index such
that am = k. By Lemma 2, we can conclude that m ≤ l. If m = n, Lemma 2
implies that n ≤ l and therefore n ≤ k ·l. Now assume that m < n. Applying
the induction hypothesis to am+1 , . . . , an and bm+1 , . . . , bn yields n−m ≤ k ·l.
Altogether we obtain n ≤ (k + 1) · l.
Lemma 4. Fix k ∈ N and a sequence of k 2 + 1 pairs of natural numbers.
Then either two of the pairs coincide, or one element of one pair is at least
as big as k.
Fix functions f, g : N → N and numbers n, m ∈ N. Set
F (n) = min {m ≤ n | ∀k ≤ n (f (m) ≤ f (k))}
G(n) = min {m ≤ n | ∀k ≤ n (g(m) ≤ g(k))}
Φ(n) = f (F (n)) + g(G(n))
Ψ(n) = max {f (G(n)), g(F (n))}
I(n) = n + Ψ(n)2 + 1
Let I m be the n-fold iteration of I. We denote a special case of Dickson’s
lemma by DL(n).
DL(n)
∃i, j (i < j ∧ j ≤ n ∧ f (i) ≤ f (j) ∧ g(i) ≤ g(j))
Note that this is a decidable formula.
Lemma 5.
∀n (DL(I(n)) ∨ Φ(I(n)) < Φ(n))
Proof. Fix n. By Lemma 4, we are in one of the following cases.
Case 1 There are i, j with n < i < j ≤ I(n) such that f (i) = f (j) and
g(i) = g(j). In this case we have DL(I(n)).
Case 2 There is j with n < j ≤ I(n) such that either f (j) ≥ Ψ(n) or
g(j) ≥ Ψ(n). Without loss of generality, assume that g(j) ≥ Ψ(n). Note
that g(F (n)) ≤ g(j). We either have f (F (n)) ≤ f (j) as well, which implies
DL(I(n)). Otherwise, f (j) < f (F (n)) which yields Φ(I(n)) < Φ(n).
Lemma 6.
∀n (DL(I n (0)) ∨ Φ(I n (0)) + n ≤ Φ(0)) .
11
Proof. For n = 0, there is nothing to prove. Now assume that the assertion
holds for n. In the case of DL(I n+1 (0)), we are done. So assume that
DL(I n+1 (0)) is false. This has two consequences:
First, applying Lemma 5 to I n (0) yields
Φ(I n+1 (0)) < Φ(I n (0)).
Second, since we have I n (0) ≤ I n+1 (0), we can conclude that DL(I n (0)) is
false as well, so the induction hypothesis yields Φ(I n (0)) + n ≤ Φ(0).
Altogether we obtain
Φ(I n+1 (0)) + n + 1 ≤ Φ(0).
Proposition 5.
DL(I f (0)+g(0)+1 (0))
Proof. Assume that
DL(I f (0)+g(0)+1 (0))
does not hold. Then Lemma 6 implies
Φ(I f (0)+g(0)+1 (0)) + f (0) + g(0) + 1 ≤ f (0) + g(0),
which is absurd.
3.2
Dickson’s lemma and Higman’s lemma are equivalent
We use the variable i for the elements of the set {0, 1} and the variables k, l
for the elements of the set N. The set {0, 1}∗ can be defined inductively by:
u ∈ {0, 1}∗
ui ∈ {0, 1}∗
() ∈ {0, 1}∗
Here () stands for the empty sequence. The length |u| of u ∈ {0, 1}∗ is given
by:
|()| = 0
|ui| = |u| + 1
We define the subword relation v on {0, 1}∗ by:
() v ()
12
uvw
uvw
u v w0
u v w1
uvw
uvw
u0 v w0
u1 v w1
Higman’s lemma is the following axiom.
HL For every function g : N → {0, 1}∗ there are k, l such that
• k<l
• g(k) v g(l)
For n ∈ N+ , Dickson’s lemma (for Nn -valued functions) is the following
axiom.
DLn For every f : N → Nn there are k, l such that
• k<l
• f (k) ≤ f (l) (component-wise)
Proposition 6.
(∀n ∈ N+ (DLn ))
⇐⇒
HL
In order to prove this proposition, we work with slightly altered but obviously
equivalent versions of the axioms.
HL’ For every function g : N → {0, 1}∗0 there are k, l such that
• k<l
• g(k) v g(l)
where {0, 1}∗0 is given by:
u ∈ {0, 1}∗
ui ∈ {0, 1}∗
0 ∈ {0, 1}∗
DL’n For every f : N → Nn+ there are k, l such that
• k<l
• f (k) ≤ f (l) (component-wise)
13
For every u ∈ {0, 1}∗0 we define λ(u) by:
λ(0) = 0
λ(ui) = i
For every u ∈ {0, 1}∗0 we define Φ(u) by:
Φ(u)
Φ(0) = 1 Φ(ui) =
Φ(u) + 1
if λ(u) = i
if λ(u) =
6 i
Φ(u) is called the weight of u. This is the number of flips in u plus one.
Define a bijection
F : {0, 1}∗0 → ∪m≥1 Nm
+
by F (0) = (1) and, for u ∈ {0, 1}∗0 ,
(k1 , . . . , km + 1) if λ(u) = i and F (u) = (k1 , . . . , km ) ,
F (ui) =
(k1 , . . . , km , 1)
if λ(u) 6= i .
Fix n ∈ N+ . Let Wn denote the elements of {0, 1}∗0 with weight n. Let Fn
be the restriction of F to Wn .
Lemma 7. Fn : Wn → Nn+ is bijective and order-preserving.
n
Define Gn : ∪m≥1 Nm
+ → N+ by
(k1 , . . . , km , 1, . . . , 1)
Gn (k1 , . . . , km ) =
(k1 , . . . , kn )
For u ∈ {0, 1}∗0 set
if m < n ,
if n ≤ m .
u!n = Fn−1 (Gn (F (u))).
Note that
• Φ(u!n ) = n,
• if Φ(u) = n then u!n = u,
• if Φ(u) < n then u!n is the shortest extension of u with weight n,
• if Φ(u) > n then u!n is the largest restriction of u with weight n,
• if Φ(u) > n then |u!n | < |u|.
Lemma 8. For all u, w in {0, 1}∗0 we have:
(2 · |u| ≤ Φ(w)) ⇒ u v w
14
Proof. In order to avoid nested induction, we show the decidable formula
∀u, w ∈ {0, 1}∗0 (|u| ≤ k ∧ |w| ≤ k ∧ (2 · |u| ≤ Φ(w)) ⇒ u v w) ,
by induction on k. For k = 1, there is nothing to prove. Now assume that
the formula holds for k ∈ N+ . Fix u, w ∈ {0, 1}∗0 such that
|u| ≤ k + 1 ∧ |w| ≤ k + 1 ∧ (2 · |u| ≤ Φ(w))
holds. We can assume that u = vi for some v ∈ {0, 1}∗0 . Set l = 2 · |v|.
Note that 2 · |v| = Φ(w!l ). Therefore, the induction hypothesis implies that
v v w!l , thus we can conclude that u v w.
Lemma 9.
Gn (F (u)) = Fn (u!n )
Lemma 10. Fix u, w ∈ {0, 1}∗0 with
Φ(u) ≤ Φ(w) ≤ n and u!n v w!n .
Then we have
u v w.
Proof. Set k = Φ(u). The assumption u!n v w!n implies Gn (F (u)) ≤
Gn (F (w)) and therefore
Fk (u) = Gk (F (u)) ≤ Gk (F (w)) = Fk (w!k ).
Thus we obtain
u v w!k .
Since k ≤ Φ(w), we obtain w!k v w and therefore u v w.
“HL’ ⇒ DL’n ”
Fix n ∈ N+ and f : N → Nn+ . Applying HL to Fn−1 ◦ f yields k < l with
Fn−1 ◦ f (k) v Fn−1 ◦ f (l),
which implies f (k) ≤ f (l).
“DL’n ⇒ HL’ ”
Fix g : N → {0, 1}∗0 . Set n = 2 · |g(0)|. Then DLn+1 yields k < l in N such
that
(Fn (g(k)!n ), Φ(g(k))) ≤ (Fn (g(l)!n ), Φ(g(l))) .
If n < Φ(g(l)) we obtain g(0) v g(l) by Lemma 8. Otherwise, we have
Φ(g(k)) ≤ Φ(g(l)) ≤ n and g(k)!n v g(l)!n .
Thus g(k) v g(l) by Lemma 10.
15
4
The fundamental theorem of asset pricing
(joint work with Gregor Svindland)
4.1
Notation
For an introduction to the constructive real numbers, we refer to [6]. To put
it in a nutshell, the constructive real numbers behave like the classical real
numbers, except that totality,
a < 0 ∨ a = 0 ∨ 0 < a (a ∈ R)
is replaced by the principle of approximate splitting,
x < y ⇒ z < y ∨ x < z (x, y, z ∈ R).
Note that totality is equivalent to LPO in Bishop-style constructivism, see
[8, p. 14]. The set of nonnegative real numbers is denoted by
R0+ = {x ∈ R | 0 ≤ x} .
The set of positive real numbers is denoted by
R+ = {x ∈ R | 0 < x} .
Fix n ∈ N+ . We set In = {1, 2, . . . , n}. If x ∈ Rn , we write x > 0 for
∀i ∈ In (xi ≥ 0) ∧ ∃i ∈ In (xi > 0) .
We set
)
(
λ ∈ Rn0+ |
Sn =
X
λi = 1
i∈In
and
)
(
S̊n =
λ ∈ Rn+ |
X
λi = 1 .
i∈In
The constructive concepts of metric space, vector space, convex set, and
Hilbert space are literally the same as in the classical setting. Let H be a
Hilbert Space. We define
SH = {x ∈ H | ||x|| = 1} .
The convex hull of x1 , . . . , xn ∈ H is given by
(
)
X
C(x1 , . . . , xn ) =
p i · xi | p ∈ S n .
i∈In
16
The relative interior of the convex hull of x1 , . . . , xn ∈ H is defined by
(
)
X
C̊(x1 , . . . , xn ) =
pi · xi | p ∈ S̊n .
i∈In
The span of x1 , . . . , xn ∈ H is defined by
)
(
X
SP(x1 , . . . , xn ) =
pi · xi | p ∈ Rn .
i∈In
4.2
The fundamental theorem of asset pricing
Our portfolio consists of m stocks, for some m ∈ N+ , and one bond. The
prices of the assets at time 0 are known and denoted by π ∈ Rm+1 , where
π1 = 1 is the price of the bond. The prices of the assets at time 1 is unknown,
but we know that there are n possible developments, for some n ∈ N+ , and
we know the prices for each case. Let cij denote the price of the i-th asset in
the case j, for i ∈ Im+1 and j ∈ In . We assume that c1j = 1 for every j ∈ In ,
which says that the future value of the bond will be 1 in any case. Moreover,
we assume that all prices are positive.
A vector p ∈ S̊n is called equivalent martingale measure if C · p = π, where
· denotes the matrix product.
A vector ξ ∈ Rm+1 is called arbitrage strategy if
ξ·π ≤0
and
ξ · C > 0.
Lemma 11. The following are equivalent:
1) there exists an arbitrage strategy ξ
2) there exists a vector µ ∈ Rm+1 such that
µ · (C | −π) > 0
Proof. 1) ⇒ 2) Just set µ = ξ.
2) ⇒ 1) There are two cases. Either there exists a column x of C such
that µ · x > 0. In this case, set ξ = µ. In the second case, we have µ · C ≥ 0
and µ · π < 0. Set
ξ = (µ1 + ε, µ2 , . . . , µm+1 ) ,
where ε is a positive number such that ξ · π ≤ 0.
17
The matrix (C | −π) is obtained when augmenting the vector −π as an
additional column to C.
The fundamental theorem of asset pricing says that there exists a martingale
measure if and only if there doesn’t exist an arbitrage strategy.
It is easy to see that there cannot be both an arbitrage strategy and a martingale measure. Another, seemingly more natural way to put the theorem
is to claim that there either exists an arbitrage strategy or else a martingale
measure. We focus on this version in the first place and discuss varieties of
it afterwards.
4.3
Equivalent conditions
We show that the disjunctive versions of a separation theorem and the fundamental theorem of asset pricing and a version of the lemma on two alternatives are constructively equivalent.
SEP Fix elements x1 , . . . , xn of a finite-dimensional Hilbert Space H. Then
we can prove one of the following alternatives:
• ∃ξ ∈ H ((hξ, x1 i, . . . , hξ, xn i) > 0)
• 0 ∈ C̊(x1 , . . . , xn )
FTAP Fix m, n ∈ N+ and a R(m+1)×n -matrix C such that
cij > 0
(i ∈ Im+1 , j ∈ In )
and
c1j = 1
(j ∈ In ) .
m+1
Furthermore, fix π ∈ R+
with π1 = 1. Then we can prove one of the
following alternatives:
• ∃ξ ∈ Rm+1 (ξ · (C | −π) > 0)
• ∃p ∈ S̊n (C · p = π)
ALT Fix m, n ∈ N+ and a Rm×n -matrix A. Then we can prove one of the
following alternatives:
• ∃ξ ∈ Rm (ξ · A > 0)
• ∃p ∈ Rn+ (A · p = 0)
18
Proposition 7. The following are equivalent: SEP, FTAP, and ALT.
Proof. SEP ⇔ ALT This is clear, since every m-dimensional Hilbert space
can be identified with Rm , for every m ∈ N+ .
ALT ⇒ FTAP
Apply ALT to the matrix (C | −π).
FTAP ⇒ ALT
Fix a Rm×n -matrix A. Define a R(m+1)×n -matrix C by
cij = a(i−1)j + c
(i ∈ {2, . . . , m + 1} , j ∈ In ) ,
and
c1j = 1
(j ∈ In ) ,
where c ∈ N is big enough such that all entries of C are positive. Now apply
FTAP to C and π = (1, c, . . . , c).
| {z }
m times
case 1 There exists p ∈ S̊n such that C · p = π. In this case we can conclude
that A · p = 0.
case 2 There exists ξ ∈ Rm+1 such that
ξ · (C | −π) > 0.
This implies that
µ·A>0
where µ = (ξ2 , . . . , ξm+1 ). Here we have used the following argument:
(a1 , . . . , an , an+1 ) > 0 ⇒ (a1 + an+1 , . . . , an + an+1 ) > 0
(a1 , . . . , an+1 ∈ R) .
4.4
One more equivalent: LPO
First we show that ALT implies LPO. Fix a ∈ R. Apply ALT to the matrix
A = (a). Then there either exists ξ ∈ R with ξa > 0, which implies |a| > 0
and therefore a > 0 or a < 0. Or there exists p ∈ R+ with ap = 0. This
implies a = 0.
In order to show that LPO implies SEP, we need some constructive linear
algebra and topology, like introduced in [9]. Let X be a metric space. We
19
cannot prove constructively that every set of positive numbers has an infimum. This has consequences for the notion of distance between a point and
a set. Let A be a subset of X. The set A is located in X if for every x ∈ X
the distance
d(x, A) = inf {d(x, a) | a ∈ A}
exists. A point x ∈ X is called a cluster point of A if there exists a sequence
in A which converges to x. The set A is closed if it contains all of its cluster
points. The closure A of A in X is given by
A = {x ∈ X | x is a cluster point of A} .
The space X is totally bounded if for every n there exists a finitely enumerable
subset S of X such that
∀x ∈ X ∃s ∈ S (d(x, s) < 1/n) .
The space X is compact if it is totally bounded and complete. Here are some
basic facts about these notions.
Lemma 12. Fix n ∈ N+ and a Hilbert space H.
a) The image of a totally bounded space under a uniformly continuous
function is totally bounded.
b) Totally bounded subsets of X are located in X.
c) If H is finite-dimensional, then SH is compact.
d) The set Sn is compact.
e) The set C̊(x1 , . . . , xn ) is convex and located, for elements x1 , . . . , xn in
H.
The constructive notion of uniformly continuous functions between metric
spaces is the same as the classical one. A proof of the following lemma can
be found in [21].
Lemma 13. The following are equivalent:
1) LPO
2) Every sequence in a compact metric space has a convergent subsequence.
Lemma 14. Assume LPO. Fix a uniformly continuous function κ : K → Y ,
where K is a compact metric space and Y is a complete metric space. Then
κ(K) is compact.
20
Proof. Lemma 12 implies that κ(K) is totally bounded. We show that κ(K)
is closed in Y . Since Y is complete, this implies that κ(K) is complete as
well. Fix a sequence (xn ) in K such that κ(xn ) converges to a y ∈ Y . By
Lemma 13, a subsequence of (xn ) converges to a x ∈ K. Thus y = κ(x).
Vectors x1 , . . . , xn of a Hilbert Space H are linearly independent if for every
λ ∈ Rn we have
X
X
|λi | > 0 ⇒ ||
λi · xi || > 0,
i∈In
i∈In
and linearly dependent if there exists λ ∈ Rn such that
X
X
|λi | > 0 ∧
λi · xi = 0.
i∈In
i∈In
The constructive notion of basis of a vector space is the same as the classical
one. However, the subtlety of the notion of linear independence carries over.
The next lemma is crucial, since it enables us to determine the dimension of
the span of finitely many vectors.
Lemma 15. The following are equivalent:
a) LPO
b) Elements x1 , . . . , xn of a Hilbert Space H are either linearly independent
or else linearly dependent.
Proof. b) ⇒ a) Consider the case of H = R, n = 1, and x1 = a for some
a ∈ R. If a is linearly independent, then |a| > 0. If a is linearly dependent,
then a = 0.
a) ⇒ b)
The set
(
D=
)
λ ∈ Rn |
X
λ2i = 1
i∈In
is compact and the map
κ : D → H, λ 7→
X
λi · xi
i∈In
is uniformly continuous. Therefore, by Lemma, 14, κ(D) is compact, in
particular, it is closed in H. If
d(0, κ(D)) = 0,
the vectors are linearly dependent. If
d(0, κ(D)) > 0,
the vectors are linearly independent.
21
The axiom SEP implies that we can decide whether 0 ∈ C̊(x1 , . . . , xn ) holds
or not, for elements x1 , . . . , xn in a Hilbert space H. It turns out that this
decidability is equivalent to LPO.
Lemma 16. The following are equivalent:
1) LPO
2) Let H be a Hilbert Space and fix x1 , . . . , xn ∈ H. Then the following
statement is decidable:
0 ∈ C̊(x1 , . . . , xn )
Proof. 1) ⇒ 2)
In view of Lemma 15, we can assume that there exists an
m ≤ n such that x1 , . . . , xm is a basis of SP {x1 , . . . , xn }. If m = n, then
0∈
/ C̊(x1 , . . . , xn ). Now suppose that m < n. There is a uniquely determined
sequence (αji )i∈Im , j∈{m+1,...,n} such that
xj =
X
αji xi (j ∈ {m + 1, . . . , n}).
i∈Im
Now 0 ∈ C̊(x1 , . . . , xn ) is equivalent to the existence of λ1 , . . . , λn > 0 such
that


X
X
X
λi +
λj αji  xi = 0 .
λ i xi =
i∈In
i∈Im
j∈{m+1,...,n}
This is equivalent to
∃a ∈ Rn−m ∀j ∈ In (ha, yj i > 0) ,
where
i
yi = −(αm+1
, . . . , αni ) (i ∈ Im )
and ym+1 , . . . , yn is the canonical basis of Rn−m . And this is decidable, in
presence of LPO.
2) ⇒ 1)
Set H = R and n = 1. Fix a ∈ R with a < 1. Note that
C̊(a, 1) = (a, 1)
and that
0 ∈ C̊(a, 1) ⇔ a < 0.
22
Lemma 17. Let H be a Hilbert Space and fix x1 , . . . , xn ∈ H. Then the set
C(x1 , . . . , xn ) is closed in H.
Proof. Apply Lemma 14 to the function
κ : Sn → H, λ 7→
X
λi · xi .
i∈In
Lemma 18. Let C be a convex and located subset of a Hilbert Space H. Fix
x ∈ H and set d = d(x, C). Then there exists a unique a ∈ C such that
||a − x|| = d. Furthermore, we have
ha − x, c − ai ≥ 0
c∈C
and therefore
ha − x, c − xi ≥ 0
c∈C .
Proof. Fix a sequence (cn ) in C such that ||cn − x|| → d. Since
||cm − cn || 2 = ||(cm − x) − (cn − x)|| 2 =
2 ||cm − x|| 2 + 2 ||cn − x|| 2 − 4 ||
|
2 ||cm − x|| 2 − d
2
cm + cn
− x|| 2 ≤
2 {z
}
≥4d2
+ 2 ||cn − x|| 2 − d2 ,
(cn ) is a Cauchy sequence and therefore converges to an a ∈ C. Since
||cn − x|| → ||a − x|| , we obtain ||a − x|| = d. Now fix b ∈ C with
||b − x|| = d. Then
||a − b|| 2 = ||(a − x) − (b − x)|| 2 =
a+b
2 ||a − x|| 2 + 2 ||b − x|| 2 − 4 ||
− x|| 2 ≤ 0 ,
2 {z
|
}
≥4d2
thus a = b.
Fix c ∈ C and λ ∈ (0, 1). Since
||a − x|| 2 ≤ ||(1 − λ)a + λc − x|| 2 = ||(a − x) + λ(c − a)|| 2 =
||a − x|| 2 + λ2 ||c − a|| 2 + 2λha − x, c − ai,
23
we obtain
0 ≤ λ ||c − a|| 2 + 2ha − x, c − ai .
Since λ can be arbitrarily small, we can conclude that
ha − x, c − ai ≥ 0 .
This also implies that
ha − x, c − xi = ha − x, c − ai + ha − x, a − xi ≥ 0 .
Lemma 19. Assume LPO. Fix a finite-dimensional Hilbert Space H and
suppose that C is a subset of H such that
• C is convex,
• C is located, and
• 0 ∈ −C.
Then we have
∃a ∈ SH ∀c ∈ C (ha, ci ≥ 0) .
Proof. Fix a sequence (xk ) in −C which converges to 0. By Lemma 18, there
exists a sequence (ak ) in C such that
||ak − xk || = d(xk , C)
(k ∈ N)
and
hak − xk , c − xk i ≥ 0
Set
bk =
ak − x k
||ak − xk ||
k ∈ N, c ∈ C .
(k ∈ N) .
By Lemma 12 and Lemma 13 there exists a subsequence (bnk ) of (bk ) that
converges to an a ∈ SH . Since
hbnk , c − xnk i ≥ 0
k ∈ N, c ∈ C ,
we obtain
ha, ci ≥ 0
24
c∈C .
Now we show that LPO implies SEP. Fix elements x1 , . . . , xn of a Hilbert
space H. In view of Lemma 15, we may assume that H = SP {x1 , . . . , xn } . By
Lemma 16, we have either 0 ∈ C̊(x1 , . . . , xn ) or 0 ∈
/ C̊(x1 , . . . , xn ). Consider
the latter case. We claim that
0 ∈ −C̊(x1 , . . . , xn ).
We prove this by showing that, for every k ∈ N+ , the vector − k1
in −C̊(x1 , . . . , xn ). First of all, note that
P
i∈In
xi lies
C̊(x1 , . . . , xn ) ⊆ C(x1 , . . . , xn )
implies
−C(x1 , . . . , xn ) ⊆ −C̊(x1 , . . . , xn ).
Therefore, it suffices to show that
1X
d(−
xi , C(x1 , . . . , xn )) > 0.
k i∈I
n
Suppose that
d(−
1X
xi , C(x1 , . . . , xn )) = 0.
k i∈I
n
By Lemma 17, the set C(x1 , . . . , xn ) is closed in H, therefore we obtain
1X
−
xi ∈ C(x1 , . . . , xn ),
k i∈I
n
so there exists λ ∈ Sn with
−
X
1X
xi =
λi · xi ,
k i∈I
i∈I
n
n
which implies 0 ∈ C̊(x1 , . . . , xn ), a contradiction which proves the claim.
Lemma 19 implies that
∃a ∈ SH ∀c ∈ C̊(x1 , . . . , xn ) (ha, ci ≥ 0) .
Since H is spanned by x1 , . . . , xn , we can conclude that
(ha, x1 i, . . . , ha, xn i) > 0.
Now we show that two natural weakenings of SEP are still nonconstructive
since they imply Markov’s principle.
25
SEP0 Fix x1 , . . . , xn ∈ Rm and assume that 0 ∈
/ C̊(x1 , . . . , xn ). Then
∃ξ ∈ Rm ((hξ, x1 i, . . . , hξ, xn i) > 0) .
SEP00 Fix x1 , . . . , xn ∈ Rm and assume that
¬∃ξ ∈ Rm ((hξ, x1 i, . . . , hξ, xn i) > 0) .
Then 0 ∈ C̊(x1 , . . . , xn ).
Bear in mind that Markov’s principle is equivalent to the following axiom [8,
p. 15, Problem 8].
a 6= 0 ⇒ a < 0 ∨ a > 0 (a ∈ R)
Lemma 20. SEP0 implies MP.
Proof. Fix a real number a with a 6= 0. Consider SEP0 in the case of n =
m = 1 and x1 = (a). Then C̊(x1 ) = {a}, therefore we can conclude that
0∈
/ C̊(x1 ). Now SEP0 yields the existence of a real number ξ with ξa > 0,
this implies that |a| > 0.
Lemma 21. SEP00 implies MP.
Proof. Fix a real number a with a 6= 0. Consider SEP00 in the case of n = 2,
m = 1, x1 = |a|, and x2 = −1. If there were a ξ ∈ R with (ξ |a| , −ξ) > 0, we
could conclude that a = 0. Now SEP00 yields the existence of a p ∈ S̊2 with
p1 |a| = p2 . This implies that |a| > 0.
Note that SEP00 corresponds to the original version of the fundamental theorem of asset pricing (no arbitrage implies the existence of a martingale
measure), whereas SEP0 corresponds to the original version of the separation
theorem (disjoint implies the existence of a separating functional). As of
today, we are neither able to show that MP implies SEP00 nor are we able to
show that MP implies SEP0 . This would be of great interest.
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