Declaración cuantitativa y cualitativa sobre riesgos de mercado

In this section we’ll apply the axiom of choice and Zorn’s lemma to show that any infinite set has the same cardinality of its Cartesian powers.

Lemma 2.26 Every infinite set contains a countably infinite subset.

Proof We shall prove that any infinite set X admits a 1-1 map f : N → X. The range f(N) will be the required countable subset. Call P0(X) the family of finite subsets

of X ; for every A∈ P0(X) the complement X − A isn’t empty and we can write

X = ∪{X − A | A ∈ P0(X)}.

The axiom of choice ensures the existence of g: P0(X) → X such that g(A) ∈ A

for every finite subset A⊂ X. Hence we can define a 1-1 map f : N → X recursively by setting

f(1) = g(∅) and f (n) = g ({ f (1), f (2), . . . , f (n − 1)}) for n > 1.



2.6 The Cardinality of the Product 37

Proof It’s not restrictive to assume A ∩ B = ∅. By Lemma2.26 there exists a countably infinite subset C ⊂ A. Note that C ∪ B is countable, i.e. it has the same cardinality of C, and so A = C ∪ (A − C) has the same cardinality of A ∪ B =

(C ∪ B) ∪ (A − C). 

Lemma 2.28 |X × N| = |X| for any infinite set X.

Proof Consider the familyA consisting of pairs (E, f ) with E ⊂ X and f : E×N → E injective. We know X contains countably infinite subsets, soA isn’t empty. We

can orderA using the extension relation, and declare (E, f ) ≤ (H, g) iff E ⊂ H and g extends f . It isn’t hard to show that any chain is bounded, and Zorn’s lemma then tells there is a maximal element(A, f ).

Now we prove|A| = |X|. By contradiction, suppose |A| < |X|, so that X − A is infinite and must contain a countably infinite subset B. Choose a 1-1 map g: B×N →

B and then define h: (A ∪ B) × N → A ∪ B by h(x, n) = f (x, n) if x ∈ A and h(x, n) = g(x, n) if x ∈ B. Then h is injective and extends f , against the maximality

of(A, f ). 

Remark 2.29 It descends from Lemma2.28that if Y is infinite and X= ∪+∞_i=1Xi is the union of a countable family where|Xi| ≤ |Y | for every i, then |X| ≤ |Y |. Let

in fact A⊂ N be the set of indices i such that Xi = ∅; if A = ∅ there’s nothing to

prove, while if A= ∅

|Y | ≤ |Y × A| ≤ |Y × N| = |Y |

and so|Y | = |Y × A|. If we choose for every i ∈ A an onto map fi: Y → Xi, it

follows that f: Y × A → X, f (y, i) = fi(y), is onto.

Theorem 2.30 Any infinite set X satisfies|X| = |X2|.

Proof The argument is very similar to that of Lemma2.28. Consider the familyA of pairs(E, f ) where E is a non-empty subset in X and f : E × E → E is one-to-one. As X contains countably infinite subsets,A = ∅. Let’s order A by extension, i.e.

(E, f ) ≤ (H, g) iff E ⊂ H and g extends f . Easily, every chain is bounded, and so

there is a maximal element(A, f ) by Zorn’s lemma. The fact that f is 1-1 tells that |A2_{| = |A|, so it is enough to prove that |A| = |X|. Again by contradiction, suppose}

|A| < |X|. Then X − A contains some B of the cardinality of A, and in particular |B| = |B × A| = |A × B| = |B × B|.

Since(A ∪ B) × (A ∪ B) = A × A ∪ A × B ∪ B × A ∪ B × B, and by Remark2.29

there is a bijection from B to A× B ∪ B × A ∪ B × B, we can extend f to A ∪ B. But this contradicts the maximality of(A, f ).  Given a set X we define S(X) to be the disjoint union of all Cartesian powers

Xn, n∈ N. When X is finite and non-empty S(X) is countable. We also remind that

Corollary 2.31 |S(X)| = |P0(X)| = |X| for any infinite set X.

Proof For any integer n> 0 we have |Xn| = |X|. In fact, if n > 1 by induction we

have|Xn| = |Xn−1× X| = |X × X| = |X|, so S(X) is a countable union of sets with the cardinality of X . The natural onto map S(X) → P0(X) − {∅} then shows

that|X| = |P0(X)|. 

Exercises

2.26 Prove that any infinite set can be written as disjoint union of countably infinite

subsets.

2.27 LetK be an infinite field. Prove that the ring K[x] of polynomials with coef-

ficients inK has the same cardinality of K.

2.28 Let X, Y be infinite sets and A ⊂ X × Y a subset such that:

1. for every x∈ X, {y ∈ Y | (x, y) ∈ A} is countable; 2. for every y∈ Y , {x ∈ X | (x, y) ∈ A} isn’t empty.

Prove that|X| ≥ |Y |. (Hint: show there exists a 1-1 map A → X × N.)

2.29 LetQ ⊂ C denote the set of complex numbers that are roots of polynomials

with rational coefficients. Show thatQ is countable and deduce the existence of tran- scendental numbers. (Hint: Exercise2.28. A number is transcendental, by definition, if it does not belong toQ.)

2.30 LetB be a basis of the vector space V over a field K. Describe how a surjective

mapP0(K × B) → V looks like and deduce that if the set K × B is infinite, then

|V | = |K × B| = max(|K|, |B|).

2.31 (K, ♥) Let I be an infinite set, K a field and B a basis for the vector space

KI_{of maps f: I → K. Prove that the cardinality of B is strictly bigger than that of} I . Deduce that every infinite-dimensional vector space cannot be isomorphic to its

algebraic dual.

References

[Ke55] Kelley, J.L.: General Topology. D. Van Nostrand Company Inc., Toronto (1955) [To03] Tourlakis, G.: Lectures in Logic and Set Theory, vol. 2. Cambridge University Press,

Cambridge (2003)

Chapter 3

Topological Structures

According to some psychologists, before the age of two-and-a-half children simply scribble. Between two-and-a-half and four they develop an understanding of sorts for topology, because they draw different pictures for open and closed figures. From four years on they can replicate all topological notions: a point inside a figure, outside it or on the border. Only between four and seven they start distinguishing simple figures (like squares, triangles) based on their size or angles.

A space is a set whose elements are called points. Putting a topological structure on a space X means being able to say, given a subset A, which are the interior points, which the exterior points and which the boundary points. Obviously this can’t be done completely arbitrarily, and there are a few requirements dictated by common sense:

1. any point in X is an interior point of X ; 2. if x is an interior point of A, then x ∈ A;

3. if x is an interior point of A and A⊂ B, x is an interior point of B;

4. if x is an interior point of both A and B, then it is an interior point of A∩ B; 5. if A◦denotes the set of interior points of A, then any point of A◦is an interior

point of A◦.

Exterior points of a subset are those that are interior to the complementary set, and boundary points are neither interior nor exterior points.

The above five conditions can be adopted as axiomatic definition of a topological structure (see Exercise3.14). Nowadays, though, the standard practice is to define topological structures by way of families of open sets, which is what we’ll do next.