CHAPTER 1-A

SET THEORY AND THE REAL LINE

1. Ordered Sets

Let A be a set • x ∈ A means “x is a member of A” • x∈ / A means “x is not a member of A” • the set with no elements is called the empty set, denoted ∅ • let B be another set: A ⊆ B means “every element of A is in B”. We say that A is a of B, and that B is a superset of A; when A is a proper subset of B we will write A ⊂ B and means that ∀x ∈ A, x ∈ B and ∃y ∈ B, y∈ / A; note that A ⊆ B and B ⊆ A if an only if A = B. A * B means that A is not a subset of B • A ∪ B is the set of elements that belong to A or belong to B • A ∩ B is the set of elements that belong to both A and B • A − B is the set of elements of A that are not members of B • Ac is the set of elements that are not in A; it is called the complement of A • The Cartesian product of A and B is

A × B = {(x, y): x ∈ A, y ∈ B}.

• the set formed by all of A including the empty set is denoted 2A. In what follows, all the sets are suppose to be nonempty.

Definition 1.1. Given sets A and B, a (binary) relation R from A to B is a subset of A × B.

Notation 1.2. We will write most often xRy for (x, y) ∈ R; x R y means (x, y) ∈/ R.

Definition 1.3. Given sets A and B and the R from A to B, the inverse relation of R is R−1 = {(y, x):(x, y) ∈ R}.

Definition 1.4. Let R be a binary relation defined on a set A. We say that R is (1) Reflexive, if ∀x ∈ A, xRx. (2) Irreflexive, if ∀x ∈ A, x R x. (3) Symmetric, if xRy implies yRx. (4) Antisymmetric, if xRy and yRx imply x = y. 1 2

(5) Transitive, if xRy and yRz imply xRz. (6) Total, if ∀x, y ∈ A, xRy or yRx or both.

Definition 1.5.

Let R be a relation defined on a set A. The relation R is an equivalence relation if it is reflexive, symmetric and transitive.

Remark 1.6. An equivalence relation R partitions A into disjoint equivalence classes such that x and y are in the same class if and only if xRy. For, writing [x] = {z : zRx}, we have x ∈ [x] , so ∪{x : x ∈ A} = A. Also, for any x, y ∈ A, either [x] = [y] or [x]∩[y] = ∅.

Definition 1.7. Let R be a relation defined on a set A. The relation R is a partial order it is reflexive, antisymmetric and transitive. It is an order if it is total, reflexive, antisymmetric and transitive.

We will use ≤ (or ≥) instead of R to denote an order relation. When x ≤ y and x 6= y, we will write x < y.

Definition 1.8. An ordered set (, or poset) is a set in which an order (a partial order) is defined.

Definition 1.9. Let (A, ≤) be a poset. Let S ⊆ A be nonempty. An element b ∈ A is a maximal element of S if for any x ∈ S, b ≤ x implies x = b.

Definition 1.10. Let (A, ≤) be a poset. Let S ⊆ A be nonempty. An element b ∈ A is a minimal element of S if for any x ∈ S, x ≤ c implies x = c.

If the order is only partial, then for a maximal (minimal) element b of A, it could be neither x ≤ b nor b ≤ x.

2 2 Example 1.11. Let (R , ≤) where for x = (x1, x2), y = (y1, y2) of R , x ≤ y if and only if x1 ≤ y1 and y1 ≤ y2. This relation is a partial order (the Pareto order). Given 2 S ⊆ R , a maximal element of S is called a Pareto optimal point of S. For instance, in 2 R , (1, 1) is Pareto optimal in (−∞, 1] × (−∞, 1], but not in (−∞, 1] × (−∞, 2]. Consider 2 now S = {(x, y) ∈ R : x ≥ 0, y ≥ 0, x + y ≤ 1}. Every point of the segment joining (0, 1) and (1, 0) is Pareto optimal (efficient frontier of S). Let T = {(x, y): y ≤ −x2}. Then {(x, −x2) : 0 ≤ x} is the set of maximals points of A and there are no minimal elements.

Example 1.12. Let R be the set of real numbers, and define the relation ≤ by x ≤ y if and only if y − x ≥ 0. It is an order relation.

Example 1.13. Let AR be the set of functions f : A → R. Define f ≤ g in AR if and only if f(x) ≤ g(x) for all x ∈ A. It is a partial order. 3

Definition 1.14. Let (A, ≤) be an ordered . Let S ⊆ A be nonempty. • An element b ∈ A is an upper bound of S if and only if for all x ∈ S, x ≤ b. • An element c ∈ A is a lower bound of S if and only if for all x ∈ S, c ≤ x. • S is upper bounded if and only if S has upper bounds. • S is lower bounded if and only if S has lower bounds. • S is bounded if and only if it is both upper and lower bounded. • An element b ∈ A is the supremum of S, b = sup S, if and only if it is the least of the upper bounds of S: (1) ∀x ∈ S, x ≤ b and (2) ∀x ∈ S, x ≤ y, implies b ≤ y. • An element c ∈ A is the infimum of S, c = inf S, if and only if it is the greatest of the lower bounds of S: (1) ∀x ∈ S, c ≤ x and (2) ∀x ∈ S, y ≤ x implies y ≤ c. • If sup S ∈ S, then the supremum is a maximum. • If inf S ∈ S, then the infimum is a minimum.

Note that to be bounded in an arbitrary ordered set is not enough to have supremum and/or infimum. See problem 3 in Problem List 1.

Definition 1.15. An order has the least upper bound property if every upper bounded set admits a supremum. An order has the greatest lower bound property if every lower bounded set admits an infimum.

It can be proved that an order has the least upper bound property iff it has the greatest lower bound property.

Definition 1.16. Given sets A and B, a relation f from A to B, written f : A −→ B, is a function provided (x, y) ∈ f and (x, z) ∈ f imply y = z. The domain of the function, domf = {x :(x, y) ∈ f}, is the set of all possible first elements of the ordered pairs in f. The image of the function, im f = {y :(x, y) ∈ f}, is the set of all possible second elements of the ordered pairs in f.

Notation 1.17. f(x) = y means (x, y) ∈ f. Given a subset E of A

f(E) = {f(x): x ∈ E} is the image of E by f. Given a subset E of B

f −1(B) = {x : f(x) ∈ B} is the inverse image of E by f. The set of all functions between A and B is denoted BA.

Definition 1.18. The function f : A −→ B is onto iff f(A) = B.

Definition 1.19. The function f : A −→ B is one-to-one iff (x1, y) ∈ f and (x2, y) ∈ f imply x1 = x2 4

Definition 1.20. The function f : A −→ B is bijective iff it is both onto and one-to-one.

Given a function f, when is the inverse relation f −1 a function?

Proposition 1.21. Given the function f : A −→ B, the inverse relation f −1 is a function iff f is bijective.

f −1 : B −→ A is called the inverse of f.

2. Finite, Countable and Uncountable Sets

Definition 2.1. We say that two sets A and B have the same and write A ∼ B, iff there is a bijective mapping between A and B.

Definition 2.2. Let n ∈ N and let Jn = {1, . . . , n}. We say that the set A is

(1) Finite, if A ∼ Jn for some n. (2) Infinite, if A is not finite. (3) Countable, if A ∼ N. (4) Uncountable, if A is neither finite nor countable. (5) At most countable, if A is finite or countable.

n Example 2.3. Z is countable: consider f : N −→ Z be defined as follows: f(n) = − 2 if n n+1 even, f(n) = 2 if n odd. It is a bijection.

Example 2.4. N × N is countable: consider f : N × N −→ N be defined as follows: f(n, m) = 2n−1(2m − 1). It is a bijection. This can be also proved by the following geometric argument. Arrange N × N as an infinite matrix (1, 1) (1, 2) (1, 3) ... (2, 1) (2, 2) (2, 3) ... (3, 1) (3, 2) (3, 3) ...... so that (n, m) is in the nth row and mth column. We can enumerate N × N without repetitions as (1, 1), (2, 1), (1, 2), (3, 1), (2, 2), (1, 3),...

The next result means that the cardinality of A is always smaller than the cardinality of the power set of A (The set formed by all subsets of A, including the empty set). The next result is intuitively obvious, but the proof requires some care.

Theorem 2.5. Every infinite subset of a countable space is countable.

The next result says that the countable union of countable sets is countable. 5

S∞ Theorem 2.6. Let {An} be a sequence of countable sets and let A = n=1 An. Then A is countable.

Corollary 2.7. Q is countable.

m Consider the sets An = {r = n : m ∈ Z} for n ∈ N. Note that An ∼ Z and ∞ Q = ∪n=1An.

Theorem 2.8. Let A be a countable set and let Bn be the set of all n-tuples (x1, . . . , xn), where xk ∈ A for k = 1, . . . , n. Then Bn is countable.

Corollary 2.9. Q is countable.

S m Another proof of this fact: = ∗ { }. Q (m,m)∈Z×Z n Next is an example of an uncountable set. Theorem 2.8 above established that the Cartesian product of finitely many copies of a countable set is countable. However, the countable copy of a set is not countable, even if the set is finite. In the following, {0, 1}N is the set of all binary sequences (or the set of all functions N −→ {0, 1}). An element of N {0, 1} is represented by (x1, x2, . . . , xi,...), where xi = 0 or 1.

Theorem 2.10. The set {0, 1}N is uncountable.

N Proof. Cantor’s diagonal process. Let f : N −→ {0, 1} and define xn as the n component in f(n). Let ( 1, if xn = 0; yn = 0, if xn = 1. and consider the binary sequence {yn}. By construction, there is no m ∈ N for which f(m) = {yn}. Thus f is not onto. As the set {0, 1}N is obviously infinite, it is uncountable. The diagonal argument of Cantor is better seeing by making the arrangement

f(1) f11 f12 f13 ... f(2) f21 f22 f23 ... f(3) f31 f32 f33 ......

Then {xn} is the sequence of diagonal elements, xn = fnn, and by construction the sequence {yn} is not in the above list. 

More examples of uncountable sets are R and intervals [a, b] of the real line with a < b. 6

3. The Real

Definition 3.1. A field is a set F with two inner operations, addition “+” and multipli- cation “·”.

See Rudin (1976) to know what properties has to satisfy an operation to be named “addition” or “multiplication”. Examples of fields are R and C. Definition 3.2. An ordered field is a field F which is also an ordered set such that addition and multiplication are consistent with the order, that is, (1) ∀x, y, z ∈ F , y ≤ z implies x + y ≤ x + z; (2) ∀x, y ∈ F , 0 ≤ x and 0 ≤ y implies 0 ≤ xy.

Theorem 3.3. There exists and ordered field, denoted R, which has the least–upper–bound property (every set bounded above has a supremum) and contain the rationals.

Automatically, any bounded above (below) subset of R has a supremum (an infimum). Remark 3.4 (Completeness Axiom). Any nonempty subset A of the real line which is bounded above (below) admits a supremum (infimum)

Remark 3.5 (Approximation property). Let A 6= ∅ be a subset of the real line. • Suppose that A is bounded above, and let b = sup A. Then

∀a < b, ∃x ∈ A such that a < x ≤ b.

• Suppose that A is bounded below, and let c = inf A. Then

∀a > c, ∃x ∈ A such that a > x ≥ c.

To prove the first assertion, note that x ≤ b for all x ∈ A. Now, if x ≤ a for all x ∈ A, then a is an upper bound of A, thus b ≤ a. This contradicts that a < b. Hence there is x ∈ A such that a < x.

Remark 3.6 (Impossibility of ordering the complex numbers). The set of complex num- √ bers, C = {a + ib : a, b ∈ R}, where i = −1, is not an ordered field. Suppose on the contrary that a compatible order with the sum and the product can be given. We will arrive to a contradiction as follows. Since i 6= 0, we have i > 0 or i < 0. If i > 0, then i2 > 0, that is −1 > 0. Adding 1 to both sides, we have 0 > 1. On the other hand, from −1 > 0, we have (−1)(−1) > 0, that is, 1 > 0, which contradicts 0 > 1. A similar reasoning invalidates the possibility i < 0.

The main properties of R are that it is Archimedean, that is, for any x, y ∈ R with x > 0, there is n ∈ N such that nx > y and that Q is dense in R, that is, for any x, y ∈ R, x < y, there is r ∈ Q such that x < r < y. 7

It can be given a nice representation of real numbers x ∈ R in terms of decimals. Let n0 be the largest integer such that n0 ≤ x. Now, let n1 be the largest integer such that k n0+n1/10 ≤ x and given k, let nk be the largest integer such that n0+n1/10+···+nk/10 ≤ x. Let n n n o E = n + 1 + ··· + k : k = 0, 1,... . 0 10 10k Then x = sup E. The decimal expansion of x is n0.n1n2n3 ....