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Math 490 Notes 3

Continuing our overview of theory, there are set operations which parallel all of the usual operations in . We previously mentioned the formation of a quotient set

A/S = {S(x) x ∈ A} relative to an S on A. If A has n elements and each equivalence (under S) has k elements, then A/S has n/k elements. Unions are es- sentially set “”; when A and B are with n and k elements, respectively, then A ∪ B has n + k elements. Set differences corresponds to subtraction.

To form set exponents, it is useful to first extend the definition of set products. Let

{Xi i ∈ I} be an arbitrary collection of sets, and let X = {f : I → X f(i) ∈ X for all i ∈ I}. i i ∈ i I X is called the of the X , denoted by X = X . Note that each func- i i i∈I tion f ∈ X is a which chooses a representative , f(i), from each set Xi. Such a function is often referred to as a “choice function”, and its existence is a consequence of the

Axiom of Choice. Also note that X = φ iff Xi = φ for some i ∈ I. In dealing with set prod- ucts, it is common to write f ∈ ∈ Xi as f =(xi)i∈I , where f(i)= xi. If I = {1, 2,...,n}, i I then f would be the n- (x1, x2,...,xn). If I = N, then f would be the sequence (xn)n∈N.

Set is simply a special case of set product. If A and I are sets, and Xi = A for all i, then ∈ Xi can be thought of as the product of A with itself ”I times”, and we write i I I I ∈ Xi = ∈ A = A . By the earlier definition of ∈ Xi, it follows that A is simply i I i I i I the set of all functions from I into A. If A and I are finite sets with n and k elements, respectively, then AI contains nk elements. Note that Rn = RI , where I = {1, 2,...,n}, and

RN is the set of all (infinite) sequences of real . p. 2

The P(X) of all of X is often written via exponential notation. Let ”2” denote the set {0, 1}, and for any A of X, let χA denote the characteristic function of A:  1, x ∈ A χ A =  . 0, x ∈ X − A  Then every f ∈ 2X is the characteristic function of some subset of X, and each subset of X is uniquely defined by its characteristic function. Thus P(X) can be identified with 2X .

Cardinal Numbers

Cardinal and ordinal numbers are not considered in Munkres, but both are frequently used in topology, so we’ll discuss them briefly here, beginning with cardinal numbers. Let A and

B be sets. Define A . B to mean that there exists a between A and a subset of

B. Also define A ∼ B to mean that there is a bijection between A and B. Then ∼ is an equivalence relation, and A ∼ B =⇒ A . B and B . A.

Schr¨oder-Bernstein Theorem If A and B are sets, then A ∼ B iff A . B and B . A.

It is natural to ask, ”On what set are the relations . and ∼ defined?”. The answer would seem to be the ”set of all sets”, but that answer is not consistent. If there were a set S of all sets, consider the subset A of S defined by

A = {B ∈ S B ∈ B}. Then by definition, if A is an element of itself, then A ∈ A, and if A is not an element of itself, then A ∈ A. This contradiction is known as Russel’s Paradox. We can avoid this paradox by saying that . and ∼ are relations on the class of all sets, denoted here by S. p. 3

The equivalence classes into which S is partitioned by ∼ are called cardinal numbers. Like

S itself, the cardinal numbers are a class which we’ll call C. The equivalence class consisting only of φ is called ”0”; the equivalence class consisting of all sets is called ”1”, etc. The equivalence class containing N is called ℵ0, and the equivalence class containing R we’ll denote by 2ℵ0 . If α and β are cardinal numbers, we define α ≤ β to mean ∃A ∈ α and

∃B ∈ β such that A . B. One can check that ≤ is a well-defined partial order on C. Note that the anti-symmetry of ≤ follows from the Schroder-Bernstein Theorem. Indeed, it can be shown that ≤ is a simple order using another equivalent version of the of Choice called the Well Ordering Theorem; more to come about that.

The smallest infinite cardinal is ℵ0. A set is countable iff its cardinal is ≤ ℵ0, and is otherwise uncountable. The distinction between countable and uncountable sets is important in analysis, so you may have seen various proofs of the facts that sets like Z and Q are countable, whereas R, R − Q, and the Cantor Set are uncountable. It is also well-known that countable unions and finite products of countable sets are countable.

We’ll conclude this discussion of cardinal numbers by briefly considering cardinal arithmetic.

For any set A, let |A| denote the cardinal number of A. Given two cardinal numbers α and

β, choose A ∈ α and B ∈ β such that A ∩ B = φ. It is not customary to define cardinal subtraction or division, but the other operations are defined as follows:

α + β = |A ∪ B|;

αβ = |A × B|;

αβ = |AB|.

For finite cardinals, cardinal arithmetic reduces to ordinary arithmetic. p. 4

Propostion N3.1 If α, β, γ are arbitrary cardinal numbers, then:

(a) Cardinal addition and are both commutative and associative;

(b) α(β + γ)=(αβ)+(αγ);

(c) α(β+γ) = αβαγ;

(d) (αβ)γ = αγβγ;

γ (e) αβγ = αβ . Propostion N3.2 Let α, β be cardinal numbers with α ≤ β and β infinite. Then:

(a) α + β = αβ = β;

(b) If α ≥ 2, then αβ > β.

Proof of (b): If α ≥ 2, then 2β ≤ αβ, so it suffices to show β < 2β. Then we know {0, 1}∈ 2, and we can choose some B ∈ β. As discussed previously, the set 2β can be regarded as (is in one-to-one correspondence with) the power set of B. The fact that |B| ≤ |2B| should be fairly clear; considering the injection from B to its singleton subsets gives us B . P(B) ∼ 2B.

Now, we show that no bijection exists between B and 2B (i.e. B ∼ 2B). Suppose, on the contrary, that there is a bijection g : B → 2B. Let A = {b ∈ B b ∈ g(b)}. Then A ∈ 2B, so ∃ a ∈ B such that g(a)= A. Now, if a ∈ g(a), then a ∈ A, and g(a)= A (a contradiction).

If a ∈ g(a), then a ∈ A, and g(a)= A (a contradiction). So no element of B can to A, and consequently g can not be a bijection. ¥

Well Ordered Sets

The theory of well ordered sets provides the foundation for the study of ordinal numbers.

A well ordered set (w.o.set) is a poset in which every non-empty subset contains a least element. p. 5

Proposition N3.3 Every well-ordered set is simply ordered.

Proof: Let X be a well ordered set, and let x, y ∈ X. Then the subset {x, y} must contain a least element, which is either x or y. If it is x, then x ≤ y. If it is y, then y ≤ x. Thus X is a poset which also satisfies the comparability axiom, and is therefore simply ordered. ¥

Every w.o.set contains a least element, but not every simply ordered set with a least el- ement is well ordered. For example, [0, 1] (with the usual ordering) has a least element

0, but the subset (0, 1) does not. Every finite simply ordered set is well ordered, as is the set N of natural numbers. If A and B are disjoint well ordered sets, then A ∪ B can be made into a w.o.set by stipulating that (a ∈ A and b ∈ B) =⇒ a < b, while leaving the orders within A and B unchanged. For instance, suppose A = {a0,a1,...,an,...} and

B = {b0, b1, . . . , bn,...}, where the elements of these sets are listed in increasing order from left to right. Then A ∪ B = {a0,a1,...,an, . . . , b0, b1, . . . , bn,...}, with elements again listed in increasing order from left to right, is a w.o.set. Furthermore, for any w.o.sets A and B, the set A×B is well ordered relative to the dictionary order. Any subset of a well ordered set with the inherited order is also well ordered. In a simply ordered set X, b is the immediate successor of a iff a < b and a

Proposition N3.4 In a w.o.set X, every element which is not the greatest element of X has an immediate successor, but every element does not necessarily have an immediate pre- decessor.

Proof: Let a be an element of X which is not the greatest element. And let b be the least element of {x ∈ A a < x}, which is is non-empty. Then b is the immediate successor of a. In the example mentioned previously involving A ∪ B, neither a0 nor b0 has an immediate predecessor. ¥ p. 6

Proposition N3.5 In a well ordered set X, every non-empty subset bounded above has a least upper bound, and every non-empty subset has a greatest lower bound.

Proof: If A = φ is bounded above, then the set U(A) of all upper bounds of A contains a least element, which is the l.u.b of A. Also, every A = φ contains a least element, which is the g.l.b. of A. ¥

An important theorem in is the Well Ordering Theorem, first proved by Zer- melo in 1904 using the . It was later discovered that the Well Ordering

Theorem is equivalent to the Axiom of Choice. We’ll state the Well Ordering Theorem here without proof.

Well Ordering Theorem Every set can be well ordered.

This theorem is clear for countable sets, but is not obvious in general. Like other results based on the Axiom of Choice, the Well Ordering Theorem asserts the existence of something without providing a means for achieving it. It is my understanding that:

If one assumes the Zermelo-Fraenkel of Set Theory and the Axiom of Choice, then it can be proved that R can be well-ordered, but no explicit formula for this well-ordering can be given;

If one assumes the Zermelo-Fraenkel Axioms as well as the “Axiom of Constructibility” (a condition stronger than the Axiom of Choice), then a complicated formula for the well- ordering of R can be given.

I believe I also remember reading that a formula for well-ordering the reals can also be given if one assumes ZF+GCH (Zermelo-Fraenkel plus Generalized Hypothesis).