NOTES - LECTURE 3

E. F. WOLD

1. Cartesian products (§5 in Munkres) 1.1. Indexed sets.

Definition 1.1. Let A be a non-empty family of sets. An indexing function for A is a surjective function f : J → A from some set J to A; the set J is called the index set. If α ∈ J we will let Aα denote the set f(α).

Before we define product space, we use the notion of indexed sets to introduce a new notation for unions and intersections. [ Aα α∈J will now be taken to mean what we have previously denoted by ∪α∈AA; the union of all the sets in A, and \ Aα α∈J will be taken to mean what we have previously denoted by ∩α∈AA; the intersection of all the sets in A (recall that our standing assumption when dealing with intersections is that J be nonempty).

If J = {1, 2, ..., n} we have that A consists of sets A1, ..., An, and in agreement with another previous notation we have [ \ Aα = A1 ∪ · · · ∪ An and Aα = A1 ∩ · · · ∩ An. α∈J α∈J Remark 1.2. Note that it is only assumed that the map f in the definition of an indexed set is surjective not injective. So for instant J could be Z+ and A = {R}, and f would be the constant map.

Date: August 31, 2020. 2010 Mathematics Subject Classification. 32E20. 1 2 WOLD

1.2. Cartesian products. We may now redefine finite Cartesian products. Let A be a family of sets and suppose that f : J = {1, 2, ..., n} → A is an indexing function. We may then form the union X = A1 ∪ · · · ∪ An and consider functions x : {1, ..., n} → X J such that x(j) = xj ∈ Aj for each j. The Cartesian product A = A1 × · · · × An is defined to n be the set of all such x. Note that if all the Aj’s are the same set A we usually write A instead of AJ .

Now, nothing stops us from considering the larger set Z+ = {1, 2, 3, ...}. If we let {A1,A2,A3, ...} be a family of sets indexed by Z+ we may form the union [ X = An, n∈Z+ and consider functions x : Z+ → X such that xn ∈ An for all n. The set of all such functions is the infinite Cartesian product

Πn∈Z+ An. ω If all the sets An are the same set A we denote this also by A .

The reason for the above notation is that the set {0} ∪ Z+ is sometimes denotet by ω, and there ω is a simple bijection f : ω → Z+ given by f(n) = n + 1. So an alternative notation for A could be AZ+ . We could now go on for any two sets X and Y to define XY .

2. Finite sets (§6 in Munkres) Definition 2.1. A set A is said to be finite if there exists a bijection

f : A → {1, 2, ..., n} = Sn+1 for some n ∈ N. Note that for n = 0 this set becomes S1 which by convention is the emptyset. We call n thee cardinality of A; in particular the emptyset has cardinality zero.

We will now consider the following question: is the cardinality of a set uniquely determined? Stated in another way: could there be a bijection fn : A → Sn and a bijection fm : A → Sm if m 6= n? Since a composition of bijections is a bijection, this could be restated: could there exist a bijection h : Sn → Sm for n 6= m. This seems of course completely unreasonable but we will never the less give a proof. We start by giving a lemma. Lemma 2.2. Let h : {0, 1} → {1, 2, ..., n} be a function. Then there exists a function g : {1, 2, ..., n} → {1, 2, ..., n} such that (i) g(h(0)) = h(1). (ii) g(h(1)) = h(0). (iii) g(j) = j if j∈ / {h(0), h(1)}. 3

Proof. To be very pedantic, consider the set A = {1, 2, ..., n} × {1, 2, ..., n}, and construct the set (rule of assignment) using the axiom of specification

r = {(j, gj) ∈ A : gj = h(1) if j = h(0), gj = h(0) if j = h(1), otherwise gj = j}. Then we have that domain r = {1, 2, ..., n} = range r. 

Proposition 2.3. Assume that f : Sn → Sm is a bijection. Then m = n.

Proof. We will prove this by induction on n ∈ Z+. Thus, we consider first the case n = 1. In that case Sn = S1 = ∅ and so clearly we cannot have m > 1, and we conclude that m = 1. So the proposition holds for n = 1.

Assume then that the proposition holds for some n ∈ Z+ with n ≥ 1; we will prove that it also holds for n + 1. So suppose f : Sn+1 → Sm is a bijection. Then clearly m ≥ 2 since Sn+1 is now non-empty. Let h : {0, 1} → Sm be the map determined by h(0) = m − 1 and h(1) = f(n + 1). Then the previous lemma furnishes a map g : Sm → Sm such that setting f˜ = g ◦ f we obtain a bijection f˜ : Sn+1 → Sm with f˜(n) = m − 1. Thus, we may restrict f˜ to get a bijection fb : Sn → Sm−1. By the induction hypothesis we have that n = m − 1 from which it follows that n + 1 = m. This completes the proof.  We will leave it as an exercise to prove the following proposition - similar in spirit.

Proposition 2.4. There is no injection f : Sn → Sm for m < n.

Corollary 2.5. Z+ is not finite.

Proof. If it were there would be a bijection f : Z+ → Sn for some n ∈ Z+. Taking the restriction ˜ we would get an injection f : Sn+1 → Sn. A contradiction.  Another result that we will leave as an exercise to prove the following proposition.

Proposition 2.6. Let A be a set, and let f : A → Sn be an injection for some n ∈ Z+. Then A is is finite. Corollary 2.7. Let A and B be finite sets. Then A × B is a finite set.

Proof. Then A is in bijective correspondence with Sn, and B with Sm, for some n, m ∈ Z+. It suffices to consider the set Sn × Sm. If either m or n equals 1 then A × B = ∅ which is finite, so we assume m, n ≥ 2. Define a map f : A = Sn × Sm → Z+ by f(k, l) = 2k · 3l.

Then clearly f(A) ⊂ S2n·3m+1, and by uniqueness of prime factorization we have that f is injective. 

E. F. Wold: Department of Mathematics, University of Oslo, Postboks 1053 Blindern, NO-0316 Oslo, Norway Email address: [email protected]