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Notes - Lecture 3 NOTES - LECTURE 3 E. F. WOLD 1. Cartesian products (x5 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 2 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 α2J will now be taken to mean what we have previously denoted by [α2AA; the union of all the sets in A, and \ Aα α2J will be taken to mean what we have previously denoted by \α2AA; the intersection of all the sets in A (recall that our standing assumption when dealing with intersections is that J be nonempty). If J = f1; 2; :::; ng 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: α2J α2J 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 = fRg, 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 = f1; 2; :::; ng ! A is an indexing function. We may then form the union X = A1 [···[ An and consider functions x : f1; :::; ng ! X J such that x(j) = xj 2 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+ = f1; 2; 3; :::g. If we let fA1;A2;A3; :::g be a family of sets indexed by Z+ we may form the union [ X = An; n2Z+ and consider functions x : Z+ ! X such that xn 2 An for all n. The set of all such functions is the infinite Cartesian product Πn2Z+ An: ! 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 f0g [ Z+ is sometimes denotet by !, and there ! is a simple bijection f : ! ! 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 (x6 in Munkres) Definition 2.1. A set A is said to be finite if there exists a bijection f : A ! f1; 2; :::; ng = Sn+1 for some n 2 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 : f0; 1g ! f1; 2; :::; ng be a function. Then there exists a function g : f1; 2; :::; ng ! f1; 2; :::; ng such that (i) g(h(0)) = h(1). (ii) g(h(1)) = h(0). (iii) g(j) = j if j2 = fh(0); h(1)g. 3 Proof. To be very pedantic, consider the set A = f1; 2; :::; ng × f1; 2; :::; ng, and construct the set (rule of assignment) using the axiom of specification r = f(j; gj) 2 A : gj = h(1) if j = h(0); gj = h(0) if j = h(1); otherwise gj = jg: Then we have that domain r = f1; 2; :::; ng = 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 2 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 2 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 : f0; 1g ! 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 2 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 2 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 2 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].
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