Week 2. This week, we covered operations on sets and cardinality.

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1 Week 2 Ths week, we covered operatons on sets and cardnalty. Defnton 0.1 (Correspondence). A correspondence between two sets A and B s a set S contaned n A B = {(a, b) a A, b B}. A correspondence from A to A s called a relaton. Defnton 0.2 (Equvalence Relaton). An equvalence relaton s a relaton R on A satsfyng the followng: Reflexvty For all a A, (a, a) R Symmetry If (a, b) R then (b, a) R. Transtvty If (a, b) R and (b, c) R then (a, c) R. We often denote (a, b) R by arb. Defnton 0.3 (Functon). A functon f : A B s a correspondence from A to B such that for each a A, there exsts a unque b B such that (a, b) f. We denote (a, b) f by f(a) = b. Defnton 0.4 (Injectve, Surjectve, Bjectve). A functon s sad to be njectve f f(a) = f(a ) mples that a = a. A functon s sad to be surjectve f for all b B, there exsts a A such that f(a) = b. A functon s sad to be bjectve f t s njectve and surjectve. Defnton 0.5 (Equvalence). We say that two sets A and B are equvalent, wrtten A B f and only f there exsts a functon f : A B whch s a bjecton. Now, on fnte sets, ths amounts to them havng the same sze (see frst homework) Defnton 0.6 (Composton of functons). If f : A B and g : B C are functons, we defne g f by g f(a) = g(f(a)). Lemma 0.1. Let f : A B be a bjecton. Then there exsts a functon g : B A such that g f(a) = a and f g(b) = b for all a A and b B. We call g the nverse of f and denote t by f 1. 1

2 Proof. Defne a correspondence g from B to A by (a, b) f f and only f (b, a) g. We wll show that g s a functon. As f s surjectve, for every b B, there exsts a A wth f(a) = b. Thus, for any b B, there exsts some a A wth (b, a) g. To see that t s unque, we note that f s njectve, and so we have f(a) = f(a ) mples a = a. Specfcally, that means that f (b, a) g and (b, a ) g then a = a. Thus, for any b B, there exsts a unque a A such that (b, a) g. So g s a functon. To see that the compostons work out, fx a A. then g(f(a)) = a because (a, f(a)) f and (f(a), a) g. The same argument works for the other composton. Proposton 0.2. s an equvalence relaton. Proof. For any set A, we have 1 A : A A the functon whch has 1 A (a) = a for all a A. Ths s a bjecton A A, and so s reflexve. If f : A B s a bjecton, then f 1 : B A s also a bjecton, and so s symmetrc. If f : A B and g : B C are bjectons, then g f : A C s a bjecton, and so s transtve. So now we have an equvalence relaton on sets. Gven a set A, we ll call A the collecton of all sets equvalent to A. We call A the cardnal number of A. For fnte sets, we use numbers. So {a, b} = 2. Ths s justfed by the fact that the arthmetc we wll descrbe shortly s exactly the usual arthmetc for natural numbers when restrcted to fnte sets, as you are to prove n your homework. Now for a couple of examples Example 0.1. We wll show that N Z. Any functon f : N A for any set A can be thought of as a sequence, or a lstng. So, to descrbe f, we must merely lst the elements of A. For A = Z, we can take f to be the lst 0, 1, 1, 2, 2, 3, 3,..., n, n,.... Ths functon s surjectve and njectve, and so s a bjecton, and demonstrates that N Z. Example 0.2. We wll also show that N R. We ll actually do t by showng that N [0, 1] R. Every element of [0, 1] can be wrtten as a =1 10. Now, assume, for contradcton, that there s a bjecton N [0, 1]. Then the number j s sent to x j = a j =1 10. We now construct a number x [0, 1] not on the lst. Let x = a +1 =1 10 wth the conventon that = 0 rather than 10. Now, x can t be on the lst, because t dffers wth every element of the lst n some poston. Thus, we ddn t have a bjecton n the frst place, and N R. As a conventon, we say that f there s an njecton f : A B that A B. If there s an njecton but no bjecton, then we say A < B. So the above example n fact says that N < [0, 1] R, and so N < R. Defnton 0.7 (Power Set). Gven a set A, defne P(A) to be the set of all subsets of A. Call ths the power set of A. 2

3 Theorem 0.3. For any set A, A < P(A). Proof. That A P(A) follows from the exstence of the njecton A P(A) gven by a A maps to {a} A. To see that there s no bjecton, we assume one exsts for contradcton. Let f : A P(A) be a bjecton. Then, n partcular, t s surjectve. So any subset of A that we can descrbe s n the mage. Look at B = {a A a / f(a)} A. Now, there must be an element of A sent to B, and we ll denote t by a 0. Here s where we encounter the problem: can a 0 be n B? If a 0 B, then a 0 f(a 0 ) and so a 0 / B, whch s a contradcton. Smlarly, f a 0 / B, then a 0 / f(a 0 ), and so by the defnton of B, we have a 0 B, whch s also a contradcton. So, f we assume that there s a bjecton, there s a contradcton, so such a functon cannot exst, and A P(A). Though we won t be usng t later, we state the followng theorem wthout proof because t s mportant to know n general: Theorem 0.4 (Cantor-Schroeder-Bernsten). Let A and B be sets. If A B and B A then A = B. All ths says s that nstead of lookng for bjectons, we can just look for njectons both ways. The proof actually uses them to construct a bjecton, but we won t go nto detal there, because t s rather complcated and not relevant at the moment. Defnton 0.8 (Operatons on Sets and Cardnal Numbers). Gven two sets A, B, we defne the followng: A B = {(a, b) a A, b B} the Cartesan product, A B = (A {0}) (B {1}) and B A =the set of all functons f : A B. We defne A B = A B, A + B = A B and B A = B A. Note that the power set of A can be dentfed wth {0, 1} A, and has cardnalty 2 A. Defnton 0.9 (Partal Orderng). A partal order on a set A s a relaton on A that s reflexve, transtve and antsymmetrc (e, f a b and b a then a = b). We call a set wth a partal orderng a poset. (Partall Ordered SET) Defnton 0.10 (Total Order). A partal orderng s a total orderng f for any a, b A we have a b or b a. A subset of a poset f called a chan f the orderng restrcts to a total order on t. Example 0.3. The usual order on the natural numbers s a lnear orderng. However, the poset of subsets of a gven set ordered by ncluson s not, as for {1, 2, 3}, nether {1, 2} or {2, 3} contans the other. 3

4 Defnton 0.11 (Well Orderng). A total orderng on S s a well orderng f every subset of S has a least element. Theorem 0.5. The followng are equvalent: Axom of Choce Gven any collecton fo nonempty sets A b ndexed by b B, there exsts a functon f : B A b such that f(b) A b for all b B. Gven any collecton of nonempty sets, ther product s nonempty. Well-Orderng Prncple Gven a set S, there exsts a partal orderng on S whch s a well orderng. Zorn s Lemma Gven a poset S, f every chan has an upper bound (that s, an element x wth a x for all a n the chan) then there exsts a maxmal element of S. (We say a s maxmal f a x mples that a = x.) Most mathematcans just take the equvalent statements above to hold, though there s some dsagreement. For our purposes, we wll assume them. Theorem 0.6. If A, B are sets, A B, then A + B = B. Proof. It s smple to check that B A + B B + B, and so t s enough to check that B = B + B for nfnte sets. We wll proceed usng Zorn s Lemma to construct a bjecton between B and B B = B {0, 1}. Look at pars (X, f) wth X B and f : X {0, 1} X a bjecton. Frst, we must check that such thngs exst. Now, every nfnte set has a subset that s equvalent to N (Exercse), so we take C B to be such a subset. Then, as N {0, 1} s equvalent to N, (To see t, sent N {0} to the even natural numbers and N {1} to the odd natural numbers) we have C {0, 1} C a bjecton. So there exst pars lke ths. Now, we order them to get a poset as follows: (X, f) (Y, g) f and only f X Y and g X = f. That s, for all x X, g(x) = f(x). That ths s a partall orderng s left as an exercse. Now take any chan wth elements (X α, f α ). Then t has an upper bound gven by X = X α and f = f α. Here s one of the rare tmes we recall that a functon s a set. Now, f wll be a functon, because once an element of X s assgned a value, t must have the same value forever after. Now, Zorn s Lemma mples the exstence of (C, g), a maxmal element. Because g s a bjecton C {0, 1} C, we note that C = C + C. We wll now show that C = B. If B \ C s nfnte, then t would contan X B \ C equvalent to N. Then we can take h : X {0, 1} X a bjecton, and have X C wth f h a bjecton from (X C) {0, 1} X C. Ths, however, contradcts C beng maxmal, and so can t happen. Thus, B \ C s fnte. So then B = C (B \ C), and so B = C + B \ C = C + n for some fnte n. As an exercse, show that for any nfnte set, addng a fnte number doesn t change the cardnalty. Wth that, we are done, C + n = C = B, and C + C = C, so B + B = B and A + B = B. 4

5 Theorem 0.7. If A, B are sets, A, B nfnte and A B, then A B = B. Proof. Here B A B B B, so all we must show s that B B = B. We wll agan use Zorn s lemma. Look at pars (X, f) wth f : X X X a bjecton. We note that ths s nonempty, because there s a subset D B wth D N, and N N N gven by (m, n) 2 m (2n + 1) s a bjecton. Applyng Zorn s lemma n the same way as above gves a maxmal element (C, g) wth C C = C. Now, for contradcton, we suppose that B\C > C. Then there s a subset D B\C such that D C. Now, D C C C C D D C D D, and all of these sets are dsjont. Ths tells us that (C D) (C D) = C C + D D + C D + D C = C + C + D + D = C + D = C D, and so there s a bjecton (C D) (C D) (C D), whch contradcts the maxmalty of C. Thus, B \ C C, and so B = C + B \ C = C, by the prevous. 5