Exercises 1 Sets and functions

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1 Exercises 1 Sets ad fuctios HU Wei September 6, Basics Set theory ca be made much more rigorous ad built upo a set of Axioms. But we will cover oly some heuristic ideas. For those iterested studets, please cosult other referece, e.g. [1]. Further, oe may wish to read some books o Mathematical logic, we will ot go through that. Exercise 1. If A, B are sets, show that A = B if ad oly if A B ad B A. This exercise is quite importat, it is almost our oly method to prove equality of two sets. Remark. Some few words about Axiom of Extesio: It is ot completely trivial. This axiom meas that whe we dealt with sets, we oly care about its extesio, i.e, cotet(elemets), scope istead of itetio, i.e., meaig, cocept. For example, the set of black US presidets is curretly equal i extesio to the set cotaiig Barack Obama as a sigle elemet, but they are differet i itetio. From: https: // math. stackexchage. com/ questios/ 85178/ what-does-extesio-mea-i-the-axiom-of-extesio. Exercise. Suppose A, B, C are sets, show that A B ad B C implies A C. Exercise 3. Show that if A is a set, the A. This may seems trivial to some of you ad weird to others. For example, suppose we have two sets, oe is of huma beigs ad the other is of flowers, the they have a subset (empty set) i commo. We uderstad this by showig its iverse is false. Actually, i mathematics, we admit law of excluded middle, i.e., either oe statemet or propositio is true or false. It is ot like dialectic. Remark. We ote that the empty set is uique, which is i essece a result of Axiom of Extesio. 1

2 Operatios Exercise 4. Show that (A B) C = A (B C). (Associative law) Solutio. Oe may proceed i the followig way. First, A A ad B B C implies A B Right had side. The C Right had side together coclude Left had side Right had side. You may show the coverse part similarly. Exercise 5. A (B C) = (A B) (A C). (Distributive law) Remark. Take the complimet of above, apply DeMorga s law, we have, A c (B c C c ) = (A c B c ) (A c C c ), the substitute A by A c, B by B c, C by C c, we have aother distributive law, Exercise 6. A (B C) = (A B) C. A (B C) = (A B) (A C). Remark. This exercise ca be uderstood as follows. B C exactly oe of B, C; A (B C) exactly oe of A, (B C) : either i A ad ot i B C, or i exactly oe of B, C ot i A either i A ad (i both B, C or ot i ay of B, C), or i exactly oe of B, C ot i A either i all A, B, C, or i A ot i both B, C, or i exactly oe of B, C ot i A either i all A, B, C, or i exactly oe of A, B, C Remark. Oe may prove that A 1 A umbers of A i. cotais those elemets that belogs to odd Remark. Oe may also prove this by cosiderig the characteristic fuctios of sets ad trasform operatios ito additio ad product of fuctios. Exercise 7. Take symmetric differece as additio, iteractio as multiplicatio, we the have a commutative algebra (rig with idetity) structure over sets. Exercise 8. A = 3 Cardiality for fiite sets At this momet, we oly cosider cardiality for fiite set. We say the cardiality of a set A is if there is a bijectio betwee A ad {1,,, }. This is well-defied sice for m, there is o bijectio betwee {1,,, } ad {1,,, m}. (Pigeohole s priciple) Exercise 9. If > m 1, prove there is o ijectio from {1,,, } to {1,,, m}.

3 Exercise 10. If A ad B are fiite sets ad A B =, show that, #(A B) = #A + #B. Exercise 11. If A is a set, S is a subset of A, show that, #(A\S) = #A #S. Exercise 1. Show that #(A B) = #A + #B #(A B). Exercise 13. Show that #(A B C) = #A + #B + #C #(A B) #(A C) #(B C) + #(A B C). Solutio. Oe ca also write #(A B C) = [A] [B\(A B)] [C\(A B C)], as disjoit uios. The, reorderig A, B, C, ad takig average, we get desired form. Remark. Oe may derive the geeral form, #( i=1a i ) = =J {1,,,} To prove this, we simply mimic last solutio. 4 Numbers Exercise 14. If a, b Z, the ( a)b = a( b) = (ab). Exercise 15. If a, b, c Z, a 0 ad ab = ac, the b = c. ( 1) #J 1 #( j J A j ). Exercise 16. Let be a positive iteger, prove the Biomial Theorem: If a, b Z, the ( ) (a + b) = a k b k. k k=0 Exercise 17. For ay positive iteger greater tha or equal to, there exists a prime p such that p divides. Exercise 18. I Exercise 7, we show take symmetric differece as additio ad itersectio as multiplicatio, sets have the commutative rig with idetity 1 structure. Now let s take all subsets of X ad ivestigate whe it is a field. Exercise 19. Z p is a field if p is prime. Remark. If ow p = ab is ot prime, the ab = 0 Z p. Therefore, a caot have multiplicative iverse. Remark. If a, b are relatively prime (1 is the oly commo divisors of them), the there are itegers x, y such that, xa + yb = 1. To prove it, oe use Euclidea divisio ad the well-orderig priciple for Z. 3

4 5 Fuctios Exercise 0. Suppose A, B, C are sets, f : A B ad g : B C are bijectios. Show that g f is a bijectio, compute (g f) 1. Exercise 1. Give f : A B, suppose there are g, h : B A such that f g = I B, h f = I A. Show that f is bijective ad that g = h = f 1. Exercise. Defie f : N Z by, Show that f is bijective. f() = { is eve, 1 is odd. Remark. This is very iterestig, we see that itegers ca be mapped oe to oe oto its proper subset. Later, we will see that this is the essetially differece betwee fiite ad ifiite sets. Exercise 3. Fid a example showig the equality dose ot hold i f(a 1 A ) f(a 1 ) f(a ). Exercise 4. Show that f(a 1 A ) = f(a 1 ) f(a ) if f is ijective. Ad if f(a 1 A ) = f(a 1 ) f(a ) holds for all subsets A 1, A, the f is ijective. Exercise 5. Let s defie two fuctios from R to N. Michelle fuctio, f(x) equals to the third digit after decimal poit. Greatest iteger fuctio, [x] equals to the largest iteger that is less tha or equal to x. Exercise 6. A is a fiite set, B is a subset of A. Show that B is also fiite ad has less or same elemets tha A. 6 Ifiite set Exercise 7. Suppose A B C, ad A, C have the same cardiality. Show that A, B has the same cardiality. Exercise 8. We give a explicit bijectio from Q + to N. For p q prime factorizatio, let s write, p q = pα1 1 p α p α, q β 1 1 q β qm βm the defie, f( pα 1 1 p α p α ) = p α 1 q β 1 1 q β qm βm 1 p α p α q β q β 1 q βm 1 m. Show that this is a bijectio. Fid the fractioal umber mapped ito 10 k. Exercise 9. Show that a subset of coutable set is either fiite or coutable. 4 i the lowest terms, by

5 Exercise 30. If A, B have the same cardiality their power sets. Exercise 31. If A 1,, A are coutable, the A 1 A A is coutable. Exercise 3. Now suppose we have coutably may coutable sets, what is cardiality of i=1 A i? 7 Axiom of Choice Exercise 33. Show that i a partially ordered set, least upper boud may ot exist. If it exists, it is uique. Axiom 1 (Axiom of Choice). For every collectio of oempty sets there exists a choice fuctio. Axiom (Hausdorff Maximality Priciple). Every partially ordered set X cotais a totally ordered subset that is maximal uder set iclusio. Axiom 3 (Zor s Lemma). If o-empty partially ordered set has the property that every o-empty totally ordered subset has a upper boud, the the partially ordered set has a maximal elemet. Axiom 4 (Well-Orderig Priciple). Every set ca be well-ordered. Namely, oe ca also itroduce a total order such that every o-empty subset has a least elemet. Theorem 1. These four axioms are equivalet to each other. Lemma. Suppose (X, R) is a o-empty partially ordered set such that every o-empty totally ordered subset has a least upper boud. If f : X X is such that x f(x) for all x X, the there is some ω such that f(ω) = ω. Refereces [1] Paul R Halmos. Naive set theory. Courier Dover Publicatios,