Seminaar Abstrakte Wiskunde Seminar in Abstract Mathematics Lecture notes in progress (27 March 2010)

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1 Seminaar Abstrakte Wiskunde Seminar in Abstract Mathematics Lecture notes in progress (27 March 2010)

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3 Contents 2009 Semester I: Elements 5 1. Cartesian product of two sets 5 2. The empty set 6 3. Subsets 7 4. Relations 7 5. Functions 8 6. Injective and surjective functions A paradox in naive set theory Semester II: Numbers Few words about mathematical logic Equivalence relations Small sets and cardinal numbers Projects Semester I: Structures Operations Monoids 22 3

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5 2009 Semester I: Elements One of the standard answers to the question What is mathematics? is Mathematics is a science which studied mathematical objects. But then one may ask what kind of objects are the mathematical objects? Of course, mathematical objects should include numbers, geometric objects like a triangle or a circle, functions, etc. A simpler and more precise answer is: mathematical objects are sets. Then, numbers are special kinds of sets, geometric objects are other kinds of sets, functions are also certain kinds of sets, and so on (details will be given later on). In other words, everything is a set. The aim of this semester is to learn to work with sets, almost like in the school we learn how to work with numbers. 1. Cartesian product of two sets Definition 1. Let x and y be two sets. Then the pair (x, y) is the set (x, y) = {{x, y}, {x}}. Lemma 1. For any two pairs (x, y) and (a, b), we have if and only if we have (x, y) = (a, b) x = a and y = b. Proof. Suppose first (x, y) = (a, b). According to the definition of a pair, this means {{x, y}, {x}} = {{a, b}, {a}}. (1) These two sets are equal to each other means that every element of one set is also an element of the other set, and vice versa. So {x, y} is an element of {{a, b}, {a}}. Then either we have {x, y} = {a, b} or {x, y} = {a}. If {x, y} = {a}, then again, since we have equality of two sets, we conclude that these sets have the same elements, so this forces x = a = y. To finish treating the case {x, y} = {a}, it remains to show y = b. Indeed, if {x, y} = {a} then by (1) we should have {x} = {a, b}, which itself implies a = x = b. This, together with the previously obtained equalities obviously gives b = a = y. The case {x, y} = {a, b} can be treated in an analogous way (see Exercise 1). Now suppose we have x = a and y = b. Then we have {x, y} = {a, b} and {x} = {a}, which gives (1). Therefore, (x, y) = (a, b). Definition 2. Let X and Y be two sets. Then the cartesian product X Y of X and Y is a new set X Y = {(x, y) x X, y Y }. In other words, X Y is the set of all pairs (x, y) where x is an element of X and y is an element of Y. 5

6 SEMESTER I: ELEMENTS Thus, for instance, if X has exactly three elements x, y, z (i.e. X = {x, y, z}, with x y, x z and y z) and Y has exactly two elements a, b, then the elements of the set X Y are the following six pairs: That is, (x, a), (x, b), (y, a), (y, b), (z, a), (z, b). X Y = {(x, a), (x, b), (y, a), (y, b), (z, a), (z, b)}. Note: above, the order in which we list the pairs is irrelevant, since the elements in a set are not ordered. Thus, for instance, {x, y} = {y, x}. Notice also the distinction between the set {x, y} and the pair (x, y), where in the latter case we would have (x, y) = (y, x) if only if x = y (as this follows from Lemma 1). Exercise 1. Complete the proof of Lemma 1. Exercise 2. If X has exactly n elements and Y has m elements, then how many elements does X Y have? (Hint: if you find it hard to work with general numbers n and m, first try to answer the question for a specific n and m, e.g. n = 2 and m = 3.) Exercise 3. Let X and Y be any two sets. Then is it true that X Y = Y X? If not always, then when is this true? 2. The empty set In order to be able to speak about sets, it seems natural that we must first begin with specifying certain objects, and then we can start constructing sets out of those objects. For instance, we can begin with the objects being all natural numbers 1, 2, 3, 4,... Then, we can form, for example, the following sets: {1, 2, 3}, {4, 5}, {1, 5}, {{4}, {3, 5}}, {{{1}}}. For instance, the last set is a set which has just one element x, where x is itself a set with just one element y, where y is the set with the only element 1. But it turns out, we do not need to begin with anything. In other words, we can begin with nothing. What is nothing? It is when we do not have any objects. Can we form a set out of it? Yes we can it is the empty set, which we will denote by. Thus, is the set which does not have elements. Now we see that we actually got something out of nothing; indeed, for is something, and we can use it to construct other sets, like we used 1, 2, 3,... For example, we can construct the following sets out of it: { }, {, { }}, {{ }}, {{{ }}}. Lemma 2. Prove that for any set X, we have X = = X. Proof. The set X is defined as the set of all pairs (x, y) where x is an element of X and y is an element of. But has no elements, so there is no such pair, which means that X also does not have any elements and therefore it is the empty set. Similar argument can be used to show X =.

7 4. RELATIONS 7 3. Subsets Definition 3. A set X is said to be a subset of a set Y, written as X Y, if every element x of X also happens to be an element of Y. Obviously, every set is a subset of itself. Lemma 3. A set X is a subset of every set if and only if X =. Proof. The empty set is a subset of any other set Y since every element of the empty set is trivially an element of Y, as there are no elements in the empty set. Conversely, if X is a subset of every set, than in particular X must be a subset of, which can only happen when X itself is the empty set (if X has at least one element x, then since x is not an element of the empty set, X cannot be a subset of ). So, this something which we created out of nothing is actually part of everything, and it is the only thing that is such. Exercise 4. Prove that if X X and Y Y then X Y X Y. Exercise 5. Prove that if X Y and Y Z, then X Z. Exercise 6. Prove that if X Y and Y X, then X = Y. Exercise 7. For a set X, the power set of X, written as P(X), is defined as the set of all subsets of X. (a) Does there exist a set X for which P(X) =? (b) If X has n number of elements, then how many elements does P(X) have? 4. Relations A relation R from a set X to a set Y, is a subset R X Y. In particular, if R =, then R is called the empty relation. We sometimes depict a relation from X to Y as an arrow of the form X Y, and then we write or R : X Y X R Y to indicate that R is a relation from X to Y. Note that the same set R could be a relation between possibly different pairs of sets (X, Y ) and (X, Y ). For instance, let R X Y and suppose X X, Y Y. Then also R X Y. In order to avoid this (since we want each relation R to remember who is X and who is Y ), we refine the definition of a relation as follows: Definition 4. A relation is a pair (R, (X, Y )), where X and Y are sets, and R is a subset R X Y. We often use the same letter R to denote the set R in the pair (R, (X, Y )), and the whole relation (R, (X, Y )) itself. The set X is said to be the domain of the relation R and the set Y is said to be the codomain of R. Given two relations X R S Y Z their composite S R is a relation S R : X Z defined by S R = {(x, z) y Y [(x, y) R (y, z) S]}

8 SEMESTER I: ELEMENTS where y Y [(x, y) R (y, z) S] means: there exists an element y in Y such that (x, y) R and (y, z) S. For a relation R, we will often use the notation xry to mean (x, y) R. Thus, xs Rz if and only if xsyrz for some y Y. For a relation R : X Y by R we denote a relation R : Y X defined by We call R the opposite relation of R. yr x xry. Remark 1. Above, the symbol stands for if and only if. Also, we will write for implies. Exercise 8. For this question we assume that we have already defined what real numbers are (we will define them later in the notes). Let R be a relation from the set R of real numbers to itself, i.e. R : R R, defined by xry x 2 + y 2 = 1. And let S be a relation S : R R defined by Then, describe the composite S R. xsy x 2 + y 2 = 2. Exercise 9. Let R be a relation from a set X to itself, i.e. we have R : X X. Show that the following conditions are equivalent to each other: (a) R R ; (b) R R; (c) R = R ; (d) For all x, y X we have xry yrx. A relation R : X X satisfying these conditions is called a symmetric relation. 5. Functions Definition 5. A function f from a set X to a set Y, written as f : X Y, is a relation f : X Y having the following property: for every element x X there exists exactly one element y Y such that (x, y) f. When f is a function f : X Y, for each x X we use the special notation f(x) for the unique element f(x) = y Y such that (x, y) f. We will use the following obvious lemma very often, when we want to establish equality of two functions: Lemma 4. Let f, g : X Y be functions from a set X to a set Y. Then f = g if and only if for each x X we have f(x) = g(x). Thus, specifying / constructing / defining a function f : X Y amounts to saying what is f(x) for each x X. Given two functions X f g Y Z we can compose them as relations to get a function X g f Y

9 5. FUNCTIONS 9 Lemma 5. Let f and g be as above. Then for each x X, the composite g f satisfies (g f)(x) = g(f(x)). Proof. By the definition of composition of relations, we have This is the same as to have g f = {(x, z) y Y [(x, y) f (y, z) g]}. g f = {(x, z) y Y [y = f(x) z = g(y)]}. So, to show (g f)(x) = g(f(x)) is the same as to show that there exists y Y such that y = f(x) and g(f(x)) = g(y). But such y clearly exists take y = f(x). Lemma 6. Composition of functions is associative; that is, for any three functions W f X g h Y Z we have h (g f) = (h g) f. Proof. By Lemma 5 we have: for each w W, (h (g f))(w) = h((g f)(x)) = h(g(f(x))). Similarly, for each w W we have ((h g) f)(w) = h(g(f(x))). So (h (g f))(w) = (h g) f)(w) for all w W and hence we can apply Lemma 4 to conclude h (g f) = (h g) f. For a set X by 1 X we denote the function 1 X : X X defined by 1 X (x) = x. Lemma 7. For any function f : X Y we have Exercise 10. Prove Lemma 7. 1 Y f = f = f 1 X. Exercise 11. Show that for any four functions V e W f X g Y the following composites are all equal to each other: h (g (f e)) ((h g) f) e h ((g f) e) (h g) (f e) (h (g f)) e Exercise 12. State and prove associativity of composition of relations. Exercise 13. (a) Prove Lemma 7. (b) State and prove the generalization of Lemma 7, which replaces the function f : X Y with a relation R : X Y. h Z

10 SEMESTER I: ELEMENTS 6. Injective and surjective functions Definition 6. A function f : X Y is said to be injective if for any x 1, x 2 X we have f(x 1 ) = f(x 2 ) x 1 = x 2. We say that f is surjective if for any y Y there exists x X such that f(x) = y. Finally, f is said to be bijective if it is both injective and surjective. Put simply, f is injective if it carries different elements to different elements; f is surjective if f fills up its codomain; and f is bijective if it defines a one-to-one correspondence between its domain and codomain. Lemma 8. A function f : X Y is injective if and only if f f = 1 X. Proof. By the definition of composition of relations, the relation f f consists of all those pairs (x, z) X X such that there exists y with f(x) = y and y = f(z). In other words, it consists of all those pairs (x, z) X X such that f(x) = f(z). So f f 1 X if and only if f is injective. Now, we always have 1 X f f. Hence, f f = 1 X if and only if f is injective. Lemma 9. A function f : X Y is surjective if and only if f f = 1 Y. Lemma 10. A function f : X Y is bijective if and only if its inverse relation f : Y X is itself a function. These three lemmas give an important characterization of bijective functions as those functions f : X Y for which there exists a function g : Y X such that f g = 1 Y and g f = 1 X. Then g is necessarily unique and it coincides with f (see Exercise 15). We call g the inverse of f. Definition 7. The image of a function f : X Y, is defined as the following set: Im(f) = {y x X f(x) = y}. Theorem 1. Any function f : X Y can be presented as a composite f = g h of a surjective function h followed by an injective function g. Proof. Define a function g : Im(f) Y by g(y) = y, and define h : X Im(f) by h(x) = f(x). Then it is trivial to check that f = g h. Exercise 14. Prove Lemmas 9 and 10. Exercise 15. Conclude from Lemmas 8, 9 and 10 that a function f : X Y is bijective if and only if there exists a function g : Y X such that f g = 1 Y and g f = 1 X. Show that when such g exists, it is necessarily unique (i.e. there is only one g satisfying the above two equalities). Exercise 16. (a) Consider a function π 1 : X Y X defined by π 1 ((x, y)) = x. When is π 1 surjective? (b) Consider a function : X X X defined by (x) = (x, x). When is injective? What is the image of?

11 7. A PARADOX IN NAIVE SET THEORY 11 Exercise 17. Complete the proof of Theorem 1. Exercise 18. (a) Prove that a composite f g is injective/surjective/bijective as soon as both f and g are injective/surjective/bijective. (b) Prove that if f g is injective then necessarily g is injective. (c) Prove that if f g is surjective then necessarily f is surjective. 7. A paradox in naive set theory Naive set theory is when one thinks of sets simply as collections of objects, and then one wants to regard sets as being themselves objects. It turns out that this leads to a paradox, for we can then say: consider the set V of all those sets Y such that Y / Y (i.e. Y is not an element of itself); if we believe that all sets are objects, then since an arbitrary collection of objects is a set, nothing is strange about the construction of V. Now, we argue: if V V then, by the definition of V, we should have V / V ; however, if V / V, then, again by the definition of V, we should have V V. So we get V V V / V which is clearly a contradiction. Thus, we should not assume that every set is itself an object. If we think of object as those things which are real, then we can say: not every set is real! In particular, the set of all real sets might not be a real set. Thus, we can only form sets of real things, and only some of these sets will turn out to be real themselves. Then the above argument merely proves that the set V of all real sets Y such that Y / Y, cannot be real (for if it is, we get a paradox). In fact, according to our intuition about sets, all real sets Y should satisfy Y / Y, so that V becomes the set of all real sets. So, indeed, the set of all real sets is not a real set. We can model the relationship between sets and real sets inside the world of real sets as follows: let X be a real set, and call its elements very real sets. Then P(X) is the set of all sets of very real sets. Elements of P(X) are real sets. The question is: can it happen that every real set is a very real set? If this can happen, then we will get that P(X) is a subset of X. In other words, we will have an injective function f : P(X) X defined by f(u) = u. But, as we will prove below (see Theorem 2), there cannot exist an injective function (of any kind) from P(X) to X. The proof of this result is in fact an immediate adaption of the argument used to establish the above paradox. All the time in the previous sections we have only been dealing with real sets. But now we should be careful: what if at some point we formed a set which was not real? To make sure this never happened we have to set out rules for constructing real sets. Here are some of these rules, which are sufficient to deduce that all sets we have been constructing in previous sections were indeed real: (a) The empty set is real. (b) If X is a real set, then any set of its elements is a real set. (c) If x and y are real sets, then the set {x, y} is also real. (d) If X and Y are real sets, then the set of all pairs (x, y) where x X and y Y, is a real set. (e) If X is a real set, then the set of all subsets of X is a real set.

12 SEMESTER I: ELEMENTS But how do we know that these rules are really correct and do not give rise to any paradoxes? We could try to invent new type of mathematics in which we would be able to prove that the above rules do not give rise to any paradoxes, but then, how would we know that this new mathematics will not have any paradoxes? Well, we could repeat the procedure and develop yet another mathematics for showing that there are no paradoxes in the previous mathematics, but then we face the same question again for the newer mathematics. Since this process continues infinitely, and since we will never be able to do infinite number of things in finite time, we will never know whether the above rules yield any paradoxes or not, unless of course we accidently discover them. Henceforth, as in previous sections, when we say set we will mean real set. Theorem 2. Let X be any set. There does not exist an injective function f : P(X) X, from the set P(X) of all subsets of X to X. Proof. Let f be a function f : P(X) X. We will find u, v such that f(u) = f(v) and u v, hence proving that f is not injective. Take v = {y X u P(X) ([f(u) = y] [y / u])}. (Here u P(X) means for any u P(X) such that....) Now, take y = f(v). Then y / v, since if y v, then we would get a contradiction with how v is defined: if y v, then, because f(v) = y, we should have y / v (set u = v above). But if y / v, then there should exist u P(X) different from v for which f(u) = f(v) = y, since otherwise, if v is the only element in P(X) with the property f(v) = y, then by definition of v we would have y v, which we already showed that cannot be true. This completes the proof.

13 2009 Semester II: Numbers The aim of this chapter is to construct natural numbers within set theory. We start by investigating a special class of relations, called equivalence relations, which will serve as a toolkit for constructing numbers. But even before this we have to say 8. Few words about mathematical logic Mathematical reasoning has very precise and strict rules, which ensures that the validity of it does not depend on the opinion of a particular person. This makes mathematics the most precise science. To study these rules is the purpose of mathematical logic. In particular, mathematical logic proposes a fixed meaning to the connectives like and, or, if... then..., etc., and then it becomes more convenient to introduce symbolic representations for them. In this section we introduce these symbols and explain their precise meaning. Throughout the section capital letters A, B,... will be used to represent arbitrary sentences, and small letters x, y,... will be used to represent sets. We will write A[x 1,..., x n ] if A is a sentence involving distinct symbols x 1,..., x n which represent arbitrary sets. Then A(y 1,..., y n ) is a new sentence obtained from A by replacing each occurrence of the symbol x i in A with the symbol y i. In this process, the symbols y 1,..., y n need not be distinct. Here is an example: if then A[x 1, x 2 ] expresses x 1 is a subset of x 2 A(y, y) expresses y is a subset of y. Clearly, the sentence A[x 1, x 2 ] is true for some pairs of sets x 1, x 2 and false for some other pairs of sets (for instance, it is false if x 2 = and x 1 = { }). While A(y, y) is always true (because any set is a subset of itself) Conjunction. Let A and B be sentences. Then A B is the following sentence obtained from A and B: A B = A and B. Mathematical logic gives us the following rule: the sentence A B is true if and only if both A and B are true. This rule can be summarized by the following table, which is similar to the multiplication table for natural numbers (think of t being 1 and f being 0): t f t t f f f f 13

14 SEMESTER II: NUMBERS Above t and f represent true and false, respectively. The logical operation is called conjunction. Note: a more accepted term for logical operation is logical connective Disjunction. Let A and B be sentences. Then A B is the following sentence obtained from A and B: A B = A or B. Mathematical logic gives us the following rule: the sentence A B is true if and only if at least one of A, B is true (or both are true). This rule can be summarized by the following table, which is similar to the addition table for natural numbers if we think of t being 1 and f being 0: t f t t t f t f The logical operation is called disjunction Negation. Let A be a sentence. Then A is the following sentence obtained from A: A = not A. The sentence A is true if and only if A is false. This rule can be exhibited by the following table: The operation is called negation. A A t f f t Exercise 19. Show that A B is true if and only if (A B) is true. Is there a similar relationship between A B and (A B)? 8.4. Implication. Let A and B be sentences. Then A B is the following sentence obtained from A and B: A B = A implies B. The sentence A B is true if and only if ( A) B is true. exhibited by the following table: This rule can be t f t t f f t f The logical operation is called implication. In other words, A B has the same meaning as if A then B. Notice that when A is false, this sentence is (by the above rule) always true (it does not depend on B). But when A is true, A B is true if and only if B is true.

15 8. FEW WORDS ABOUT MATHEMATICAL LOGIC Universal quantifier. Let A[x] be a sentence about a set x. Then y A(y) is the following sentence obtained from A: y A(y) = for all y, A(y). The sentence y A(y) is true if and only if A(y) is true for any set y. The quantifier is called the universal quantifier. Exercise 20. Give several examples of sentences A[x] such that y A(y) is true and also give examples when y A(y) is false Existential quantifier. Let A[x] be a sentence about a set x. Then y A(y) is the following sentence obtained from A: y A(y) = there exists y such that A(y). The sentence y A(y) is true if and only if A(y) is true for at least one set y. The quantifier is called the existential quantifier. There are the following relationships between existential and universal quantifiers: y A(y) is true if and only if ( y (A(y))) is true; y A(y) is true if and only if ( y (A(y))) is true Equivalent sentences. Two sentences A and B are said to be equivalent when A is true if and only if B is true. In this case we write A B. Notice that we can regard A B as a sentence, and then becomes a logical operation. It is easy to see that A B is true if and only if (A B) (B A) is true. Above we have given rules which say whether a sentence A made up of sentences B 1,..., B n using the quantifiers, and connectives,,, is true or false, depending on the truth and falsity of B 1,..., B n. In other words, if we know exactly which of the B 1,..., B n is true and which is false, we will know whether A is true or false, and it will not depend on opinion, since our rules are very strict and they cover all possible cases (of truth and falsity of B 1,..., B n ). However, we might not always know if, say, B 1 is true or not. For example, if B is a sentence about a set x saying x =, then we do not know whether B is true or false, since the truth or falsity of B depends on the choice of x: if x is the empty set, then B is true, and if x is nonempty, then B is false. In general we will always have this problem, when B is a sentence about certain sets x 1,..., x n. Note however, that as soon as we take either the sentence C 1 = x (x = ) or the sentence C 2 = x (x = ) this problem disappears: C 1 and C 2 no longer talk about a particular set x. Clearly, C 1 is false (because it is not true that any set is empty) and C 2 is true (since there does indeed exist a set which is empty the empty set itself). To distinguish this situation with the situation we had above for B, we introduce the following concept: Definition 8. A variable x is said to be free in a sentence A if x does not appear in the subscript of either the universal or the existential quantifier. Our first remark now is that when we introduced notation A[x 1,..., x n ] for the case when A is a sentence about distinct sets x 1,..., x n, under the word about we meant that x 1,..., x n are all free variables in A. Indeed, consider the example of C 1. The variable x is not free in C 1. Then, if we still write C 1 as C 1 [x], then we

16 SEMESTER II: NUMBERS are allowed to form a sentence y C 1 (y) which clearly would not make sense since it would be expressing there exists y such that for all y, y =. We make further adjustment in terminology: from now on we will use the term formula instead of a sentence, and we will call sentences only those formulas which do not have free variables. The justification for using the term formula in the case when there are (possibly) free variables is as simple as this: when we have a free variable x, we can assign to it different values, just like in the usual formulas that we know from school. Now it seems that when B is a sentence (i.e. a formula with no free variables) there should no longer be a problem of not being able to determine whether B is true or false. But it turns out that there is still a problem, since the set theory we work with (and hence all of mathematics) is an abstract theory and it might have many difference concrete instances: in certain instances a given sentence might be true, and in certain instances it might be false! It is a difficult task, and too complex at this stage, to give examples of different instances of set theory (and in some sense it is not even possible). So, to be able to understand the above phenomenon let us replace set theory with a simpler theory, say, theory of numbers. Here we have two instances in mind one is the system of real numbers, and another is the system of complex numbers. So now for us, the variables in sentences represent numbers, rather than sets. In the abstract theory of numbers (suitably introduced) we can formulate a sentence B = x y (y 2 = x). Clearly there are no free variables in B, so B is indeed a sentence in the above sense. However, in general we do not know whether B is true or false. Indeed, in the case of the system of real numbers, B is false since for x = 1 there does not exist y such that y 2 = 1. On the other hand, in the system of complex numbers B is true. So we should live with the fact that there might be some sentences which are neither true nor false! Let us give an example of such a sentence for set theory: let B be the sentence stating that there exists a set X such that there are injections N X R (where N is the set of natural numbers and R is the set of real numbers) but X is not bijective either to N or to R. In other words, the sentence says that there is a set which is bigger than the set of natural numbers, but smaller than the set of real numbers (the set of rational numbers turns out to be as big as the set of natural numbers, as we will see in this chapter). This sentence is known in mathematics as continuum hypothesis. 9. Equivalence relations In the definition below we work with any given set X, and variables x, y, z in quantifiers represent elements of X. We could also avoid such remark by writing, say, x X instead of x. We will do so in the future. Definition 9. A relation R : X X is said to be reflexive, if x xrx; symmetric, if x y (xry yrx); transitive, if x y z ([xry yrz] xrz);

17 9. EQUIVALENCE RELATIONS 17 an equivalence relation, if it is reflexive, symmetric and transitive at the same time. Exercise 21. (a) Show that the composite of two reflexive/symmetric relations is reflexive/symmetric. (b) Given an example of two transitive relations R and S such that R S is not transitive. (c) Show that if a relation R is transitive, then also R R is transitive. In the future we will abbreviate, say, x X y X as x,y X. Definition 10. A partition P of a set X is a set of subsets of X (i.e. P P(X)) having the following properties: (a) Every element of P is nonempty: U P (U ); (b) Every two different elements of P are disjoint: U,V P [(U V ) (U V = )]; (c) Every element of X is an element of an element of P : x X U P (x U). Lemma 11. For any set P the following conditions are equivalent: (a) P satisfies 10(a) and 10(b); (b) U,V P [(U V ) (U V = )]. Proof. First we prove (a) (b): Suppose P satisfies 10(a) and 10(b). Let U, V P. We must show (U V ) (U V = ), which amounts to showing (U V ) (U V = ) and (U V = ) (U V ). Let us first show (U V ) (U V = ). But this follows trivially from 10(b). Let us now show (U V = ) (U V ). Suppose U V =. If U = V, then U = U V = which contradicts with 10(a). So U V. This completes the proof of (a) (b). The proof of (b) (a) is left as an exercise. Exercise 22. Complete the proof of Lemma 11. Exercise 23. Show that a partition of a set X is empty if and only if X itself is empty. Exercise 24. List all possible partitions of a set X = {x, y, z} where x, y, z are three distinct elements. Let E be an equivalence relation on a set X. For an element x of X we write [x] E to denote the following set: [x] E = {y xey}. This set is called the equivalence class of x (under the equivalence relation E). Exercise 25. Prove the following: x X y X (xey [x] E = [y] E ). Theorem 3. There is a bijection F : Eq(X) Part(X) from the set Eq(X) of all equivalence relations on X, to the set Part(X) of all partitions of X, defined as follows: F (E) = {[x] E x X}.

18 SEMESTER II: NUMBERS The inverse of F is given by F 1 (P ) = {(x, y) U P (x U y U)}. Thus, every equivalence relation produces a partition, and each partition arises in this way from exactly one equivalence relation. This correspondence will provide a very important tool for constructing numbers: we will defined numbers as elements of the partition corresponding to a suitable equivalence relation on a suitable set. Proof of Theorem 3. First of all, let us check that for each equivalence relation E on X, the set F (E) is indeed a partition of X. Note that since [x] E P(X) for each x X, we have F (E) P(X) (recall that P(X) denotes the set of all subsets of X). So to see that F (E) is a partition of X we must check that P = F (E) satisfies conditions (a), (b) and (c) in Definition 10. (a)&(c): Let U F (E). Then U = [x] E for some equivalence relation E on X. By reflexivity of E, we have x [x] E. So U in nonempty. This proves (a). Since x [x] E F (E) for any element x of X, we also get (c). (b): Let U, V F (E). Suppose U = [x] E and V = [y] E have a common element z. Then xez and yez. By symmetry of E, zey, and by transitivity xey. This implies that [x] E = [y] E (see Exercise 25). So if U V then U and V cannot have a common element and hence are disjoint (i.e. U V = ). This proves (b). Thus, the assignment E F (X) indeed defines a function F : Eq(X) Part(X). The rest of the proof is left as an exercise (see Exercise 26). Exercise 26. Complete the proof of Theorem Small sets and cardinal numbers From this point on we assume that in our universe of sets we have a set S having the following properties: S contains the empty set and is closed under formation of products and power sets (i.e. if X, Y S then X Y S and P(X) S). S is closed under all intersections, i.e. for any function f : X S, where X is any set, the set {y x X y f(x)} is an element of S. S is closed under (binary) unions, i.e. if X, Y S, then the set X Y = {x (x X) (x Y )} is also in S. Many things we will do from this point on depends on the choice of such set S, so henceforth we assume to have one such S fixed. We call the elements of this fixed S the small sets. Definition 11. Consider the following relation E on the set of small sets: XEY when there exists a bijection f : X Y ; in this case we say X is bijective to Y. This relation is an equivalence relation, and the elements of the partition corresponding to E (i.e. the equivalence classes for the equivalence relation E) will be called cardinal numbers.

19 11. PROJECTS 19 Exercise 27. Verify that E is an equivalence relation. 11. Projects Project 1. Introduce cardinal arithmetic, i.e. define addition and multiplication of cardinal numbers and establish their basic properties. Project 2. Use the principle of mathematical induction as a definition of natural numbers and establish basic properties of natural numbers. Project 3. Define rational numbers as equivalence class of pairs of natural numbers, for a suitably chosen equivalence relation. Project 4. Define real numbers.

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21 2010 Semester I: Structures In the previous chapter we defined natural numbers and saw that the arithmetic of natural numbers obeys certain general principles. For example, addition of natural numbers is commutative, i.e. we have n + m = m + n for any two natural numbers n and m. Such general principle is called an axiom on the operation +. The pair (N, +), where N is the set of natural numbers, and + is the operation of addition of natural numbers, is a mathematical structure. In general, a mathematical structure is a system consisting of sets, operations on these sets, relations, etc. Modern mathematics is full of various different kinds of mathematical structures. In this chapter we will look at some of the most important structures. Before we can start learning about mathematical structures, we have to say more precisely what is an operation. 12. Operations Let X and Y be sets. By X Y we denote the set of all functions from X to Y. For a natural number n we write X n to denote X Y where Y = {1,..., n}. In particular, if n = 0 then Y = and hence X 0 has only one element. When n 1, a function f : {1,..., n} X can be represented by an ordered list (in other words, a sequence) f 1,..., f n of elements of X. In particular, f i in this list is simply f(i). Thus, elements of X n are lists of n elements of X (with possibility of repetitions in a list). Definition 12. An n-ary operation on a set X is a function p : X n X. When n = 0, the set X 0 has only one element, so to specify a 0-ary operation amounts to specify an element in X. For example, the natural number 0 can be regarded as a 0-ary operation on the set N of all natural numbers. Similarly, the natural number 1 can be regarded as a 0-ary operation on N. When n = 2, we often write x 1 px 2 instead of p(x 1, x 2 ). For instance, when X = N and p is the addition + of natural numbers, we always write x 1 + x 2 rather than +(x 1, x 2 ). An example of a 1-ary operation is the operation on N defined by the assignment x x + 1. If we denote this operation by s (for successor ), then we can write s(x) = x + 1. Exercise 28. Give examples of operations of various arities on the set of natural numbers. 21

22 SEMESTER I: STRUCTURES We say nullary for 0-ary, unary for 1-ary, binary for 2-ary, ternary for 3-ary. 13. Monoids Definition 13. A monoid is a mathematical structure (X, m, e), where X is a set, m is a binary operation on X and e is an element of X (regarded as a nullary operation on X), such that m is associative, i.e. for all x, y, z X we have m(m(x, y), z) = m(x, m(y, z)), e is a unit for m, i.e. for all x X we have m(x, e) = x = m(e, x). The system (N, +, 0), where N is the set of natural numbers, + is addition of natural numbers (and 0 is usual natural number zero), is of course an example of a monoid. Exercise 29. Give other examples of monoids. Exercise 30. Prove that in any monoid, the following holds: m(m(m(x, y), z), t) = m(x, m(y, m(z, t))). Definition 14. A monoid (X, m, e) is said to be commutative if the operation m is commutative, i.e. m satisfies for all x, y X. m(x, y) = m(y, x) Of course, (N, +, 0) is a commutative monoid. Exercise 31. Let X be any set. Show that the system (X X,, 1 X ), where X is any set (and X X is the set of all functions from X to itself), is the composition of functions and 1 X is the identity function, is a monoid. Show that this monoid is commutative if and only if X is either the empty set or has exactly one element. Exercise 32. Show that all monoids (X, m, e) where X has at most two elements are commutative. Describe all three-element monoids (i.e. all monoids (X, m, e) where X has exactly three elements).