# Introducing Functions

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1 Functions 1

2 Introducing Functions A function f from a set A to a set B, written f : A B, is a relation f A B such that every element of A is related to one element of B; in logical notation 1. (a, b 1 ) f (a, b 2 ) f b 1 = b 2 ; 2. a A. b B.(a, b) f. The set A is called the domain and B the co-domain of f. If a A, then f(a) denotes the unique b B st. (a, b) f. 2

3 Comments If the domain A is the n-ary product A 1... A n, then we often write f(a 1,..., a n ) instead of f((a 1,..., a n )). The intended meaning should be clear from the context. Recall the difference between the following two Haskell functions: f :: A -> B -> C -> D f :: (A,B,C) -> D Our definition of function is not curried. 3

4 Image Set Let f : A B. For any X A, define the image of X under f to be f[x] def = {f(a) : a X} The set f[a] is called the image set of f. 4

5 Example Let A = {1,2,3} and B = {a, b, c}. Let f A B be defined by f = {(1, a),(2, b),(3, a)}. A B a b c The image set of f is {a, b}. The image of {1,3} under f is {a}. 5

6 Example Let A = {1,2,3} and B = {a, b}. Let f A B be defined by f = {(1, a),(1, b),(2, b),(3, a)}. This f is not a well-defined function. A a b B 6

7 Examples The following are examples of functions with infinite domains and co-domains: 1. the function f : N N N defined by f(x, y) = x + y; 2. the function f : N N defined by f(x) = x 2 ; 3. the function f : R R defined by f(x) = x + 3. The binary relation R on the reals defined by x R y if and only if x = y 2 is not a function 7

8 Cardinality Let A B denote the set of all functions from A to B, where A and B are finite sets. If A = m and B = n, then A B = n m. Sketch proof For each element of A, there are n independent ways of mapping it to B. You do not need to remember this proof. 8

9 Partial Functions A partial function f from a set A to a set B, written f : A B, is a relation f A B such that just some elements of A are related to unique elements of B: (a, b 1 ) f (a, b 2 ) f b 1 = b 2. The partial function f is regarded as undefined on those elements which do not have an image under f. We call this undefined value (pronounced bottom). A partial function from A to B is a function from A to (B + { }). 9

10 Examples of Partial Functions 1. Haskell functions which return run-time errors on some (or all) arguments. 2. The relation R = {(1, a),(3, a)} {1,2,3} {a, b}: A a b B Not every element in A maps to an element in B. 3. The binary relation R on R defined by x R y iff x = y. It is not defined when x is negative. 10

11 Properties of Functions Let f : A B be a function. 1. f is onto if and only if every element of B is in the image of f: b B. a A. f(a) = b. 2. f is one-to-one if and only if for each b B there is at most one a A with f(a) = b: a, a A. f(a) = f(a ) implies a = a. 3. f is a bijection iff f is both one-to-one and onto. 11

12 Example Let A = {1,2,3} and B = {a, b}. The function f = {(1, a),(2, b),(3, a)} is onto, but not one-to-one: A a b B We cannot define a one-to-one function from A to B. There are too many elements in A for them to map uniquely to B. 12

13 Example Let A = {a, b} and B = {1,2,3}. The function f = {(a,3),(b, 1)} is one-to-one, but not onto: a b It is not possible to define an onto function from A to B. There are not enough elements in A to map to all the elements of B. 13

14 Example Let A = {a, b, c} and B = {1,2,3}. The function f = {(a,1),(b, 3), c,2)} is bijective: A a b c B 14

15 Example The function f on natural numbers defined by f(x, y) = x + y is onto but not one-to-one. To prove that f is onto, take an arbitrary n N. Then f(n, 0) = n + 0 = n. To show that f is not one-to-one, we need to produce a counter-example: that is, find (m 1, m 2 ),(n 1, n 2 ) such that (m 1, m 2 ) (n 1, n 2 ), but f(m 1, m 2 ) = f(n 1, n 2 ). For example, (1,0) and (0,1). 15

16 Examples 1. The function f on natural numbers defined by f(x) = x 2 is one-to-one. The similar function f on integers is not. 2. The function f on integers defined by f(x) = x + 1 is onto. The similar function on natural numbers is not. 3. The function f on the real numbers given by f(x) = 4x + 3 is a bijective function. The proof is given in the notes. 16

17 The Pigeonhole Principle Pigeonhole Principle Let f : A B be a function, where A and B are finite. If A > B, then f cannot be a one-to-one function. Example Let A = {1,2,3} and B = {a, b}. A function f : A B cannot be one-to-one, since A is too big. It is not possible to prove this property directly. The pigeonhole principle states that we assume that the property is true. 17

18 Proposition Let A and B be finite sets, let f : A B and let X A. Then f[x] X. Proof Suppose for contradiction that f[x] > X. Define a function p : f[x] X by p(b) = some a X such that f(a) = b. There is such an a by definition of f[x]. We are placing the members of f[x] in the pigeonholes X. By the pigeonhole principle, there is some a X and b, b f[x] with p(b) = p(b ) = a. But then f(a) = b and f(a) = b. Contradiction. 18

19 Proposition Let A and B be finite sets, and let f : A B. 1. If f is one-to-one, then A B. 2. If f is onto, then A B If f is a bijection, then A = B. Proof Part (a) is the contrapositive of the pigeonhole principle. For (b), notice that if f is onto then f[a] = B. Hence, f[a] = B. Also A f[a] by previous proposition. Therefore A B as required. Part (c) follows from parts (a) and (b).

20 Composition Let A, B and C be arbitrary sets, and let f : A B and g : B C be functions. The composition of f with g, written g f : A C, is a function defined by g f(a) def = g(f(a)) for every element a A. In Haskell notation, we would write (g.f) a = g (f a) It is easy to check that g f is indeed a function. The co-domain of f must be the same as the domain of g. 20

21 Example Let A = {1,2,3}, B = {a, b, c}, f = {(1, a),(2, b),(3, a)} and g = {(a,3),(b, 1)}. Then g f = {(1,3),(2,1),(3,3)}. A B A 1 a 2 b

22 Associativity Let f : A B, g : B C and h : C D be arbitrary functions. Then h (g f) = (h g) f. Proof Let a A be arbitrary. Then (h (g f))(a) = h((g f)(a)) = h(g(f(a))) = (h g)(f(a)) = ((h g) f)(a) 22

23 Proposition Let f : A B and g : B C be arbitrary bijections. Then g f is a bijection. Proof It is enough to show that 1. if f, g are onto then so is g f; 2. if f, g are one-to-one then so is g f. 23 Assume f and g are onto. Let c C. Since g is onto, we can find b B such that g(b) = c. Since f is onto, we can find a A such that f(a) = b. Hence g f(a) = g(f(a)) = g(b) = c. Assume f and g are one-to-one. Let a 1, a 2 A such that g f(a 1 ) = g f(a 2 ). Then g(f(a 1 )) = g(f(a 2 )). Since g is one-to-one, f(a 1 ) = f(a 2 ). Since f is also one-to-one, a 1 = a 2.

24 Identity Let A be a set. Define the identity function on A, denoted id A : A A, by id A (a) = a for all a A. In Haskell, we would declare the function id :: A -> A id x = x 24

25 Inverse Let f : A B be an arbitrary function. The function g : B A is an inverse of f if and only if for all a A, for all b B, g(f(a)) = a f(g(b)) = b Another way of stating the same property is that g f = id A and f g = id B. 25

26 Examples 1. The inverse relation of the function f : {1, 2} {1, 2} defined by f(1) = f(2) = 1 is not a function. When an inverse function exists, it corresponds to the inverse relation. 2. Let A = {a, b, c}, B = {1,2,3}, f = {(a,1),(b, 3),(c,2)} and g = {(1, a),(2, c),(3, b)}. Then g is an inverse of f. 26

27 Proposition Let f : A B be a bijection, and define f 1 : B A by f 1 (b) = a whenever f(a) = b The relation f 1 is a well-defined function. It is the inverse of f (as shown in next proposition). Proof Let b B. Since f is onto, there is an a such that f(a) = b. Since f is one-to-one, this a is unique. Thus f 1 is a function. It satisfies the conditions for being an inverse of f. 27

28 Proposition Let f : A B. If f has an inverse g, then f must be a bijection and the inverse is unique. Proof To show that f is onto, let b B. Since f(g(b)) = b, it follows that b must be in the image of f. To show that f is one-to-one, let a 1, a 2 A. Suppose f(a 1 ) = f(a 2 ). Then g(f(a 1 )) = g(f(a 2 )). Since g f = id A, it follows that a 1 = a 2. To show that the inverse is unique, suppose that g, g are both inverses of f. Let b B. Then f(g(b)) = f(g (b)) since g, g are inverses. Hence g(b) = g (b) since f is one-to-one. 28

29 Example Consider the function f : N N defined by f(x) = x + 1, x odd = x 1, x even It is easy to check that (f f)(x) = x, considering the cases when x is odd and even separately. Therefore f is its own inverse, and we can deduce that it is a bijection. 29

30 Definition Cardinality of Sets For any sets A, B, define A B if and only if there is bijection from A to B. Proposition Relation is reflexive, symmetric and transitive. Proof Relation is reflexive, as id A : A A is a bijection. To show that it is symmetric, A B implies that there is a bijection f : A B. By previous proposition, it follows that f has an inverse f 1 which is also a bijection. Hence B A. The fact that the relation is transitive follows from previous proposition. 30

31 Example Let A, B, C be arbitrary sets, and consider the products (A B) C and A (B C). There is a natural bijection f : (A B) C A (B C): f : ((a, b), c) (a,(b, c)) Also define the function g : A (B C) (A B) C: Function g is the inverse of f. g : (a,(b, c)) ((a, b), c) 31

32 Example Consider the set Even of even natural numbers. There is a bijection between Even and N given by f(n) = 2n. Not all functions from Even to N are bijections. The function g : Even N given by g(n) = n is one-to-one but not onto. To show that Even N, it is enough to show the existence of such a bijection. 32

33 Example Recall that the cardinality of a finite set is the number of elements in that set. Let A = n. There is a bijection c A : {1,2,..., n} A. Let A and B be two finite sets. If A and B have the same number of elements, we can define a bijection f : A B by f(a) = (c B c 1 A )(a). Two finite sets have the same number of elements if and only if there is a bijection between then. 33

34 Cardinality Given two arbitrary sets A and B, then A has the same cardinality as B, written A = B, if and only if A B. Notice that this definition is for all sets. 34

35 Exploring Infinite Sets The set of natural numbers is one of the simplest infinite sets. We can build it up by stages: 0 0,1 0,1,2... Infinite sets which can be built up in finite portions by stages are particularly nice for computing. 35

36 Countable A set A is countable if and only if A is finite or A N. The elements of a countable set A can be listed as a finite or infinite sequence of distinct terms: A = {a 1, a 2, a 3,...}. 36

37 Example The integers Z are countable, since they can be listed as: 0, 1,1, 2,2, 3,3,... This counting bijection g : Z N is defined formally by g(x) = 2x, x 0 = 1 2x, x < 0 The set of integers Z is like two copies of the natural numbers. 37

38 Example The set N 2 is countable: Comment The rational numbers are also countable

39 Uncountable Sets Cantor showed that there are uncountable sets. An important example is the set of reals R. Another example is the power set P(N). We cannot manipulate reals in the way we can natural numbers. Instead, we use approximations: for example, the floating point decimals of typefloat in Haskell. For more information, see Truss, section

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