Violencias conectadas

1. Which of the following relations are equivalence relations on the given set S?

(i) S = R, and a ∼ b ⇔ a = b or −b.

(ii) S = Z, and a ∼ b ⇔ ab = 0.

(iii) S = R, and a ∼ b ⇔ a²+a = b²+b.

(iv) S is the set of all people in the world, and a ∼ b means a lives within 100 miles of b.

EQUIVALENCE RELATIONS 153 (v) S is the set of all points in the plane, and a ∼ b means a and b are the same distance from the origin.

(vi) S = N, and a ∼ b ⇔ ab is a square.

(vii) S = {1,2,3}, and a ∼ b ⇔ a = 1 or b = 1.

(viii) S = R×R, and (x,y) ∼ (a,b) ⇔ x²+y²=a²+b².

2. For those relations in Exercise 1 that are equivalence relations, describe the equivalence classes.

3. By producing suitable examples of relations, show that it is not possible to deduce any one of the properties of being reflexive, symmetric or transitive from the other two.

4. Prove that if S is a set and S1, . . . ,Sk is a partition of S, then there is a unique equivalence relation ∼ on S that has the Sias its equivalence classes.

5. (a) How many relations are there on the set {1,2}?

(b) How many relations are there on the set {1,2,3} that are both reflex-ive and symmetric?

(c) How many relations are there on the set {1,2,...,n}?

6. Let S = {1,2,3,4}, and suppose that ∼ is an equivalence relation on S.

You are given the information that 1 ∼ 2 and 2 ∼ 3.

Show that there are exactly two possibilities for the relation ∼, and de-scribe both (i.e., for all a,b ∈ S, say whether or not a ∼ b).

7. Let ∼ be an equivalence relation on Z with the property that for all m ∈ Z we have m ∼ m+5 and also m ∼ m+8. Prove that m ∼ n for all m,n ∈ Z.

8. Critic Ivor Smallbrain has made his peace with rival Greta Picture, and they are now friends. Possibly their friendship will develop into some-thing even more beautiful, who knows. Ivor and Greta are sitting through a showing of the latest Disney film, 101 Equivalence Relations. They are fed up and start to discuss how many different equivalence relations they can find on the set {1,2}. They find just two. Then on the set {1,2,3}

they find just five different equivalence relations.

Have they found all the equivalence relations on these sets? How many should they find on {1,2,3,4} and on {1,2,3,4,5}? Investigate further if you feel like it!

Chapter 19

Functions

Much of mathematics and its applications is concerned with the study of func-tions of various kinds. In this chapter we give the definition and some elemen-tary examples, and introduce certain important general types of functions.

DEFINITION LetS and T be sets. A function from S to T is a rule that assigns to each s ∈ S a single element of T, denoted by f (s). We write

f : S → T

to mean that f is a function from S to T. If f (s) = t, we often say f sendss → t.

If f : S → T is a function, the image of f is the set of all elements of T that are equal to f (s) for some s ∈ S. We write f (S) for the image of

f . Thus

f (S) = { f (s)|s ∈ S} . Example 19.1

(1) Define f : {1,2,3} → Z by f (x) = x²− 4 for x ∈ {1,2,3}. The image of f is {−3,0,5}.

(2) Define f : R → R by f (x) = x² for all x ∈ R. The image of f is f (R) = {y|y ∈ R,y ≥ 0}.

(3) A body is dropped and falls under gravity for 1 second. The dis-tance travelled at timet is ¹₂gt². If we call this distances(t) and write I = {t ∈ R|0 ≤ t ≤ 1}, then s is a function from I to R defined by s(t) =¹₂gt². The image ofs is the set of reals between 0 and ^g₂.

(4) Define f : N×N → Z by f (m,n) = m−n for all m,n ∈ N. The image of f is Z.

(5) Let S = {a,b,c} and define functions f : S → S and g : S → S as follows:

f sends a → b, b → c, c → a, g sends a → b, b → c, c → b . 155

156 A CONCISE INTRODUCTION TO PURE MATHEMATICS

Then f (S) = S, while g(S) = {b,c}.

(6) LetS be any set, and define a functionι_S: S → S by ι_S(s) = s for all s ∈ S .

This functionι_S is called theidentity function of S.

We now define certain important types of functions.

DEFINITION Let f : S → T be a function.

(I) We say f is onto if the image f (S) = T; i.e., if for every t ∈ T there existss ∈ S such that f (s) = t.

(II) We say f is one-to-one (usually written simply as 1-1) if whenever s1,s2∈ S with s16= s2, then f (s1)6= f (s2); in other words, f is 1-1 if f sends different elements ofS to different elements of T. Another way of putting this is to say that for alls1,s2∈ S,

f (s1) = f (s2)⇒ s1=s2 .

This is usually the most useful definition to use when testing whether functions are 1-1.

(III) We say f is a bijection if f is both onto and 1-1.

Functions that are onto are often called surjective functions, or surjections;

and functions that are 1-1 are often called injective functions, or injections.

You will find these terms in many books, but I prefer to stick to the slightly more descriptive terms “onto” and “1-1.”

Let us briefly discuss which of these properties the functions in Exam-ple 19.1 possess.

The function in Example 19.1(1) sends 1 → −3, 2 → 0, 3 → 5, so it is 1-1.

It is clearly not onto.

The function in Example 19.1(2) is not onto and is not 1-1 either, since it sends 1 and −1 to the same thing.

On the other hand, the function s : I → R in Example 19.1(3) is 1-1, since, for t1,t2∈ I,

s(t1) =s(t2)⇒1 2gt¹²=1

2gt²²⇒ t1=t2. Also, s is not onto.

The function in Example 19.1(4) is onto but is not 1-1 since, for example, it sends both (1,1) and (2,2) to 0.

In Example 19.1(5), the function f is a bijection, while g is neither 1-1 nor onto. Finally, the identity function in Example 19.1(6) is a bijection.

FUNCTIONS 157 Here is a quite useful result relating 1-1 and onto functions to the sizes of sets.

PROPOSITION 19.1

Let f : S → T be a function, where S and T are finite sets.

(i) If f is onto, then |S| ≥ |T|.

(ii) If f is 1-1, then |S| ≤ |T|.

(iii) If f is a bijection, then |S| = |T|.

PROOF (i) Let |S| = n and write S = {s1, . . . ,sn}. As f is onto, we have

T = f (S) = { f (s1) , . . . ,f (sn)} .

Hence|T | ≤ n. (Of course |T | could be less than n, as some of the f (si)’s could be equal.)

(ii) Again let |S| = n and S = {s1, . . . ,sn}. As f is 1-1, the elements f (s1), . . . ,f (sn) are all different and lie inT. Therefore |T| ≥ n.

(iii) If f is a bijection, then |S| ≥ |T| by (i) and |S| ≤ |T| by (ii), so

|S| = |T |.

Part (ii) of Proposition 19.1 implies that if |S| > |T|, then there is no 1-1 function from S to T. This can be phrased somewhat more strikingly in the following way:

If we put n + 1 or more pigeons into n pigeonholes, then there must be a pigeonhole containing more than one pigeon.

(For if no pigeonhole contained more than one pigeon, the function sending pigeons to their pigeonholes would be 1-1.)

The above statement is known as the pigeonhole principle, and it is surpris-ingly useful. As a very simple example, in any group of 13 or more people, at least two must have their birthday in the same month (here the people are the

“pigeons” and the 12 months are the “pigeonholes”). As another example, in any set of 6 integers, there must be two whose difference is divisible by 5: to see this, regard the 6 integers as the pigeons, and their remainders on division by 5 as the pigeonholes.

Here’s a slightly more subtle example.

Example 19.2

Prove that ifn+1 numbers are chosen from the set {1,2,...,2n}, there will always be two of the chosen numbers that differ by 1. (This is not necessarily true if we choose onlyn numbers — for example, we could choose then numbers 1,3,5,...,2n−1.)

158 A CONCISE INTRODUCTION TO PURE MATHEMATICS Answer This becomes easy when we make the following cunning choice of what the pigeonholes are. Define the pigeonholes to be the n sets

{1,2}, {3,4},...,{2n − 1,2n}.

Since we are choosing n + 1 numbers, each of which belongs to one of these pigeonholes, the pigeonhole principle tells us that two of them must lie in the same pigeonhole. These two will then differ by 1. Pretty neat, eh?

More examples of the use of the pigeonhole principle can be found in Exer-cise 5 at the end of the chapter.