# Sets and Mappings by T. S. Blyth, E. F. Robertson (auth.)

By T. S. Blyth, E. F. Robertson (auth.)

IT, because it is frequently stated, arithmetic is the queen of technology then algebra is definitely the jewel in her crown. during its colossal improvement during the last half-century, algebra has emerged because the topic during which you can discover natural mathe matical reasoning at its top. Its attractiveness is matched in simple terms by way of the ever-increasing variety of its functions to a very wide variety of subject matters in parts except 'pure' arithmetic. the following our target is to offer, within the kind of a chain of 5 concise volumes, the basics of the topic. on the whole, we've got lined in the entire now conventional syllabus that's present in first and moment 12 months college classes, in addition to a few 3rd yr fabric. extra learn will be on the point of 'honours options'. The reasoning that lies at the back of this modular presentation is easy, particularly to permit the scholar (be he a mathematician or no longer) to learn the topic in a manner that's extra applicable to the size, content material, and quantity, of a few of the classes he has to take. even supposing we've taken nice pains to incorporate a large selec tion of illustrative examples, we've not incorporated any exer cises. For an appropriate better half selection of labored examples, we might refer the reader to our sequence Algebra via perform (Cambridge college Press), the 1st 5 books of that are acceptable to the cloth lined here.

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**Example text**

But ( n+ O 1) = 1 E IN and 1 ( n+ ) = 1 E IN n+1 ' while for 1 ~ r ~ n we have, by (1), Then (;) E IN and V~ 1 ) E IN give (n~ 1 ) E IN as required. 0 Example Let E be a set with lEI= n. Then E contains (;) subsets A with IAI = r. We prove this by induction. The result is clearly true for n = 0 and n = 1. Suppose that it holds for all sets having no more than n elements where n ~ 1. AI = r. Observe that the result holds for r = n + 1 and r = 0, so suppose that 1 ~ r ~ n. Let x E E and consider E\ {x}.

There are infinitely many left inverses of f. ~ 1; if n = 0. Then for every n E IN we have so each gP is a left inverse of f. Example f : IN -+ IN given by f(n) = n 2 is injective, for n 2 = m2 with n, m ~ 0 gives n = m. -+ IN given by f(n) = n2 is not injective. For example, we have /(1) = /(-1). Example f : IR-+ IR x IR given by f(x) = (x, 0) is injective. VOLUME 1 : SETS AND MAPPINGS 26 Example f: E-> P(E) given by l(x) = {x} is injective. Example I: (-> ( given by l(x + iy) = x- iy is injective.

1) = fJ[I- {1}] = XJ-{1} =I, and so {} o 1-' is the identity mapping on 2A. Also, it is clear that for every BE P(A) we have xii{1} = B, and therefore so that 1-'o{} is the identity mapping on P(A). We conclude that {}is a bijection with fJ- 1 ='"''so P(A) and 2A are equipotent. The important feature of a bijection is that it allows us to compare two sets. It is intuitively clear that if A, B are finite sets then they are equipotent if and only if they have the same number of elements. The notion of a bijection can also be used 36 VOLUME 1 : SETS AND MAPPINGS to compare infinite sets, though we must postpone a discussion of this until later when we shall be able to tackle properly the question of what is meant by an 'infinite' set and how to compare the 'sizes' of two such sets.