 By Anthony N. Michel, Charles J. Herget

ISBN-10: 0817647066

ISBN-13: 9780817647063

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Extra info for Algebra and Analysis for Engineers and Scientists

Example text

RELATIONS AND EQUIVALENCE RELATIONS Throughout the present section, X denotes a non-void set. We begin by introducing the notion of relation, which is a generalization of the concept of function. 1 Deftnition. Let X and Y be non-void sets. 2. Example. Let A = u{ , v, ,x y) and B = a{ , b, c, d). Let ~ = ({ u, a), (v, b), (u, c), (x, a»). Then ~ is a relation from A into B. ). 3. Example. Let X = Y = R, the set of real numbers. The set y) E R x R: :x ::;;; y) is a relation in R. Also, the set ({ ,x y) E R x R: x = sin y) is a relation in R.

This implies that (X I ,f(X I» and (X 2,f(X 2» E f and so (f(x l ), IX ) and (f(x 2), x 2) E g. Since f(x l ) = f(x 2) and g is a function, we must have IX = X 2. Therefore,fis injective. _ The above result motivates the following definition. 9. Definition. 8 the inverse off Hereafter, we will denote the inverse of fbyf- I . 10. Theorem. Y Then fis a one-to-one mapping of X onto fSt(f); f- I is a one-to-one mapping of fSt(f) onto ;X for every X E ,X f- l (f(X » = ;X and for every y E al(f),f(f- I (y» = y.

We do so by contradiction. Suppose that f(J ) I. Since f(J ) c I, this implies that 1- f(J ) 0. L e t q be the smallest integer in 1 - f(J ) . Then q f(1) because f(l) E f(J ) , and so q > f(I). This implies that 1 n J q _ . 0. Since In J q _ . is non- v oid and finite, we may find the largest integer in this set, say r. q But r < q implies that r E f(J ) . This means there is an s E J such that r = f(s). q Hence, q E f(J ) and we have arrived at a contradition. Thus, f is surjective. This completes the proof.