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A subset of a set is a set all of whose elements belong to the original set. Thus A is a subset of B, written as A ~ B, if and only if x EA => x E B. If, as this definition allows, B contains elements that do not belong to A then A is called a proper subset of lJ. This is denoted by A C B. For the number sets listed above, f'\J c "1L C QC IR. To sec that the subsets are pro_ger, note that -1 E 7L but -1 t1. N, E O! but ~ fl. 1). Sets can be manipulated using standard operations. For sets A and B the union, intersection and dlfTerence, denoted respectively by AU H, An B and A - B, are defined by i A U B = {x : x A n E A or x E B = {x: x EA and x A - B = {x : x E B} E B} A and x fl.

This fact is expressed as 272 < ~, which is read as ·i,_ is less than ~·- ll now seems reasonable, though sadly wrong, to suppose that every point on the number line can be labelled with a rational. To sec why this intuition is wrong, consider a point to the right of the point labelled 0 whose distance L from 0 satisfies 0 < L < 1. Since there arc infinitely many. infinitely closely packed rntionals between 0 and 1, surely one of them must coincide with L? But the rationals 0, ·1. , i, ~. ~, ~' ft;, ...

P(B) ~ H) (1i(A U B) Show by example that equality need not hold in part (b ). 8. -0 if x < 0 1 Find formulae for f g and g "f. Explain briefly why none off, g and f o g possess an inverse. Find the inverse of g c f. 0 PROBLEMS I 9, 39 Let A be any set and let f: A - A be a function satisfying(' =id A (that is, f ~ f f is the identity function on A). Prove that f is bijective. Suppose now that A= {x e R: x 4=0, 1} and that /:A - A is given by f(x) = 1 - 1/x. Calculate / 1 and hence deduce that f is bijective.

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