374 ANSWERS AND SOLUTIONS TO SELECTED EXERCISES
- (c) Proof Assume [x] = [y]. Then y E [x], so that x E y. Conversely,
suppose x E y and let w E [y]. Then y E w. Since x E y, we have x E w, so that
w E [x] and [y] G [x]. The reverse inclusion is proved in identical fashion.
Article 7.4
- (c) 2 is a lower bound and 60 an upper bound for {4,6, lo), which has
neither a greatest nor a least element. - (a) M v N is simply the union M u N, by Exercise 5(a).
- (a) (iii) lub C, = {1,3,5,7,.. .), which is also the greatest element of C,.
Also, glb C, = 0. Note that C, has no least element. - (b) (i) Proof Given the hypotheses [MI] = [M,] and [N,] = [N,], suppose
[Ml] 5 [N1 1. Then n(M,) = n(M,) I n(N,) = n(N2), so that n(M2) I n(N,), as
desired. The implication = follows in a similar manner.
Article 8.1
- (c) (ii) r, is not a function since the ordered pair (0, y) E r,, for any y E R.
- (e) j is not one to one since j(0) = j(1) = j(- 1) = 0.
- (c) Proof Note first that f is a mapping of A into B, and ix is a
mapping of X into A, so that f 0 i, is a mapping of X into B. By definition
of restriction, f/, is a mapping of X into B, also. Hence f 0 ix and f/, have
the same domain and codomain. To prove equality of mappings, we need
only prove that (f 0 i,)(x) = (f/,)(x), for each x E X. Letting x be an
arbitrary element of X, we note that (f 0 i,)(x) = f(i&)) = f(x) = (f/,)(x), as
desired. 0 - (e) Proof By definition of composition, dorn (g 0 f) = {x E dorn f I f (x) E dorn 9).
Since dorn f = A and dorn g = C, this translates to {x lx E A and f (x) E C).
Since mg f r C, we have that x E A implies f (x) E C, so that {x 1 x E A
and f (x) E C) = {X ( x E A) = A. Hence dorn (g 0 f) = A, as desired. Suppose
now that z E mg (g 0 f), so that z = (g 0 f)(x) for some x E A. Hence
z = g( f (x)) for some x E A. Since f (x) E rng f and rng f E C, then f (x) E C
and z = g(c) for some e E C [namely; c = f (x)]. Since g maps C into D,
then z = g(c) E D. Hence mg (g 0 f) E D. - (a) Proof If g: C + B is an extension off: A + B, then f E g, by definition.
To prove A c C, let a be an arbitrary element of A. Since A = dorn f,
then (a, f (a)) E f. Since f E g, we have (a, f (a)) E g, so that a E dorn g.
Since C = dorn g, we conclude a E C, as desired. 0
Article 8.2
- (b) Since f (x) = (x2 - x - 12)/(x - 4) = x + 3, for all x E dorn f, then f is
clearly injective [recall Exercise qb), Article 5.21. However, since 7 4 irn f,
then im f c R, so that f is not surjective, and hence is not bijective.
(k) tan-' is defined as the inverse of the tangent function, restricted to