366 ANSWERS AND SOLUTIONS TO SELECTED EXERCISES
- Hx): X is a field, q(x): X is a ring, r(x): X is an integral domain; this argument
has the form: P E Q, Q n R # 0, R n P' # 0, therefore Q n P' # 0. The
argument is invalid. Let U = {1,2, 3,4, 51, P = {1,2}, Q = {1,2},
R = (2, 3,4}.
Article 4.1
- (a) Proof Assume X c Y and X c 2. To prove X E Y n Z, let x be an
arbitrary element of X. We must prove x E Y n Z; that is, x E Y and x E Z.
Now since x E X and X E Y, we have x E Y. Similarly, since x E X and
X c 2, we have x E Z. Hence x E Y and x E Z SO x E Y n Z, as desired. - (b) Proof Assume B' E A'. We must prove A s B. Now, by 7(a), since
B' G A', then A" E B". But since A = A" and B = B [by Exercise 6(a)(iii)],
this leads to A G B, our desired conclusion. - (a) Proof =+ Assume A n B = U. We must prove A = U and B = U. For
the first of these, note that A E U is always true so that we need only establish
the reverse containment U G A. To do this, let x E U. Since U = A n B, then
x E A n B. Since x E A n.B, then x E A and x E B so that, in particular, x E A.
An identical argument verifies that U E B, so that U = B, as well. -= Assume
A = U and B = U. To prove A n B = U, let x E U; we must show x E A n B;
that is, x E A and x E B. Now since x E U and U = A, then x E A. Since x E U
and U = B, then x E B. Hence x E A and x E B, as desired.
(Note: An alternative presentation of the argument is "assume A = U and
B = U. Then A n B = U n U = U.")
Article 4.2
- and 2. (b) union = N, intersection = (1) = A,; increasing. (c) union =
N = A,, intersection = 0; decreasing. (j) union = R, intersection =
(- GO, 11 = A,; increasing. (k) union = (- GO, - 1) = A,, intersection = @;
decreasing. - (b) Proof Assume {A,) k = 1,2,.. .} is a decreasing family. We may prove
the desired equality by proving mutual inclusion, where we note that 2
follows immediately from Exercise 4(a). To prove E, let x be an arbitrary
element of u,"=, A,; we must prove x E A,. Now since x E u,"= , A,, then
x E Aj for some j E N. 1fj = 1, we are finished. If j > 1, then A, c A,, by
the "decreasing" hypothesis. In that case x E Aj and Aj c A, together imply
x E A,, our desired conclusion. - (e) (i) Proof Let x E n,"!, A,. To prove x E ng, A,, we must prove x E A,
for every h 2 n. So assume h E N and h 2 n. Since m < n I h, we have
h > m. Since x E ngrn A,., then x E A, for every k 2 m; in particular x E A,.
Since h was an arbitrarily chosen integer greater than or equal to m, we
have established x E n,"=, A,, as desired. - (a) Yes. Let A, = (0, 1 + llk) for each k = 1,2,.... Then nr=, A, = (0, 11.
Note that this is not possible in a finite collection of sets.