372 ANSWERS AND SOLUTIONS TO SELECTED EXERCISES
positive integer M, such that Ix, - x) < E whenever n 2 M. Hence, for the
desired m, simply choose a positive integer larger than both N and M, and note
that Ix,,, - XI < e, as desired. For uniqueness, suppose x' is another cluster
point of {x,). Let e > 0 be given. We will show Ix - x'l < e, so that x = x',
by Example 5, Article 6.2. Since x, -* x, then corresponding to this E, divided
by 2, 3N E N such that Ix, - xl < ~/2 whenever n 2 N. Since x' is a cluster
point of (x,), then corresponding again to ~/2 a& the preceding N, there
exists p 2 N such that Ix, - x'l < ~12. Note that (x, - xl < ~/2 since p 2 N.
Hence IX - x'I = IX - xP + xP - x'I I IX - xPI + IxP - x'I < 612 + 812 = E, as
desired.- (c) Proof Assume S, and S, are both bounded above in R. To prove that
S, u S, is bounded above in R, we must show that there exists M E R such
that x I M for all x E S1 u S,. Now we know that 3M, E R such that
x r; M, for all x E S1 and 3M2 E R such that x 5 M, for all x E S,. Let
M = max {MI, M,). If x E S1 u S,, then either x E S1 (in which case
x r; M, 5 M, as desired), or else x E S, (so that x I M, r; M, again, as
desired). - (a) See the proof of Theorem 1, Article 9.3.
(e) Proof Assume u = lub S and u # S. We must show that every
neighborhood N(u; E) of u contains a point of S other than u itself. So let
E > 0 be given. By (ii) of the definition of lub, there exists y E S such that
y > u - E, SO that u - E < y I u and y E N(u; E). Clearly u # y since u q! S and
YES. 0
For the example, let S = [I, 21 u (3). Then 3 = lub S, but 3 is clearly not a
point of accumulation of S.
Article 7.1
- (a) R, contains the ordered pairs (4, -$), (8, -2), (g, O), (4,3, and (n, n),
among many others. - (a) R is transitive if and only if
(Vx)(Vy)(Vz)[((x, y) E R A (y, z) E R) ((x, z) E R)]; R is antisymmetric if and
only if (Vx)(Vy)[((x, y) E R A (y, x) E R) (x = y)]. (b) R is not reflexive on
A if and only if (3x E A)(& x) 4 R); R is not symmetric if and only if
(3x)(3y)[(x, y) E R A (y, x) & R]; R is not transitive if and only if
(3x)(3y)(3z)[(x, y) E R A (y, z) E R A (x, z) 4 R]; R is not antisymmetric if and
only if (34(3y)[(x, Y) E R A (Y, x) E R A (x Z Y)]. - (a) R,, is symmetric only (if y = llx, then x = l/y). It is not reflexive (2 # $,
e.g.), not transitive (since 3 = l/(j) and (i) = 3, but 3 # i), and clearly not
antisymmetric. (b) R,, is reflexive (since Ix - xl = 0 S 1 for any x E R),
symmetric (following from the fact that Ix - yl = ly - XI), not transitive (since,
e.g., 12.5 - 21 r; 1 and 13.2 - 2.5) s 1, but 13.2 - 2) > I), and not antisymmetric
[since (3,2) and (2,3) are both in R,,, but 3 # 2). - (c) Proof =+ Assume R is symmetric. We prove R = R- ' by mutual inclusion.
First, let (x, y) E R. By symmetry, (y, x) E R so that (x, y) E R - ' and R G R - '.
The reverse inclusion is proved in an identical manner. C= Conversely, suppose