Math 116c Solutions to Homework # 7
Due: 05-23-02
1. (a) We have rank(cfw_x) = supcfw_rank(y ) + 1 : y cfw_x = rank(x) + 1.
(b) Let = maxcfw_rank(x), rank(y ). Since x, y x y we have rank(x y ) . and since
both x and y V , we have rank(x y ) .
2. We know
Math 116c Solutions to Homework # 6
Due: 05-16-02
1. That + and are countable follows from the fact that the disjoint union and direct
product of countable sets are countable (proved earlier). For proceed by induction on .
If is countable, then so is +1 =
Math 116c Solutions to Homework # 5
Due: 05-09-02
1. Let a A be a non-nal element, so there is a b A with a b. Let S = cfw_c A : c a Then
S is a proper initial segment, so there is some x A such that S = Sx = cfw_y A : y x.
Hence a x and since a is a maxi
Math 116c Solutions to Homework # 4
Due: 05-02-02
1. Suppose A,
has the l.u.b. property. Let = B A be bounded below, and let C be the
set of lower bounds for B , i.e. C = cfw_c A : b B (c b). Then C = and C is bounded
above by any element of B . Hence C h
Math 116c Solutions to Homework # 3
Due: 04-25-02
1. We will show (in Z) that (a) (c) (b) (a).
(a) (c): Let g be a surjection from x to y , and let be a choice function for x. Dene
h by:
h(w) = (cfw_z x : g (z ) = w)
This is dened since cfw_z x : g (z )
Math 116c Solutions to Homework # 2
Due: 04-18-02
1. (a) Let A = be a bounded subset of R. Let x = A. We rst check that x is a real, i.e. a
cut. Since A = we have x = , and since A is bounded we have x = R. If p x and
q < p then p y for some y A, so q y a
Math 116c Solutions to Homework # 1
Due: 04-11-02
1. Note that if two sets are disjoint, then their union is the same as their symmetric dierence.
Let x and y be two sets. Then x \ y = x (xy ), and x y is the disjoint union of x \ y and
y , so
x y = y (x
Math 116c Solutions to Final Homework
Due: 6-14-02
1. (a) Note that we showed this in an earlier exercise for = 0 since H(0 ) = HF and
h(0 ) = cfw_x : TC(x) is nite. So we may assume > 0 .
Let x h(), so |TC(x)| < . Let y TC(x). Since TC(x) is transitive,
Ma116c, Homework 3 Solutions
(1) For a subset A cfw_1, 2, . . . , let
FS(A) = cfw_
a : F A, F nite, nonempty
aF
consist of all nite sums of distinct elements of A. Prove Folkmans
Theorem: Given positive integers m, n, there is a positive integer N
such th
Math 116c Solutions to Homework # 8
Due: 05-30-02
1. (a) If is a successor cardinal then it is regular, so cof() = = 2 > . So assume
cf
is a limit cardinal. Let = cof(), and let f : . Let I = and let = |f ( )|
for < . Let = for all < , so < for all . The