We show < and > 1 implies < by induction on > . For
= + 1 we have +1 = > since > 1. If < then
< +1 . Finally if < for all < < , limit, then
< sup< = .
We show + = , by induction on . Obvious for = 0. Assume
+ = . Then
+( +1) = ( + )+1 = + = = +1 .
Ma116b 2010, Homework 6 Solutions
1) On ordinal arithmetic
Below , , , . . . , , , . . . range over ordinals.
Solution.
A)
( + ) = + holds since it holds for order types.
If < then + is an initial segment of + .
Induction on starting with = + 1, in whi
Ma116b 2010, Homework 5 Solutions
1a) (AC) Show that a relation
is wellfounded i it has no innite descending chains.
Solution. Let be a binary relation on X . (): If x2 x1 x0 is an innite
descending chain then cfw_x0 , x1 , x2 , . . . is nonempty and has
Ma116b 2010, Homework 4 Solutions
1) Find ordered sets A, B such that
A
B, B
A, but A B .
=
Solution. A = + 1 and B = .
2*) Prove that if A Q, then at least one of A or Q \ A contains a subset X such that
X, < Q, < .
=
Solution. If A is dense in Q then A,
Ma116b 2010, Homework 3 Solutions
(1) Two sets x, y are called almost disjoint if their intersection is nite. Show that there
is a collection A of 20 subsets of which are almost disjoint.
Solution. For each f : 2, let If = cfw_f |n : n 2< where f |n =
(f
Ma116b 2010, Homework 2 Solutions
n
X be a function and x0 a xed element of X . Prove that
(1) Let g :
n X
there is a unique function f : X such that
f (0) = x0 ;
f (n) = g (f |n, n), if n > 0.
Here f |n is the restriction to n = cfw_0, 1, . . . , n 1.
S
Ma116b 2010, Homework 1 Solutions
(1) Let m, n be natural numbers. Show that m < n m n.
Solution. () Assume m < n. The relation < is a linear ordering on , so m = n
since linear orderings are non-reexive. By denition, m < n means that m n,
which implies m
+ is not continuous since the cut determined by the rst copy of R has no
l.u.b., hence + = .
(2) a) Z, < = + is obvious and can be proved by a picture. + has a least
element while + does not. 1 + has a least element but + 1 does not.
b) Obvious.
(3)
a) A
Ma116b 2010, Homework 7 Solutions
1) A) Enumerate = cfw_n nN and inductively dene the sequence cfw_n nN by 0 :=
0 , and n+1 = the least ordinal less than such that n+1 > maxcfw_n , 0 , . . . , n .
This sequence is clearly an increasing sequence of ordinal