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
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 s
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 c
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
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. Fo
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
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-reex
+ 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
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 .
Thi