MATH 329: Exam 2 Review Solutions
Monday, April 18, 2011
1. Let S = cfw_ < 3 | lim() cof () = . Suppose |S | < 3 . 3 is a successor cardinal
and thus is regular. Hence |S | is bounded in 3 by some < 3 . However, + > and
cof ( + ) = . Thus + S , contradict
21-329 HOMEWORK 9
GREGGO JOHNSON
1) Enumerate A = cfw_ai | i < and B = cfw_bi | i < . We will construct by induction
a series of partial isomorphisms fi | i < such that
(a) for i < j , fi fj ;
(b) for any a, a dom fi , we have aEa i f (a)Ef (a );
(c) ai
21-329 HOMEWORK 8
GREGGO JOHNSON
5.38) Since u T , u s . We know that s s , so u s and
A = cfw_ < | s ( ) = u( ) is nite. Set v = s ( A) cfw_(, u( ) | A.
Then cfw_ < | s ( ) = v ( ) = A is nite so v T and v = u.
1 - 5.39) We know that u s . Set v = u
0 an
21-329 HOMEWORK 7
GREGGO JOHNSON
5.27)
1. is a strategy for player I in G i is a function whose domain is and
A
strings of the form s0 , n0 , s1 , . . . , sk , nk for some k < , where each si < 2
and each ni and whose range is in < 2.
2. is a winning stra
21-329 HOMEWORK 6
GREGGO JOHNSON
5.1) Note that both I and S are nonempty.
1. No. Let x I . Suppose we had an s < such that x Ns I . However,
s Ns I , contradiction.
0
2. Yes. Let A = cfw_Ns | s < and there is n + 1 < (s) such that s(n) = (s) 1 .
Then A i
21-329 HOMEWORK 5
GREGGO JOHNSON
4.6) Fix an ordinal . Dene a sequence n | n < by 0 = +1 > and n+1 = n .
Now set = supn< n . Then, since the n s are conal in , cf = and
= supcfw_ | < = supcfw_n | n < =
4.8) (a) For every a < b R, there is a c Q so a <
21-329 HOMEWORK 4
GREGGO JOHNSON
4.1*) 1.) Clearly < 2 is innite because it contains cfw_(i, 0) | i < n for each
n < . Then
0 |< 2| |
n
n<
|n 2| =
2|
n<
2n
n<
0 = 0 0 = 0
n<
2.) Given x = y 2, we have that there is some n0 so x(n0 ) = y (n0 ).
This mean
21-329 HOMEWORK 3
GREGGO JOHNSON
Claim: type A type B i there is an order preserving injection from A to B
3.4)
1) By denition of ordinal addition,
0 + = type (0 )
= type [(cfw_0 0) (cfw_1 ), ]
= type [cfw_1 , ]
where, for all , , (1, ) (1, ) i < . Thus,
21-329 HOMEWORK 2
GREGGO JOHNSON
3.1 Proof by contradiction. So, for all y A, there is S = pred(A, A ) y . For
each y , there are two possibilities:
There is an x A so x S and x pred(A, A ) y . This means that
y x. Since x S , we can conclude that y S .
21-329 HOMEWORK 1
GREGGO JOHNSON
1.2 1
The only function from to B is the empty function, f = . In general,
a function g from A to C is a subset of A C such that for every a A,
there is a unique c C such that (a, c) g . In this case, B = , so
the only pos
Full Name:
MATH 329: Exam 2
Wednesday, April 20, 2011
Prove only 3 of the following 5 problems. If you prove more than 3 then only the rst 3
problems will be graded. I wont even look at the other ones.
Problem Points Score
1
33
2
33
3
33
4
33
5
33
Your N
MATH 329: Exam 1 Solutions
1. (a)
(b) 3
(c) 3 3
(d) 2 + + 3
= |cfw_0 cfw_1 |
= | |
= | |
2. (a)
(b) Dene H : ( ) by H (f ) = g where g is the function dened by g (m, l) =
f (m)(l) for m and l .
3. (a)
0
(b) 22
(c) 1
(d) 0
(e) 2
4. (a) V3
(b) 1
(c) 1
MATH 329: Exam 1 Solutions
1. (a)
(b) 3
(c) 3 3
(d) 2 + + 3
= |cfw_0 cfw_1 |
= | |
= | |
2. (a)
(b) Dene H : ( ) by H (f ) = g where g is the function dened by g (m, l) =
f (m)(l) for m and l .
3. (a)
0
(b) 22
(c) 1
(d) 0
(e) 2
4. (a) V3
(b) 1
(c) 1
Full Name:
MATH 329: Exam 2
Wednesday, April 20, 2011
Prove only 3 of the following 5 problems. If you prove more than 3 then only the rst 3
problems will be graded. I wont even look at the other ones.
Problem Points Score
1
33
2
33
3
33
4
33
5
33
Your N
MATH 329: Final Exam Solutions
1. (a) Let D be a club on . Since cf () > , we know C D is club on (intersection of 2
clubs is a club). Since S is stationary, we know S (C D) = and thus (S C ) D = .
Since D was an arbitrary club on , this implies S C is st
21-329 HOMEWORK 1
GREGGO JOHNSON
1.2 1
The only function from to B is the empty function, f = . In general,
a function g from A to C is a subset of A C such that for every a A,
there is a unique c C such that (a, c) g . In this case, B = , so
the only pos
21-329 HOMEWORK 2
GREGGO JOHNSON
3.1 Proof by contradiction. So, for all y A, there is S = pred(A, A ) y . For
each y , there are two possibilities:
There is an x A so x S and x pred(A, A ) y . This means that
y x. Since x S , we can conclude that y S .
21-329 HOMEWORK 3
GREGGO JOHNSON
Claim: type A type B i there is an order preserving injection from A to B
3.4)
1) By denition of ordinal addition,
0 + = type (0 )
= type [(cfw_0 0) (cfw_1 ), ]
= type [cfw_1 , ]
where, for all , , (1, ) (1, ) i < . Thus,
21-329 HOMEWORK 4
GREGGO JOHNSON
4.1*) 1.) Clearly < 2 is innite because it contains cfw_(i, 0) | i < n for each
n < . Then
0 |< 2| |
n
n<
|n 2| =
2|
n<
2n
n<
0 = 0 0 = 0
n<
2.) Given x = y 2, we have that there is some n0 so x(n0 ) = y (n0 ).
This mean
21-329 HOMEWORK 5
GREGGO JOHNSON
4.6) Fix an ordinal . Dene a sequence n | n < by 0 = +1 > and n+1 = n .
Now set = supn< n . Then, since the n s are conal in , cf = and
= supcfw_ | < = supcfw_n | n < =
4.8) (a) For every a < b R, there is a c Q so a <
21-329 HOMEWORK 6
GREGGO JOHNSON
5.1) Note that both I and S are nonempty.
1. No. Let x I . Suppose we had an s < such that x Ns I . However,
s Ns I , contradiction.
0
2. Yes. Let A = cfw_Ns | s < and there is n + 1 < (s) such that s(n) = (s) 1 .
Then A i
21-329 HOMEWORK 7
GREGGO JOHNSON
5.27)
1. is a strategy for player I in G i is a function whose domain is and
A
strings of the form s0 , n0 , s1 , . . . , sk , nk for some k < , where each si < 2
and each ni and whose range is in < 2.
2. is a winning stra
21-329 HOMEWORK 8
GREGGO JOHNSON
5.38) Since u T , u s . We know that s s , so u s and
A = cfw_ < | s ( ) = u( ) is nite. Set v = s ( A) cfw_(, u( ) | A.
Then cfw_ < | s ( ) = v ( ) = A is nite so v T and v = u.
1 - 5.39) We know that u s . Set v = u
0 an
21-329 HOMEWORK 9
GREGGO JOHNSON
1) Enumerate A = cfw_ai | i < and B = cfw_bi | i < . We will construct by induction
a series of partial isomorphisms fi | i < such that
(a) for i < j , fi fj ;
(b) for any a, a dom fi , we have aEa i f (a)Ef (a );
(c) ai
MATH 329: Exam 2 Review Solutions
Monday, April 18, 2011
1. Let S = cfw_ < 3 | lim() cof () = . Suppose |S | < 3 . 3 is a successor cardinal
and thus is regular. Hence |S | is bounded in 3 by some < 3 . However, + > and
cof ( + ) = . Thus + S , contradict
21-329 HOMEWORK 9
GREGGO JOHNSON
1) Enumerate A = cfw_ai | i < and B = cfw_bi | i < . We will construct by induction
a series of partial isomorphisms fi | i < such that
(a) for i < j , fi fj ;
(b) for any a, a dom fi , we have aEa i f (a)Ef (a );
(c) ai
21-329 HOMEWORK 8
GREGGO JOHNSON
5.38) Since u T , u s . We know that s s , so u s and
A = cfw_ < | s ( ) = u( ) is nite. Set v = s ( A) cfw_(, u( ) | A.
Then cfw_ < | s ( ) = v ( ) = A is nite so v T and v = u.
1 - 5.39) We know that u s . Set v = u
0 an