Math M114S and Phil M134, Homework 2 Solutions
x2.3. We know P (N) =c R from Theorem 2.26 together with Lemmas
2.24 and 2.25. Now we show that for all n 2, =c n . (Since R =c ,
this gives Rn =c n , and then we are done by Proposition 2.2.) The
bijection b
Math M114S and Phil M134, Homework 1 Solutions
x1.2. First, let x C \(A B ). Then x belongs to C but not to
A B ; so x belongs to neither A nor B ; thus, x belongs to C \ A and to
C \ B , i.e., x (C \ A) (C \ B ). For the opposite direction, suppose
x (C
Math M114S and Phil M134, Homework 3 Solutions
x3.1. We check that X E = Y E for all X and Y from E . To see this,
let x X E . Then x belongs to every U E . Therefore, x Y E as
well. The converse hold by the same argument, and so X E = Y E .
Second, we ch
Math M114S and Phil M134, Homework 4 Solutions
x4.2. First we check that
cfw_0, x = cfw_0, x = x = x ,
by assuming that cfw_0, x = cfw_0, x and considering two cases.
Case 1. x = 0. Now cfw_0, x = cfw_0, and so cfw_0, x is a singleton, i.e.,
cfw_0, x
Math M114S and Phil M134, Homework 4 Solutions
1. (i) If there are no atoms (as was specied during the exam), the statement
is true. Assume that A is transitive. Assume y x P (A). Since x A,
y A. Since there are no atoms, y is a set. Since A is transitive
MATHEMATICS 114S PHILOSOPHY 134
WINTER 2012
2nd Midterm Examination
Solutions
1. Let and be ordinal numbers. Prove by transnite induction that
( ON) ( + ) + = + ( + ).
Do this directly from the denition of ordinal addition.
Solution. Let be an ordinal num
Math M114S and Phil M134, Formulas and Deniteness Axioms
This note describes the alternative way sketched in class to state axioms for
denite conditions.
Let us mean by a variable any upper or lower case roman letter, with or
without a numberical subscrip
Math 114S and Phil 134, Ordinal Numbers
Denite operations. On pages 26-27, the book gives axioms for denite
conditions and denite operations together. In Supplement 1, we used the
notion of a formula to give a single axiom for denite conditions. Instead
o
Math 114S and Phil 134 Supplement 3: Cardinal Numbers
A (von Neumann) cardinal number is an ordinal number such that
( < ) =c .
Theorem ( and Cardinals). Every natural number (i.e., every member
of ) is a cardinal number, and is a cardinal number.
Proof.
EXTRA PROBLEMS FOR MATH M114S, PHIL M134
#1. True or false? (And prove your answer.)
(a) cfw_ = cfw_.
(b) cfw_, cfw_ = cfw_.
Solution. Both parts depend in the fact that = cfw_, which is true because
has no members while cfw_ has a member.
(a) is false,
M114S, Solutions to HW #7 x7.10. The fact that U o V is a linear ordering is quite trivial. If = X U o V , let y = min{y V | (x U )[(x, y) X ]},
V
x = min{x U | (x, y ) X }
U
and check easily that (x , y ) is the least element of X . x7
M114S and M134, solutions to HW #6
x6.2. Suppose that S is a chain in the function space (A E). We need to find a least upper bound for S. For each a A, let S(a) = {f(a) | f S}. Then we show that for each a A, S(a) is a chain in E. Let f(a) and
M114S, Solutions to HW #5
x5.3. Both properties of exponentiation are proved by induction on k. First we show that n (m+k) = n m n k . If k = 0, then n (m+k) = n m , and n k = n 0 = 1. Now it is a general fact that for all p, p 1 = p. (Proof: By (
M114S, Solutions to HW #4
x4.11. The associativity of cardinal addition is shown on p. 43 and Exercise 4.27. Commutativity boils down to the assertion that =c . This commutativity property of disjoint unions holds for all sets, not just cardinals
M114S, Solutions to HW #3
x3.4. First note that if W is a connection of A and B, then W P(A B). Thus W P(P(A B). So the set (A, B) of connections of A and B is a subset of P(P(AB); it exists by the Separation Axiom (since the conditions in the d
M114S, Solutions to HW #2
x2.3. We know P(N) =c R from Theorem 2.26 together with Lemmas 2.24 and 2.25. Now we show that for all n 2, =c n . (Since R =c , this gives Rn =c n , and then we are done by Proposition 2.2.) The bijection between and n
M114S, Solutions to HW #1
x1.1. Suppose that x A (B C ). Then either x A (in which case x belongs to both A B and A C and hence to their intersection), or else x B C (so, again, x (A B) (A C ). This proves half of what we want. For the o
SAMPLE PROBLEMS FOR THE SECOND M114S MIDTERM
Forget about the numbers-the problems come from previous midterms and finals and I have not bothered to change their numbers Problem 3. Consider the following binary relation on the set (N N) of all func