4158 Fall 2012 Test 3
Do any two of the following problems.
1. Prove that 2 = .
Proof. Recall that every nite ordinal k is either even (k = 2n for some
n ) or odd (k = 2n + 1 for some n ), but not both. Hence there
is a bijection f : 2 dened by f (m, n) =
4158 Fall 2012 Test 2
Do any two of the following problems.
1. Let X, Y and Z be counting systems. Prove that if X is isomorphic to Y
and Y is isomorphic to Z, then X is isomorphic to Z.
Proof. Suppose X = (X, x0 , s), Y = (Y, y0 , t) and Z = (Z, z0 , u).
4158 Fall 2012 Test 1
Do any two of the following problems.
1. Prove that, for any sets A and B , A \ (A \ B ) = A B .
Proof. First suppose that x A \ (A \ B ). Then x A \ B , so either x A
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or x B . But x A, so x B as well, and therefore x A B .
Conve
The foundations of mathematics
Notes for MATH 4158
R.A. Litherland
August 16, 2012
These notes cover most of the material in the classic texts by Halmos [1]
and Landau [2]; the course will cover at most whats here, and possibly less
depending on how thing
1. Exercise 15.1. Prove Theorem 15.7: If A and B are nite then A B is
nite and #(A B ) = (# A)(# B ).
Proof. Let m = # A and n = # B . Then A m and B n, so
A B m n. By denition of ordinal multiplication, mn = ord(m n)
(where m n is given the reverse lexic
1. Exercise 13.1: Let X be a non-empty well-ordered set and E any set of
well-ordered sets order-isomorphic to X . Show that there is a well-ordered
set Y X with Y E. [Hint: dene Y by replacing the least element of
/
X by a suitably chosen set.]
Proof. Le
1. Prove parts (3) and (4) of Theorem 11.5:
For all x in R:
(3) x + 0 = x;
(7) x + (x) = 0.
Proof. (3) For x = [x1 , x2 ],
x + 0 = [x1 , x2 ] + [1, 1] = [x1 + 1, x2 + 1] = [x1 , x2 ] = x.
The next to last equality holds because (x1 + 1) + x2 = (x2 + 1) +
1. Prove parts (4) and (7) of Theorem 9.5:
For all a, b and c in Q+ :
(4) (ab)c = a(bc);
(7) (a + b)c = ac + bc.
Proof. In both parts, we let a = [a1 , a2 ], b = [b1 , b2 ] and c = [c1 , c2 ].
(4) We compute, using Theorem 8.20,
(ab)c = [a1 b1 , a2 b2 ][c
6.1. Prove that if a poset has the greatest lower bound property, it has the
least upper bound property.
Proof. Suppose X is a poset with the greatest lower bound property. Let A
be a non-empty subset of X that is bounded above. Then the set B of upper
bo
3.4. Let R be a relation on a set X . Prove that R is transitive i R R is a
subset of R.
Proof. First suppose that R is transitive, and suppose that (x, z ) R R.
Then there is some y X with xRy and yRz . By transitivity, xRz , or
(x, z ) R. Thus R R R.
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