## Sample Exam II Questions

Click on question number to see the answer

The following definition applies to the first 4 questions.

We define the sum of two cardinal numbers a and b as follows:
If A is a set with |A| = a and B is a set, disjoint from A, with |B| = b then we define

a + b = | AB |.

1. Show that 2 + 0 = 0.

2. Show that 0 + 0 = 0.

3. A famous theorem (the Cantor-Schröder-Bernstein Theorem) states that if there exists an injection from A to B and another injection from B to A, then A and B are equivalent. Use this result and the fact that the open interval (0,1) is equivalent to to prove that (0,1) is equivalent to. (i.e., c + 0 = c where c = | |.)

4. Prove that c + c = c.

5. If S is a finite set, prove that every subset of S is a finite set. [You may use the fact that every subset of one of the a special sets, m, is a finite set.

6. Discuss the meaning and implications of the statement 1 = c.

7. If A and B are denumerable sets, show that A × B is a denumerable set.

Give it an "A" if the statement and proof are correct.
Give it a "C" if the statement is correct but the proof is wrong.
Give it an "F" if the statement and the proof are wrong.

Statement: If f,g and f-1 are functions on A, then g = f-1o (g o f).

Proof: f-1o (g o f) = f-1o (f o g) by commutativity,
= (f-1 o f) o g by associativity,
= IA o g (the composition of a function and its inverse function is the identity),
= g ( the identity composed with anything is the anything).
Q.E.D.

9. Let f: be given by f(x) = [2x + 1] . ( [ .. ] denotes the floor function, i.e., round down to the nearest integer, so for example: [4.325] = 4, [3.999] = 3, [-2.34] = -3, [16] = 16 ).

a) Show that f is onto .
b) Show that f is not a one-one function.

10. Given that f: A B is one-to-one and that g: B C is one-to-one, prove that g o f : A C is one-to-one.

11. The old rules for forming a proper 3-digit area code were:
a) The 3 digit number can not start with a 0 or a 1.
b) The middle digit must be a 0 or a 1.
How many different area codes are possible under these rules?

Let A = {a, b} and B =. We will show that AB is denumerable. To do this we need to find a bijection f: AB. Define f(1) = a, f(2) = b and f(i) = i - 2 for all i > 2. It is easy to see that f is onto AB since a comes from 1, b comes from 2 and any natural number n comes from n + 2. It is also easy to see that no two natural numbers can be mapped to the same place, so f is one-to-one. The function f is thus a bijection and so, AB is denumerable.

Let A = {even natural numbers} and B = {odd natural numbers}. The functions f:A given by f(n) = 2n and g:B given by g(n) = 2n - 1, are both bijections, so A and B are denumerable. AB = , thus proving the statement.

Consider the identity map f: (0,1) given by f(x) = x. This is clearly an injection. Since (0,1) is equivalent to there exists a bijection g:(0,1). Now change the codomain of this function to (0,1). The function g is no longer a bijection since it is now not onto, but it is still one-to-one. Having found one-to-one maps in both directions, by the Cantor-Schröder-Bernstein theorem, the two sets are equivalent.

Since every interval of real numbers (open, closed, half-open) is equivalent to we can take A = (0, 1) and B = [1, 2) and these disjoint sets both have cardinality c. Their union is the interval (0, 2) which also has cardinality c, proving the result.

Since S is a finite set, it is equivalent to some m by a bijection f. Let A be any subset of S. B = {f(x) | x A} is a subset of m, which we know is a finite set. This means that there is a bijection g: Bk, for some k. Now, the composition g o f maps A to k, and being the composition of two bijections, it is itself a bijection. Thus, A is a finite set.

This statement is known as the continuum hypothesis. It claims that the second largest infinite cardinal number is c, the cardinality of the reals. The continuum hypothesis is known to be independent of the Zermelo-Frankel set theory axioms. That is, it can not be proved or disproved from these axioms. One may assume that it is true, i.e., take it as an additional axiom of set theory, and obtain what is now called "standard" set theory, or assume that it is not true and obtain a "non-standard" set theory. Neither assumption leads to a contradiction.

Recall that the set x is denumerable, as the function f:x given by f(m, n) = 2m-1(2n - 1) is a bijection. Since A is denumerable there is a bijection g: A and since B is denumerable there is a bijection h: B. Now, define a function j: A x B x by j(a, b) = (g(a), h(b)). If we show that j is a bijection, then the map f o j : A x B will be a bijection, thus proving that A x B is denumerable. To show that j is onto, let (n, m) be an arbitrary element of x. Since g is onto, there exists an aA so that g(a) = n. Since h is onto, there exists a bB so that h(b) = m. Therefore, j(a, b) = (g(a), h(b)) = (n, m) and so, j is onto. Now, suppose that j(a, b) = j(c, d). Then, (g(a), h(b)) = (g(c), h(d)). But this means that g(a) = g(c) and h(b) = h(d). Since both g and h are injections, we see that a = c and b = d, i.e., (a, b) = (c, d) and so, j is one-to-one.

F. In the first line of the proof, composition of functions is not generally commutative. The statement is false, for instance, with f(x) = x + 1 and g(x) = x2, we would have f-1 = x - 1 and f-1o (g o f) = (x + 1)2 - 1 = x2 + 2x, which is not g(x).

a) Let y. For x = (y-1)/2 we have f(x) = f((y-1)/2) = [ 2(y-1)/2 + 1] = [y] = y (since y is an integer.) Thus, any element in the codomain is in the range and f is onto.

b) f(.3) = 1 and f(.4) = 1, so f is not a one-one function.