## Sample Final Exam 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. Define the following terms (do not just indicate notation):

1. Given a set S with a partial order R, and a subset B of S, an upper bound of B is ...

2. A partition of a set S is ...

3. A function f: A B is a ...

4. A relation R on a set S has the Transitive Property if ...

5. A partial order on a set S is a ...

6. The Power Set of a set A is ...

7. A relation R on a set S is said to be antisymmetric if ...

8. A relation R on a set S is reflexive if ...

9. Given an equivalence relation R on a set S, the equivalence class containing x S is ...

10. The binomial coefficient C(n,k) counts the number of ...

8. Let A. Prove that sup(A) is unique if it exists.

9. Let A and B be subsets of . Prove that if inf(A) and inf(B) exist, then inf(AB) exists and inf(AB) = min{inf(A), inf(B)}.

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.

Claim: If the relation R is symmetric and transitive, it is also reflexive.

"Proof": Since R is symmetric, if (x,y)R, then (y,x)R. Thus (x,y) R and (y,x)R, and since R is transitive (x,x) R. Therefore, R is reflexive.

11. Let A be a set. Consider the partial order on P(A), the power set of A. If C and D are subsets of A, prove that the least upper bound of {C,D} is C D.

12. Let f:{0,1,2,3} be defined by f(x) = remainder after dividing x by 4, and define a relation, R, on by x R y means f(x) = f(y). Show that R is an equivalence relation, and describe the elements of the equivalence class which contains 5.

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.

1. Given a set S with a partial order R, and a subset B of S, an upper bound of B is an element u of S such that b R u for all elements b of B.

2. A partition of a set S is a collection of non-empty subsets of S whose union is all of S and which are mutually disjoint.

3. A function f: A B is a relation between A and B which has the property that for each element a in A there is a unique ordered pair in f having a as its first coordinate.

4. A relation R on a set S has the Transitive Property if whenever (x,y) and (y,z) are in R, (x,z) is in R for every x, y and z in S.

5. A partial order on a set S is a relation on S which is reflexive, antisymmetric and transitive.

6. The Power Set of a set A is the set of all subsets of A.

7. A relation R on a set S is said to be antisymmetric if whenever x R y and y R x then x = y.

8. A relation R on a set S is reflexive if for every x in S, x R x.

9. Given an equivalence relation R on a set S, the equivalence class containing x S is the set of all y in S so that (x,y) is in R.

10. The binomial coefficient C(n,k) counts the number of k element subsets of an n element set.

Suppose that x and y are both least upper bounds of the set A. Since x is a least upper bound and y is an upper bound, we have that xy. Since y is a least upper bound and x is an upper bound, we have that yx. This implies that x = y. [ If x y, then the two conditions say that x < y and y < x, which contradicts the law of trichotomy.]

Every element of AB is either in A or in B and so is greater than or equal to either inf(A) or inf(B), and so, clearly greater than or equal to the smaller of these two numbers. Thus, AB is bounded from below and inf(AB) exists. If min{ inf(A), inf(B)} is not the greatest lower bound of the union, then there exists a lower bound which is larger than it, call it L. Now L must be larger than at least one of inf(A) or inf(B), and since these are the greatest lower bounds for their sets, there will be an element in A which is less than L if L is greater than inf(A), or an element in B which is less than L if L is greater than inf(B). In either case, there is an element of the union which is less than L, so L can not be a lower bound of the union. This contradiction proves that min{ inf(A), inf(B)} is the greatest lower bound of the union.

Grade = F. The statement is false. Consider the set {a, b, c} and the relation {(a, a), (b, b), (a, b), (b, a)}. This relation is symmetric and transitive, but it is not reflexive since (c, c) is missing. The problem with the "proof" is that it is never shown that an ordered pair exists in the relation which has an arbitrary element of the set as a first coordinate.