Mat 220S
SOLUTIONS: Extra practice problems for Exam 2
1. Outline a proof for each theorem. If there is a standard form of argument for moving the
outline further inward, in either direction, use it. Don’t give reasons, just show the statements
which would form the outline of the proof.
Let A, B, C, D represent sets.
(a) Theorem: A  B
Proof: Let x be an arbitrary element of A.
...
...
Therefore, x is in B.
(b) Theorem: x   Ai
iI
Proof: Cite a specific element j in I.
...
...
Deduce that x is in A j .
(c) Theorem: (A  B)  C
Proof: Let x be an arbitrary element of A union B.
By definition, x is in A or x is in B.
Case 1: x is in A.
.....
.....
So, x is in C.
Case 2: x is in B.
.....
.....
So, x is in C.
In either case, x is in C.
(d) Theorem: A = B.
Proof: First, assume x is an arbitrary element of A.
...... deduce x is in B.
Conversely, assume x is an arbitrary element of B.
...... deduce x is in A.
(e) Theorem: A  B.
Proof: Find a specific element x in A which is not in B, OR a specific element x in B which
is not in A.
(f) Theorem: The empty set is a subset of A.
Proof: For the sake of contradiction, assume phi is not a subset of A.
.....
Contradiction! Therefore, phi is a subset of A.
(g) Theorem: x is in (A\B)c.
Proof: We need to prove that x is not in both A and Bc.
That is, we must prove that either x is not in A or x is in B.
(How we do this, exactly, will depend on what we know about x.)
....
....
Therefore, either x is not in A or x is in B.
2. Prove with an element argument: For all sets A and B, ( A  B) c  Ac  B c .
Proof:
First, let x be an arbitrary element of ( A  B) c .
So x is not in A intersect B.
Equivalently, either x is not in A or x is not in B.
Case 1: x is not in A.
Then x is in the complement of A.
This implies x is in Ac  B c .
Case 2: x is not in B.
Then x is in the complement of B.
This also implies x is in Ac  B c .
In either case, x is in Ac  B c .
Since x was an arbitrary element of ( A  B) c , this proves ( A  B) c  Ac  B c .
Conversely, let x be an arbitrary element of Ac  B c .
By definition, x is in A complement or x is in B complement.
Case1: x is in A complement.
That is, x is not in A.
Since x is not in both A and B, x is not in A intersect B.
Thus, x is an element of ( A  B) c .
Case 2: x is in B complement.
Similarly to case 1, since x is not in B, x cannot be in A intersect B.
Thus, again, x is an element of ( A  B) c .
In either case, x is in ( A  B) c .
Since x was an arbitrary element of Ac  B c , this proves Ac  B c  ( A  B) c .
We have shown containment in both directions, so the two sets are proven to be equal.
3. Use definitions and negation rules to translate the following into statements involving
quantifiers and/or logical connectives. Leave no set operations except , . Do not leave a
negation as a prefix for a compound sentence. (A, B etc. denote sets.)
a. y  C  D if and only if y is not in C or y is not in D
b. y   Bn if and only if for some n in N, y is in Bn.
nN
c. z  A \ D if and only if z is in A and z is not in D.
d. z   C i if and only if for all i in I, z is in Ci.
iI
e. A  P(S ) if and only if for all x in A, x is in S.
f. B is not a subset of A if and only if there exists an x in B such that x is not in A.
4. Let A  {1, 2, 3}, B  {2, 4, 5} . Find each of the following.
(a) A  B = {1,2,3,4,5}
(b) A  B = {2}
(c) A \ B = {1, 3}
(d) P( A) = {  ,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}} (8 elements)
(e) P ( ) = {  } (1 element)
(f) If the universe is all natural numbers, the complement of A = {4,5,6,7,...}
(g) If the universe is all natural numbers, describe P( A c ) in plain English: all subsets of the
natural numbers which do NOT contain 1, 2, or 3.