HomeWork Answers
Section 2.1
7. Determine whether these statements are true or false:
a) 0  
False – the empty set has no elements
b)   {0}
False – the empty set is a subset of {0}, but is not an element of it.
c) {0}  
False – no set can be a proper subset of the empty set since, by
definition, that would require the empty set to contain at least one element.
d)  {0}
True – the empty set is a subset of every set, so  {0}. In addition the
set {0} has one element, which is not contained in the empty set.
e) {0}  {0}
False – the set {0} is a subset of itself, but is not an element of itself.
f) {0}  {0}
False –{0}  {0} since {0}  {0}. But, as noted above, the proper subset
relation requires the larger set to have at least one element not in the other
one.
g) {0}  {0}
True, as noted above. By Theorem 1, every set is a subset of itself.
8. Determine whether these statements are true or false:
a)   {
True – the set { has exactly one element, namely .
b)   {
True, the empty set is one of the two elements of that set.
c) {  {
False – this is just like 7(e). A set is a subset of itself, but is not an
element of itself.
d) {  {{
True – here the set which has the empty set as its only element is, in
turn, the only element of the set on the right.
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e) {  {{
True – the set on the left has the empty set as its only element and the
empty set occurs as an element of the set on the right, which also contains a
second element, namely {}.
f) {{  {{
True – the set on the left has {} as its only element, which occurs as
one of the two elements of the set on the right.
g) {{  {{{
False – although it appears that the set on the right has cardinality of 2,
it has, in fact, cardinality of 1, since the same element occurs in it twice. Thus
we see that the set on the left and the set on the right are equal. So neither
can be a proper subset of the other.
9. Determine whether these statements are true or false:
a) x  {x
True – the set {x has exactly one element, namely x.
Note similarity to 8(a).
b) {x}  {x
True, every set is a subset of itself.
c) {x  {x
False – this is just like 7(e). A set is a subset of itself, but is not an
element of itself.
d) {x  {{x
True – here the set which has x as its only element is, in turn, the only
element of the set on the right.
e)   {x
True – the empty set is a subset of every set.
f)   {x
False, the only element of the set {x} is x.
15. Let A  B and B  C. Show that A  C.
By definition 4, from the first statement we have:
x (x  A  x  B)
x (x  B  x  C)
x (x  A  x  C)
and, again by definition, A  C.
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16. Find two sets A and B such that A  B and A  B.
Let:
A=
B = {}
Then, by exercise 8(a), A  B
And A  B, since by Theorem 1,  is a subset of every set.
19. Find the power set of each set, where a and b are distinct elements.
a) P({a}) = {, {a}}
b) P({a, b}) = {, {a}, {b}, {a, b}}
c) P({, {}}) = {, {}, {{}}, {, {}}}
Note, by comparison:
P({a, {a}}) = {, {a}, {{a}}, {a, {a}}}
20. Show: If P(A) = P(B), then A = B.
Proof: By contradiction
Let:
P(A) = P(B), but A  B.
Then:
There is an x such that x  A but x  B, or vice versa.
Suppose:
x  A but x  B.
Then:
{x}  P(A) but {x}  P(B).
But this implies P(A)  P(B). 
A similar result would occur if we assumed x  B but x  A.
21. How many elements does each set have, if a and b are distinct elements?
a) P({a, b, {a,b}})
Since the set has 3 elements, the power set has 23 = 8.
b) P({, a, {a}, {{a}}})
Since the set has 4 elements, the power set has 24 = 16.
c) P(P())
Since the empty set has 0 elements, P() has20 = 1 element.
And since P() has 1 element, P(P()) has 21 = 2 elements.
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This is also verified in Example 14, p. 117 where we have:
P() = {}, and
P({) = {}.
22. Determine whether each of these sets is the power set of a set, where a and b are
distinct elements.
a) 
No, A power set cannot be empty, since it includes at least one element
– the empty set itself.
b) {, {a}}
Yes, P({a}) = {, {a}}.
c) {, {a}, {, {a}}
No, this set has cardinality of 3, which is not a power of 2.
d) {, {a}, {b}, {a,b}}
Yes, P({a, b}) = {, {a}, {b}, {a,b}}.
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