Additional Problem Set 3 (Sets) Solutions

1
.
Let A = {x∈N:3≤x≤13}, B = {x∈N:x is even}, and C = {x∈N:x is Odd}
1
.
1
.
A⋂B ={4,6,8,10,12}
1
.
2
.
A⋃B = {x∈: x is even or 3≤x≤13}
1
.
3
.
B⋂C= ∅
1
.
4
.
B⋃C=
2
.
The power set of {1,2,3,4} is:
{{},{1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}}
3
.
No,
Assume there are such sets A and B we proceed by contradiction
|A⋃B| = |A| +|B| -|A⋂B|,
so if |A| = |B|, and |A⋂ B|=5then
|A⋃B]=2|A| -5
which is an odd number so it cannot be 10.
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4
.
Let A = {2, 4, 6, 8}. Suppose B is a set with |B| = 5.
4
.
1
.
The largest value of |A⋃B| is 9. We can get this if A and B are disjoint. The least value is 5 if A⊆B.
4
.
2
.
The largest value of |A⋂B| is 4, which we can get if A⊆ B. The least value is 0 if A and B are disjoint.
4
.
3
.
The largest value of |AB| is 20, the smallest is 20. Cartesian product size only depends on the size of the sets.
5
.
Let A, B, and C be sets:
5
.
1
.
Yes. Let x∈A, then by A⊆B we have that x∈B. Then by B⊆C, we have x∈C.
5
.
2
.
No, we show this with a counter example:
Let A={1}, B={{1}}, C={{{1}}}. Note that {1} is not an element of C. The only element of C is a set containing {{1}}.
6
.
Because some cards are BOTH red and face cards. Let C = the set of cards in a deck. F = the set of face cards in a deck, and R the set of red cards in a deck. So even though |R|=26 and |F|=12 we have that |R⋃F| ≤ |R| + |F|. In fact |R⋃F| = |R|+|F|-|R⋂F| , and the intersection has size 6.
7
.
We proceed by cases:Assume A is a set, and proceed by contradictionWe have two case, either |A| ≠2 or |A|=2 If |A|≠2, then A= {2,|A|} so |A|=2, which is a contradiction. If |A|=2, then A={2,2}={2} so |A|=1, which is a contradiction.