Mathematics 220 Workshop 1 - Set Theory
Note: Your goal is to practice working with notation and definitions. Some problems will
require knowledge of definitions - as you encounter a definition that you have to look up, then
do so and immediately learn it by heart.
1. Which of the following sets are subsets of others on the list? Consider all 12 possibilities.
• A = {n ∈ N : n is odd}
• B = {n ∈ N : n is prime}
• C = {4n + 1 : n ∈ N}
• D = {x ∈ R : x2 − 39x + 74 = 0}
Solution:
• A 6⊆ B, since 9 is odd but not prime (9 ∈ A but 9 6∈ B.)
• A 6⊆ C, since 7 is odd but 7 = 4n + 1 ⇒ n =
3
2
∈
/ N (7 ∈ A but 7 6∈ C.)
• A 6⊆ D, since 1 is odd but 12 − 39(1) + 74 6= 0 (1 ∈ A but 1 6∈ D.)
• B 6⊆ A, since 2 is prime but not odd (note this is the only counterexample.)
• B 6⊆ C, since 7 is prime but 7 = 4n + 1 ⇒ n =
3
2
∈
/ N.
• B 6⊆ D, since 3 is prime but 32 − 39(3) + 74 6= 0.
• C ⊆ A, since m ∈ C ⇒ ∃m ∈ N such that m = 4n + 1 = 2(2n) + 1, ⇒ m is
odd.
• C 6⊆ B, since 9 = 4(2) + 1, but 9 is not prime.
• C 6⊆ D, since 5 = 4(1) + 1, but 52 − 39(5) + 74 6= 0.
• D 6⊆ A, since 22 − 39(2) + 74 = 0, but 2 is not odd.
• D ⊆ B, since x ∈ D ⇒ x2 −39x+74 = 0 ⇒ (x−2)(x−37) = 0 ⇒ x ∈ 2, 37 ⊂ B,
since 2 and 37 are prime.
• D 6⊆ C, since 22 − 39(2) + 74 = 0, but 2 = 4n + 1 ⇒ n =
the only counterexample, since 37 = 4(9) + 1.)
1
4
∈
/ N (note this is
• A ⊆ A, since a ∈ A ⇒ a ∈ A. Similarly, B ⊆ B, C ⊆ C, D ⊆ D.
2. Let
A = {n ∈ N : n = 4k for some k ∈ N, 1 ≤ k ≤ 10},
B = {n ∈ N : n = 3k for some k ∈ N, 1 ≤ k ≤ 30}.
Calculate the following:
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Mathematics 220 Workshop 1 - Set Theory
(a) |A| and |B|
(b) |A ∩ B|
(c) |A ∪ B|
(d) Put together your thought process for items (c), and (d) and guess a formula for
|A ∪ B| in terms of |A|, |B|, and |A ∩ B|. Draw Venn Diagrams to help you.
(e) Guess a similar formula for three sets.
Solution: (a). Since there are 10 integers k satisfying 1 ≤ k ≤ 10, there are also
10 elements of the form 4k, made from these values of k. Thus, |A| = 10. Similarly,
|B| = 30.
(b) Note that the set A could be written differently:
A = {n ∈ N : n ≤ 40, and 4 divides n}.
Similarly,
B = {n ∈ N : n ≤ 90, and 3 divides n}.
(make sure you understand why this is true). Then A ∩ B consists of the natural
numbers that are divisible both by 3 and by 4, and they have to be not greater than
40 (and not greater than 90 as well, but this becomes redundant). We will prove
later (but it is probably obvious to you now) that if an integer is divisible both by
3 and 4, then it is divisible by 12. There are only 3 natural numbers divisible by 12
that are not greater than 40: they are 12, 24, and 36. Then A ∩ B = {12, 24, 36} and
|A ∩ B| = 3. (Note: you could also have just counted the numbers that are both in
A and B).
(c) and (d). One can solve (c) by counting, but this is tedious. Let us think about
it instead. As we go through the first few elements of A and B, they are all different
numbers, and so it is temping to say that |A ∪ B| = |A| + |B| (since you put together
all elements of A and all elements of B). This is wrong though, because when we get
to the number 12, it would appear both in A and in B, and this when we compute
|A| + |B|, we would count it twice, but it should only be counted once as an element
of A∪B. This is a very common situation, called overcounting: we have conted every
element of A ∩ B twice instead of counting it once. Then
|A ∪ B| = |A| + |B| − |A ∩ B|.
This is a general formula (and the proof we gave does not refer to any special properties of A and B, so it is general). In our special case, we get |A∪B| = 10+30−3 = 37.
(e) Answer:
|A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|.
To prove this formula, again start by counting all the elements of A, B, C, and then
think of which ones were counted at least twice? The answer: the ones in A ∩ B,
A∩C, and B ∩C. So we should subtract the cardinalities of these sets. Now, however,
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Mathematics 220 Workshop 1 - Set Theory
the elements of A ∩ B ∩ C have not been counted at all: they were at first counted 3
times, and then subtracted 3 times, so now they are not accounted for, and we should
add their number back.
This formula is called the inclusion-exclusion formula. If you are using Venn diagram
for illustrations, you should count the number of circles covering a particular area,
and it will help you understand this formula. In particular, here the elements of
A ∩ B ∩ C are covered by all 3 circles, so they were counted 3 times at first.
A
C
B
The areas with a single colour represent the elements counted just
once when you compute |A|+|B|+|C|, the areas with two colours
represent the elements counts twice, and the intersection of all three
sets has three colours, so it is initially counted three times, but then it
also gets subtracted three times.
.
3. Let A and B be subsets of a universal set U. Prove that A ∪ B = A ∩ B.
Do this by showing that each side is a subset of the other side. A Venn diagram may help
convince you this is true but is not a proof.
Solution: We need to prove two inclusions of sets: A ∪ B ⊆ A ∩ B, and A ∪ B ⊇
A ∩ B. Both will be proved using the definition of containment of sets. You always
have to prove these two statements separately when proving equality of sets.
Let us prove the first statement first. We need to show that each element contained
in A ∪ B is also contained in A ∩ B. To talk about an element, we need to give it a
name: call it x. So, we start with: Let x ∈ A ∪ B. We need to prove that x ∈ A ∩ B.
Proof: Since x ∈ A ∪ B, by definition of the complement, x ∈
/ A ∪ B. This means,
and x ∈
/ A and x ∈
/ B, by definition of the union of sets. Thus, x ∈ A and x ∈ B (by
definition of the complement). Then x ∈ A ∩ B by definition of the intersection.
Now we need to prove the converse: A∩B ⊆ A ∪ B. Let x ∈ A∩B. Then by definition
of the intersection, x ∈ A and x ∈ B. Then by definition of the complement, x ∈
/A
and x ∈
/ B. But then x ∈
/ A ∪ B, which means that x ∈ A ∪ B, and the proof is
completed.
Sets in geometry - practice using a picture or a graph to help you solve the problem.
However, keep in mind: a picture is not a solution! Mainly, it is a guide for yourself to
obtain intuition about the solution
4. Let C be the circle of radius 1 centered at the origin in the xy-plane. Express this set in
terms of its points (x, y), and some property p(x, y).
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Mathematics 220 Workshop 1 - Set Theory
Solution: This circle is described by the equation x2 + y 2 = 1, so the points of the
circle are given by
C = {(x, y) ∈ R2 : x2 + y 2 = 1}
5. Let a, b, c, d be real numbers with a < b < c < d. Express the set [a, b] ∪ [c, d] as the
difference of two sets.
Solution: the answer is [a, d] − (b, c).
Power sets
6. Compute the power set P ({{1}, 1, 2})
Solution: The answer is:
P ({{1}, 1, 2}) = {∅, {{1}}, {1}, {2}, {{1}, 1}, {{1}, 2}, {1, 2}, {{1}, 1, 2}} .
(Pay close attention to the braces).
7. Binomial coefficients and the power set. Let A be a set of cardinality n. In this
question we get to see how the number of k-element subsets of A adds up to 2n as k runs
through the numbers from 0 to n.
(a) Warm-up: let A = {1, 2, 3, 4, 5}. How many 3-element subsets does it have?
Solution: Note that to make a 3-element subset we need to select three numbers
from the set {1, 2, 3, 4, 5}; which is equivalent to sleeting the two numbers that
we do not pick into our subset. So let us compute the number of ways to select
two elements that would be “left behind”. There are 5 options for the first one,
and 4 options for the second one (after we picked one number, only 4 remain).
So it seems that the answer should be 5 × 4 = 20. However, this way we have
counted every subset twice: for example, the set {2, 3} was counted once when
we first selected 2 and then selected 3, and then again, when we first selected 3
and then selected 2. Thus, we have overcounted by a factor of 2, and the correct
answer is 10. You could also just list all the possible 3-elements subsets to see
that this is true.
(b) Consider the expressions
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2 b + 3ab2 + b3
...
(a + b)n = an + C1 an−1 b + C2 an−2 b2 + · · · + Cn−1 abn−1 + bn .
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Mathematics 220 Workshop 1 - Set Theory
We are interested in these coefficients C1 , . . . Cn−1 (we can also write C0 = Cn = 1),
called binomial coefficients. We do not yet know a formula for these coefficients,
but we already can prove something about them!
Prove that C0 + C1 + · · · + Cn = 2n .
Hint: consider the expression (1 + 1)n .
Solution: By definition of the coefficients, when we plug in a = b = 1 into
the formula (a + b)n = an + C1 an−1 b + C2 an−2 b2 + · · · + Cn−1 abn−1 + bn , we get
2n = 1 + C1 · 1 + C2 · 1 + . . . Cn−1 · 1 + 1 = C0 + · · · + Cn .
(c) Prove that for every k = 0, 1, ...n, the coefficient Ck equals the number of k-element
subsets of A.
Solution: First consider small n and small k, and convince yourself that this
is true: for k = 0, there is only one subset with 0 elements: it is the empty
set. For k = 1, there are n subsets: {1}, {2}, ... {n}. Now let us consider the
expression
(a + b)n = (a + b)(a + b)(a + b)....(a + b).
Let us number these factors: the first (a + b), the second (a + b), and so on, all
the way to the n-th (a + b). Now, when we are multiplying it all out, from each
(a + b) we get to pick either a or b. If we get the product ak bn−k as a result of
this picking of a’s and b’s, then it means that we picked a precisely k times, and
we picked b the rest of the time. Thus, the coefficient Ck is the number of ways
to select k a’s (and the rest n − k factors then have to be b’s). Let us mark the
factors from which we picked a. We get a k-element subset of A = {1, . . . , n}.
Thus we have matched up the ways to pick k factors a, and k-element subsets
of A. Then there are precisely Ck subsets of k elements.
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