MATH 2200 Learning Celebration #2 Solutions
Thomas Goller
February 12, 2013
General Comments:
• To prove that two sets are equal, you should almost always prove both containments,
namely (⊆) and (⊇). This was the easiest way to prove Claims 1, 3, and 6. If you
don’t understand what this means, then look at the solutions below, and come talk
to me if it still doesn’t make sense. This method of proving set equality is essential –
without it, you’ll have a hard time writing a rigorous proof that two sets are equal!
• The symbols (⊆) and (⊇) do not mean the same thing as ( =⇒ ) and ( ⇐= ).
• Labeling the elements of a set (as in “Write A = {a1 , a2 , . . . , a|A| }”) is completely
useless except in the case when A is finite. That’s because it is impossible to label the
elements of a general (infinite) set! My apologies for making the method of labeling
seem better than it is!
• Do not ever try to prove a statement by assuming the conclusion is true and then trying
to deduce something true. That doesn’t work! For instance, starting with 1 = 2, which
is false, I can multiply both sides by 0 and obtain the true statement 0 = 0, but that
doesn’t make 1 = 2 any less false! For Claim 1, don’t assume A = (A − B) ∪ (A ∩ B)
and then show that A = A. That doesn’t prove anything!
• To give a counterexample, define any relevant sets and functions (“Let A = {ψ, φ},
...”; you get to choose them!), and briefly state why the assumptions of the claim are
met and why the conclusion fails.
• Surjective and injective are not opposites! They are simply two important types of
functions, which together make a function invertible (as we will see this week).
• The empty set is denoted ∅. This is not the same as {∅}, which is a set containing the
empty set as an element!
• Stick to the notation we introduced in class. For instance, writing A − C is better than
writing A ∩ C c because C c does not make sense unless you specify in which “universal”
set you are taking the complement.
• The maximum score was 60. The mean was 34 and the median was 32. The highest
score was 60.
1
Problems
Claim 1. Let A and B be sets. Then A = (A − B) ∪ (A ∩ B).
Proof. (⊆): Let a ∈ A. Then either a ∈ B or a ∈
/ B. If a ∈ B, then a ∈ A ∩ B, so
a ∈ (A − B) ∪ (A ∩ B). If a ∈
/ B, then a ∈ A − B, so a ∈ (A − B) ∪ (A ∩ B).
(⊇): Let a ∈ (A − B) ∪ (A ∩ B). Then a ∈ A − B or A ∈ A ∩ B, namely (a ∈ A and
a∈
/ B) or (a ∈ A and a ∈ B). In each case, a ∈ A.
f
! B and B
Claim 2. Let A, B, and C be sets, and let A
is surjective and f is injective, then g is injective.
g
C be functions. If g ◦ f
!
This claim is false! Counterexample: Let A = {λ}, B = {α, β}, and C = {ξ}. Let
f
g
! B be defined by λ +→ α, and let B
! C be defined by α +→ ξ and β +→ ξ. Then
A
f is injective and g ◦ f is surjective, but g is not injective since g(α) = g(β).
Claim 3. Let A, B, and C be sets. Then (A − C) ∪ (B − C) = (A ∪ B) − C.
Proof. (⊆): Let a ∈ (A − C) ∪ (B − C). Then a ∈ A − C or a ∈ B − C. If a ∈ A − C, then
a ∈ A and a ∈
/ C. Then a ∈ A ∪ B, so a ∈ (A ∪ B) − C. If a ∈ B − C, then a ∈ B and
a∈
/ C. Then a ∈ A ∪ B, so a ∈ (A ∪ B) − C.
(⊇): Let a ∈ (A ∪ B) − C. Then a ∈ A ∪ B and a ∈
/ C. So a ∈ A or a ∈ B. If a ∈ A,
then a ∈ A − C. If a ∈ B, then a ∈ B − C. So a ∈ (A − C) ∪ (B − C).
Claim 4. Let A, B, and C be sets. Then A ∪ (B ∩ C) = (A ∩ B) ∪ (A ∩ C).
This claim is false! Let A = {Zugzwang}, B = ∅, and C = ∅. Then A ∪ (B ∩ C) =
{Zugzwang}, but (A ∩ B) ∪ (A ∩ C) = ∅.
Claim 5. Let A, B, and C be sets, and let A
is surjective, then g is surjective.
f
!
B and B
g
!
C be functions. If g ◦ f
Proof. Let c ∈ C be arbitrary. Since g ◦ f is surjective, there exists an a ∈ A such that
(g ◦ f )(a) = c, namely g(f (a)) = c. Then g maps f (a) to c, so g is surjective.
Claim 6. Let A, B, and C be sets. Then A − (B ∪ C) = (A − B) ∩ (A − C).
Proof. (⊆): Let a ∈ A − (B ∪ C). Then a ∈ A and a ∈
/ B ∪ C, so a ∈
/ B and a ∈
/ C. Thus
a ∈ A − B and a ∈ A − C, so a ∈ (A − B) ∩ (A − C).
(⊇): Let a ∈ (A − B) ∩ (A − C). Then a ∈ A − B and a ∈ A − C, so a ∈ A, a ∈
/ B, and
a∈
/ C. So a ∈
/ B ∪ C, which implies a ∈ A − (B ∪ C).
2