MAT235 Fall 2015 - Assignment 6
Section 3.4 - Proofs Involving Conjunctions and Biconditionals
3.4.3 . Suppose A ⊆ B. Prove that for every set C, C \ B ⊆ C \ A.
Givens
A⊆B
Goal
C \B ⊆C \A
Expanding using the definition of subset gives
Givens
∀x(x ∈ A → x ∈ B)
Goal
∀x(x ∈ (C \ B) → x ∈ (C \ A))
We can now let x be arbitrary in both the given and goal and bring the left
hand side of the goal to the givens
Givens
x∈A→x∈B
x∈C \B
Goal
x∈C \A
Finally by set difference we have
Givens
x∈A→x∈B
x ∈ C ∧ x 6∈ B
Goal
x ∈ C ∧ x 6∈ A
Separating the and statements gives
Givens
x∈A→x∈B
x∈C
x 6∈ B
Goal
x∈C
x 6∈ A
Thus we will prove each goal separately. First, since x ∈ C the first goal is
clearly satisfied. Second, since x 6∈ B by modus tollens since x ∈ A → x ∈ B
gives x 6∈ A which proves the second goal.
Formally,
Theorem. Suppose A ⊆ B. Then for every set C, C \ B ⊆ C \ A.
Proof. Let x be an arbitrary element with x ∈ C \ B. Then x ∈ C and x 6∈ B.
Since x 6∈ B and A ⊆ B we know that x 6∈ A. But then since x ∈ C and x 6∈ A
we can conclude that x ∈ C \ A and because x was arbitrary that C \ B ⊆ C \ A.
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3.4.17 . Suppose F and G are families of sets.
a. Prove that ∪(F ∩ G) ⊆ (∪F) ∩ (∪G).
Givens
F and G are families
Goal
∪(F ∩ G) ⊆ (∪F) ∩ (∪G)
Expanding using subset and letting x be arbitrary gives
Givens
F and G are families
x ∈ ∪(F ∩ G)
Goal
x ∈ (∪F) ∩ (∪G)
Which by the definition of intersection and family union gives
Givens
F and G are families
∃A ∈ (F ∩ G)(x ∈ A)
Goal
∃B ∈ F(x ∈ B) ∧ ∃C ∈ G(x ∈ C)
Letting A = A0 in the given and separating the goal gives
Givens
F and G are families
(A0 ∈ (F ∩ G)) ∧ (x ∈ A0 )
Goal
∃B ∈ F(x ∈ B)
∃C ∈ G(x ∈ C)
Finally applying the definition of intersection and separating the givens gives
Givens
F and G are families
A0 ∈ F
A0 ∈ G
x ∈ A0
Goal
∃B ∈ F(x ∈ B)
∃C ∈ G(x ∈ C)
Now we can easily prove the first goal since x ∈ A0 and A0 ∈ F so letting
B = A0 implies that ∃B ∈ F(x ∈ B) or x ∈ ∪F. Similarly since x ∈ A0 and
A0 ∈ G letting C = A0 we know that ∃C ∈ G(x ∈ C) or x ∈ ∪G. Since x ∈ ∪F
and x ∈ ∪G we know x ∈ (∪F) ∩ (∪G).
Formally,
Theorem. Suppose F and G are families of sets. Then ∪(F ∩G) ⊆ (∪F)∩(∪G).
Proof. Let x be an arbitrary element with x ∈ ∪(F ∩ G). Since x ∈ ∪(F ∩ G)
let A0 be a set such that x ∈ A0 and A0 ∈ F ∩ G. Then A0 ∈ F and A0 ∈ G.
Since x ∈ A0 and A0 ∈ F we have that ∃A ∈ F(x ∈ A) which gives x ∈ ∪F.
Similarly since x ∈ A0 and A0 ∈ G we have that ∃B ∈ G(x ∈ B) which gives
x ∈ ∪G. Because x ∈ ∪F and x ∈ ∪G we can conclude that x ∈ (∪F) ∩ (∪G)
and since x is arbitrary that ∪(F ∩ G) ⊆ (∪F) ∩ (∪G).
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b. What is wrong with the following proof that (∪F) ∩ (∪G) ⊆ ∪(F ∩ G).
Proof. Suppose x ∈ (∪F) ∩ (∪G). This means that x ∈ ∪F and x ∈ ∪G, so
∃A ∈ F(x ∈ A) and ∃A ∈ G(x ∈ A). Thus, we can choose a set A such that
A ∈ F, A ∈ G, and x ∈ A. Since A ∈ F and A ∈ G, A ∈ F ∩ G. Therefore
∃A ∈ F ∩ G(x ∈ A), so x ∈ ∪(F ∩ G). Since x was arbitrary, we can conclude
that (∪F) ∩ (∪G) ⊆ ∪(F ∩ G).
The problem with this proof is in the selection of the same existential instantiation A for ∃A ∈ F(x ∈ A) and ∃A ∈ G(x ∈ A). The existence statements
mean there are sets for each family, but they are not necessarily the same sets.
Thus if the element came from different sets in F and G then those sets would
not be in F ∩ G.
c. Consider the families F = {{1}} and G = {{1, 2}}. Then ∪F = {1}, ∪G = {1, 2}
and hence (∪F) ∩ (∪G) = {1}. But clearly F ∩ G = ∅ since F and G contain no
sets in common. This gives ∪(F ∩ G) = ∅ =
6 {1}.
Section 3.5 - Proofs Involving Disjunctions
3.5.2 . Suppose A, B, and C are sets. Prove that (A ∪ B) \ C ⊆ A ∪ (B \ C).
Givens
A, B, and C are sets
Goal
(A ∪ B) \ C ⊆ A ∪ (B \ C)
Applying the definition of subset to the goal gives
Givens
A, B, and C are sets
Goal
∀x(x ∈ (A ∪ B) \ C → x ∈ A ∪ (B \ C))
Letting x be arbitrary (universal instantiation) and bringing the left hand side
of the conditional to the givens gives
Givens
A, B, and C are sets
x ∈ (A ∪ B) \ C
Goal
x ∈ A ∪ (B \ C))
Expanding the given using the definition of set difference (separating the and)
and the goal using the definition of union then gives
Givens
A, B, and C are sets
x∈A∪B
x 6∈ C
Goal
x ∈ A ∨ x ∈ (B \ C))
Lastly we can expand the second given using the definition of set union giving
the disjunction
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Givens
A, B, and C are sets
x∈A∨x∈B
x 6∈ C
Goal
x ∈ A ∨ x ∈ (B \ C))
Now we will consider the two cases from the disjunction in the given:
Case 1: Suppose x ∈ A. Then clearly the first part of the goal is satisfied (and
x ∈ A ∪ (B \ C)).
Case 2: Suppose x ∈ B. Then since x 6∈ C gives x ∈ B \ C which is the second
part of the goal (and again x ∈ A ∪ (B \ C).
Formally,
Theorem. Suppose A, B, and C are sets. Then (A ∪ B) \ C ⊆ A ∪ (B \ C).
Proof. Suppose that x is arbitrary and x ∈ (A ∪ B) \ C. Then x ∈ A ∪ B and
x 6∈ C. Since x ∈ A ∪ B we know that either x ∈ A or x ∈ B so consider two
cases:
Case 1: Suppose x ∈ A. Then clearly x ∈ A ∪ (B \ C).
Case 2: Suppose x ∈ B. Then since x 6∈ C we have that x ∈ B \ C and thus
x ∈ A ∪ (B \ C).
Thus in either case x ∈ A ∪ (B \ C) and since x was arbitrary that (A ∪ B) \ C ⊆
A ∪ (B \ C).
3.5.27. Consider the following putative theorem.
Theorem? For any sets A, B, and C, if A\B ⊆ C and A 6⊆ C then A∩B 6= ∅.
Is the following proof correct? If so, what proof strategies does it use? If not,
can it be fixed? Is the theorem correct?
Proof. Since A 6⊆ C, we can choose some x such that x ∈ A and x 6∈ C. Since
x 6∈ C and A \ B ⊆ C, x 6∈ A \ B. Therefore either x 6∈ A or x ∈ B. But we
already know that x ∈ A, so it follows that x ∈ B. Since x ∈ A and x ∈ B,
x ∈ A ∩ B. Therefore A ∩ B 6= ∅.
The proof (and thus the theorem) is correct. First since A 6⊆ C means ∃x(x ∈
A ∧ x 6∈ C and thus it employs existential instantiation to assert that an x can
be chosen (although using x0 may have been a better choice of symbol). Then
it uses a modus tollens argument since x 6∈ C but A \ B ⊆ C (i.e. ∀x(x ∈
A \ B → x ∈ C) to assert that x 6∈ A \ B. Using the definition of set difference
and applying DeMorgan’s law then asserts that x 6∈ A ∨ x ∈ B. We can then
use a variant of modus ponens (since x 6∈ A ∨ x ∈ B ⇒ x ∈ A → x ∈ B and
x ∈ A) to assert that x ∈ B. Finally using the definition of intersection gives
that x ∈ A and x ∈ B means x ∈ A ∩ B and therefore A ∩ B 6= ∅.
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Section 3.6 - Existence and Uniqueness Proofs
3.6.6. Let U be any set.
a. Prove that there is a unique A ∈ P(U ) such that for every B ∈ P(U ), A∪B = B.
Existence. Let A = ∅. Clearly since A ∈ P(U ) (meaning A ⊆ U ), for A = ∅
gives ∅ ⊆ U which is true for all U . Furthermore, for every set B ∈ P(U ),
A ∪ B = ∅ ∪ B = B.
Uniqueness. Now we must show that for any other set C that C = ∅. Assume
C ∈ P(U ) such that for all B ∈ P(U ) that C ∪ B = B. Let B = ∅ which clearly
is ∈ P(U ). This implies that C ∪ ∅ = ∅ but we know that C ∪ ∅ = C. Thus
C = ∅.
b. Prove that there is a unique A ∈ P(U ) such that for every B ∈ P(U ), A∪B = A.
Existence. Let A = U . Clearly since A ∈ P(U ) (meaning A ⊆ U ), for A = U
gives U ⊆ U which is true for all U . Furthermore, for every set B ∈ P(U )
(meaning B ⊆ U ) giving ∀x(x ∈ B → x ∈ U ). Therefore A ∪ B = U ∪ B = U .
Uniqueness. Now we must show that for any other set C that C = U . Assume
C ∈ P(U ) such that for all B ∈ P(U ) that C ∪ B = B. Let B = U which clearly
is ∈ P(U ). This implies that C ∪ U = C but we know that C ∪ U = U . Thus
C = U.
3.6.10. Suppose A is a set, and for every family of sets F, if ∪F = A then A ∈ F.
Prove that A has exactly one element.
We will tackle this proof through contradiction by assuming that A does not
have exactly one element. This means that it either has no elements (existence)
or multiple elements (uniqueness).
Existence. Assume A = ∅ and ∪F = ∅. Then since ∪F = A gives that A ∈ F,
i.e. ∅ ∈ F, and hence F 6= ∅. However if ∪F = ∅ then F = ∅ which is a
contradiction and thus A must have at least one element.
Uniqueness. Assume A has more than one element, specifically let A = {x, y}
giving x ∈ A and y ∈ A with x 6= y. Furthermore, let F = {{x}, {y}}. Clearly
then ∪F = A so A ∈ F. Yet A 6∈ F which is a contradiction and hence A
cannot have more than one element.
Therefore if A must have at least one element but A cannot have more than
one element, then A must have exactly one element.
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