SET EQUALITY PROBLEM, LAY 2.1.21
Before we do the proof, let us recall that if A, B are sets in the same
universe U ,
(1)
A \ B = A ∩ Bc.
Let us also remember De Morgan’s laws, which explain how complements
interact with unions and intersections:
(2)
(A ∪ B)c = Ac ∩ B c
(A ∩ B)c = Ac ∪ B c .
Lastly, let us remember that union distributes over intersection and vice
versa:
(3) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ B)
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Problem (Lay 2.1.21). Prove or give a counterexample:
A \ (A \ B) = B \ (B \ A).
This one is true. We will give two proofs. The first proof is an “elementchasing” proof and is the type of thing you will be responsible for on your
quiz on Thursday.
Element chasing. (⊆) Suppose x ∈ A \ (A \ B). Then x ∈ A and x 6∈ A \ B.
The second statement implies that x 6∈ A or x ∈ B.1 However, since x 6∈ A
is false (we already know x ∈ A), it must be that x ∈ A and x ∈ B. In
other words,
x ∈ A ∩ B.
Now, we must show x 6∈ B \ A to complete the proof. This is equivalent to
(x 6∈ B or x ∈ A), which is a true statement (we know x ∈ A). All together,
x ∈ B and x 6∈ B \ A,
which has the same meaning as x ∈ B \ (B \ A).
(⊇) Above, we showed the inclusion A \ (A \ B) ⊆ B \ (B \ A) for arbitrary
sets A, B. Therefore, interchanging A and B, we have the opposite inclusion,
i.e. A \ (A \ B) ⊇ B \ (B \ A).
The second proof is not as “low level,” i.e. there is no element-chasing.
It just requires remembering a few laws.
Proof. Using (in order) the definition of set complement (1), De Morgan’s
laws (2), and the distributive laws (3), we have the set equalities
A \ (A \ B) = A ∩ (A ∩ B c )c = A ∩ (Ac ∪ B) = (A ∩ Ac ) ∪ (A ∩ B) = A ∩ B.
Date: September 27, 2016.
1¬(x ∈ A \ B) ⇐⇒ ¬(x ∈ A ∧ x 6∈ B) ⇐⇒ x 6∈ A ∨ x ∈ B
1
2
SET EQUALITY PROBLEM, LAY 2.1.21
The same analysis with A and B interchanged shows that
B \ (B \ A) = A ∩ B.
Therefore, A \ (A \ B) = B \ (B \ A).