A Probability Course for the Actuaries A Preparation for Exam P/1

24 BASIC OPERATIONS ON SETS
Proof.
We prove part (a) leaving part(b) as an exercise for the reader.
(a) Let x ∈ (A ∪ B) c . Then x ∈ U and x ∉ A ∪ B. Hence, x ∈ U and (x ∉ A
and x ∉ B). This implies that (x ∈ U and x ∉ A) and (x ∈ U and x ∉ B).
It follows that x ∈ A c ∩ B c .
Conversely, let x ∈ A c ∩ B c . Then x ∈ A c and x ∈ B c . Hence, x ∉ A and
x ∉ B which implies that x ∉ (A ∪ B). Hence, x ∈ (A ∪ B) c
Remark 2.3
De Morgan’s laws are valid for any countable number of sets. That is
and
(∪ ∞ n=1A n ) c = ∩ ∞ n=1A c n
(∩ ∞ n=1A n ) c = ∪ ∞ n=1A c n
Example 2.9
Let U be the set of people solicited for a contribution to a charity. All the
people in U were given a chance to watch a video and to read a booklet. Let
V be the set of people who watched the video, B the set of people who read
the booklet, C the set of people who made a contribution.
(a) Describe with set notation: “The set of people who did not see the video
or read the booklet but who still made a contribution”
(b) Rewrite your answer using De Morgan’s law and and then restate the
above.
Solution.
(a) (V ∪ B) c ∩ C.
(b) (V ∪ B) c ∩ C = V c ∩ B c ∩ C = the set of people who did not watch the
video, did not read the booklet, but did make a contribution
If A i ∩ A j = ∅ for all i ≠ j then we say that the sets in the collection
{A n } ∞ n=1 are pairwise disjoint.
Example 2.10
Find three sets A, B, and C that are not pairwise disjoint but A∩B ∩C = ∅.
Solution.
One example is A = B = {1} and C = ∅