The study guide provides 3 methods for proving that the power set of a set A with n elements has 2^n elements: 1) Using the product rule of counting, 2) Applying the binomial theorem, and 3) Using induction and defining a function between subsets of A with and without a particular element x.

1. Math 189 Final Exam Study Guide
Jason Smith
May 7, 2014
1 Counting
Use the Product Rule for counting to prove that
| P (A) |= 2n
for a set A with | A |= n.
Hint: Let A = {a1, a2, a3, . . . , an}
For B ⊆ A, B = {−, −, −, . . . , −}.
How many ways to choose a subset {−, −, −, . . . , −} of A?
2 Binomial Thereom
Use the Binomial Theorem to prove that
| P (A) |= 2n
for a set A with | A |= n.
Binomial Theorem: (x + y)n = Σn
k=0
n
k xn−kyk.
Hint: Let x = y = 1 in Binomial Theorem to obtain:
2n
=
n
0
+
n
1
+ · · · +
n
n − 1
+
n
n
3 Induction and functions
Use Induction to prove that
| P (A) |= 2n
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2. (a) Let x ∈ A.
Let S = P (A − {x}) = {B ⊆ A | x /∈ B}
Let S = {B ⊆ A | x ∈ B}.
Define f : S → S by the rule f(B) = B ∪ {x}.
Prove that f is 1-1 and onto.
Use (a) in the induction step to finish the proof.
Hypothesis: If | B |= n, then | P (B) |= 2n, for any set B.
Let | A |= n + 1, x ∈ A. Use hypothesis on S = P (A − {x})
hint: P (A − {x}) = P (A) ∪ S
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