The document provides an introduction to set theory, focusing on the power set and its properties, including the theorem that a set with n elements has 2^n subsets. It discusses the proof of this theorem using mathematical induction and presents bit-vector representations of subsets and their applications in programming. Additionally, it includes exercises to test understanding of set operations and relationships between power sets.

1. Introduction to set theory and to methodology and philosophy of
mathematics and computer programming
Power sets
An overview
by Jan Plaza
c 2017 Jan Plaza
Use under the Creative Commons Attribution 4.0 International License
Version of March 5, 2017

2. A family of sets – a set of sets.
{{0, 1, 2}, {1, 2, 3}, {5}} is a family of sets.
{∅} is a family of sets.
∅ is a family of sets.

3. Power Set Axiom
∀X ∃Y ∀u (u ∈ Y ↔ u ⊆ X)
For any set X,
there is a family Y of sets, s.t.
u ∈ Y ↔ u ⊆ X, for every u.
For any set X,
there is a family Y of all subsets of X.
Such a family Y is called the power set of X , and is denoted P(X) .
Fact
x ∈ P(X) iff x ⊆ X.

4. Example
P({0, 1, 2}) = {∅, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, {0, 1, 2}}.
P({0, 1}) = {∅, {0}, {1}, {0, 1}}.
P({0}) = {∅, {0}}.
P(∅) = {∅}.
Theorem
If X has n elements, then P(X) has 2n elements.

5. Proof
Consider the following property p(n) of a natural number n:
every set with n elements has 2n subsets.
We will prove that every natural number has this property.
Proof by induction on n.
Base case.
Goal: p(0), i.e. every set with 0 elements has 20 subsets.
Take any set X with 0 elements. X = ∅.
∅ has only 1 subset, namely ∅. So X has 20 subsets.

6. Proof, continued.
Inductive case.
Take an arbitrary natural number n.
Assume p(n): every set with n elements has 2n subsets; this is the inductive hypothesis.
Goal: p(n + 1), i.e. every set with n + 1 elements has 2n+1 subsets.
Take any set X with n + 1 elements.
As n ∈ N, this implies that X has at least one element.
Select any x ∈ X.
Let Y = X − {x}.
Y has n elements.
By inductive hypothesis, P(Y ) has 2n elements.

7. Proof, continued.
Now we will count the subsets of X.
First, count the subsets of X that do not contain x:
these are just the subsets of Y ,
so there are 2n such subsets.
Second, count the subsets of X that contain x:
every such subset X0 can be written as Y0 ∪ {x} for exactly one Y0 ⊆ Y ;
so their number equals the number of subsets of Y , i.e. 2n.
Notice that in this way every subset of X was counted exactly once.
So, the total number of subsets of X is 2n + 2n = 2n+1.

8. Bit-vector representation of subsets
The subsets of {x0, x1, x2} can be represented as bit-vectors v0, v1, v3 .
vi = 1 iff xi is in the subset.
Subsets of {x0, x1, x2} Bit-vectors
∅ 0, 0, 0
{x0} 1, 0, 0
{x1} 0, 1, 0
{x2} 0, 0, 1
{x0, x1} 1, 1, 0
{x0, x2} 1, 0, 1
{x1, x2} 0, 1, 1
{x0, x1, x2} 1, 1, 1
Works also for any set {0, 1, ..., n − 1}.
There are 2n bit-vectors of length n, and there are 2n subsets of an n-element set.
This is used in programming when subsets of a fixed U = {x0, x1, ..., xn−1} are needed.
Set operations ∪, ∩, c are then implemented as bit-wise operations and, or, not.

9. Exercise
1. True or false? P(X) ∩ P(Y ) = P(X ∩ Y ).
2. True or false? P(X) ∪ P(Y ) ⊆ P(X ∪ Y ).
3. Disprove: P(X) ∪ P(Y ) ⊇ P(X ∪ Y ).