Selected items from the Introduction to set theory and to methodology and philosophy of mathematics

1. Introduction to set theory and to methodology and philosophy of
mathematics and computer programming
Families ordered by inclusion
An overview
by Jan Plaza
c 2017 Jan Plaza
Use under the Creative Commons Attribution 4.0 International License
Version of March 10, 2017

2. A Hasse diagram of P({1, 2, 3}) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
{1, 2, 3}
Z
Z
Z
Z
{1, 2} {1, 3} {2, 3}
Z
Z
Z
Z
Z
Z
Z
Z
{1} {2} {3}
Z
Z
Z
Z
∅
A Hasse diagram of a finite family of sets. Nodes - the members of the family.
Edges correspond to the ⊂ relation, with the smaller set below the bigger set,
provided that there is no set between the two.

3. Definition
Let X be a family of sets.
x is smallest/least in X (with respect to ⊆) if x ∈ X and ∀y∈X x ⊆ y
– x ∈ X, and every set in X is “greater than or equal to” x.
x is greatest/largest/biggest in X (w.r.t. ⊆) if x ∈ X and ∀y∈X y ⊆ x
– x ∈ X, and every set in X is “smaller than or equal to” x.
x is minimal in X (with respect to ⊆) if x ∈ X and ∀y∈X y ⊂ x
– x ∈ X, and no set in X is strictly “smaller” than x.
x is maximal in X (with respect to ⊆) if x ∈ X and ∀y∈X x ⊂ y
– x ∈ X, and no set in X is strictly “greater” than x.

4. Example
P({1, 2, 3}) − {∅}.
{1, 2, 3}
Z
Z
Z
Z
{1, 2} {1, 3} {2, 3}
Z
Z
Z
Z
Z
Z
Z
Z
{1} {2} {3}
No smallest set;
Exactly three minimal sets: {1}, {2}, {3};
Exactly one greatest set: {1, 2, 3};
Exactly one maximal set: {1, 2, 3}.

5. Example
Family {{1}, {2}, {3}}.
The Hasse diagram of this family has three nodes and no edges:
{1} {2} {3}
No smallest set;
Exactly three minimal sets: {1}, {2}, {3};
No greatest set;
Exactly three maximal sets: {1}, {2}, {3}.

6. Example
The family of all the non-empty sets of natural numbers.
No smallest set;
The minimal sets are {0}, {1}, {2}, ...;
Exactly one greatest set: N;
Exactly one maximal set: N.

7. Exercise
Produce examples of families of sets with properties as in this chart.
0/2 stands for: exactly 0 smallest sets and exactly 2 minimal sets, etc.
Family size Number of smallest sets and minimal sets
0 0/0
1 1/1
2 1/1 or 0/2
3 1/1 or 0/2 or 0/3
...
...
n 1/1 or 0/2 or 0/3 or ... or 0/n
...
...
infinity 0/0 or 0/1 or 1/1 or 0/2 or 0/3 or ... or 0/infinity

8. Definition
Sets x and y are comparable (with respect to ⊆) if x ⊆ y or y ⊆ x.
Sets x and y are incomparable (with respect to ⊆) if they are not comparable.
Example
Sets {1, 2} and {1, 2} are comparable.
Sets {1} and {2} are incomparable.
Sets {1, 2} and {1, 3} are incomparable.
Exercise
Give an example of four sets s.t. every two different sets are incomparable.

9. Two minimal sets
Proposition
1. ...
2. If in a family there are two or more different minimal sets,
then there is no smallest set.
Family size Number of smallest sets and minimal sets
0 0/0
1 1/1
2 1/1 or 0/2
3 1/1 or 0/2 or 0/3
...
...
n 1/1 or 0/2 or 0/3 or ... or 0/n
...
...
infinity 0/0 or 0/1 or 1/1 or 0/2 or 0/3 or ... or 0/infinity

10. Two minimal sets, proof
Proposition
1. Any two different minimal sets in a family are incomparable.
2. If in a family there are two or more different minimal sets,
then there is no smallest set.
Proof
1. Assume that x1 and x2 are two different minimal sets and they are comparable.
Our goal is to obtain contradiction.
As x1 and x2 are comparable, x1 ⊆ x2 or x2 ⊆ x1.
We will consider two cases.
Case: x1 ⊆ x2. As x2 is minimal, we must have x1 = x2 – contradiction.
Case: x2 ⊆ x1. As x1 is minimal, we must have x1 = x2 – contradiction.
2. Assume that x1 and x2 are two different minimal sets.
Assume s is a smallest set. Goal: contradiction.
As s is smallest, s ⊆ x1 and s ⊆ x2.
As x1, x2 are minimal, s ⊂ x1 and s ⊂ x2.
So, s = x1 and s = x2. So, x1 = x2 – a contradiction.

11. Uniqueness of a smallest set
Proposition
No family of sets has two different smallest sets.
(There can be only 1 or 0 smallest sets in a family.)
Family size Number of smallest sets and minimal sets
0 0/0
1 1/1
2 1/1 or 0/2
3 1/1 or 0/2 or 0/3
...
...
n 1/1 or 0/2 or 0/3 or ... or 0/n
...
...
infinity 0/0 or 0/1 or 1/1 or 0/2 or 0/3 or ... or 0/infinity

12. Uniqueness of a smallest set, proof
Proposition
No family of sets has two different smallest sets.
(There can be only 1 or 0 smallest sets in a family.)
Proof
Assume that a family of sets has two different smallest sets x1 and x2.
Our goal is to obtain a contradiction.
As x1 is smallest, x1 ⊆ x2.
As x2 is smallest, x2 ⊆ x1.
As x1 ⊆ x2 and x2 ⊆ x1, we have x1 = x2 – contradiction.

13. Smallest vs. minimal
Proposition
If there exists a smallest set in a family,
it is also a minimal set,
and it is the only minimal set.
Family size Number of smallest sets and minimal sets
0 0/0
1 1/1
2 1/1 or 0/2
3 1/1 or 0/2 or 0/3
...
...
n 1/1 or 0/2 or 0/3 or ... or 0/n
...
...
infinity 0/0 or 0/1 or 1/1 or 0/2 or 0/3 or ... or 0/infinity

14. Smallest vs. minimal, proof
Proposition
If there exists a smallest set in a family,
it is also a minimal set,
and it is the only minimal set.
Proof
First we will prove that if x is smallest then x is minimal. (Later we will still need to
prove that there are no minimal sets other than x.)
Assume that x is smallest.
To prove that x is minimal, take any y and assume that y ⊂ x.
The goal is to obtain contradiction.
As y ⊂ x, we obtain that y ⊆ x and y =x.
As x is smallest, x ⊆ y.
As x ⊆ y and y ⊆ x we obtain that x = y – contradiction.

15. Smallest vs. minimal, proof, continued
Now, we will prove that if x is smallest and z is minimal then x = z.
Assume that x is smallest and z is minimal.
The goal is to show that x = z.
We proved above that x is minimal.
As x and z are minimal, by point 2, they are incomparable.
So, x z – this contradicts the assumption that x is smallest.

16. Definition
Let X be a family of sets and Y ⊆ X.
Y is downward closed in X if for every y ∈ Y , if x ∈ X and x ⊆ y then x ∈ Y .
A Hasse diagram. Family X – all the dots. If Y is downward closed in X and
if the two black dots belong to Y , then all the gray dots must belong to Y .

17. Minimal set in a downward closed subfamily
Fact
Let m ∈ Y ⊆ X, and let Y be downward closed in X.
If m is minimal in Y then m is minimal in X.

18. Lemma
Let X be a finite family of sets.
Then every member of X has a subset minimal in X.
Proof
Consider the following condition:
(*) every member of the family has a subset minimal in the family.
Assume that there exists a finite family of sets that violates (*).
The goal is to obtain a contradiction.
As there exists a finite family that violates (*); among all such families
there is a family X that has the smallest number of elements.
X =∅, because the empty family satisfies (*).
As X =∅, take a set x ∈ X that has no subset minimal in X.
As x is its own subset, x is not minimal in X.
As x is not minimal in X, there exists in X a proper subset y0 of x:
y0 ∈ X and y0 ⊂ x.
Let Y = {y ∈ X : y ⊆ y0}.

19. Proof, continued
Notice that:
Y ⊆ X;
Y ⊂ X, because x ∈ X − Y0;
Y satisfies (*), because it has fewer elements than X;
y0 ∈ Y ;
y0 has a subset m minimal in Y ;
Y is downward closed in X;
m is minimal in X, by the fact above;
m ⊆ y0 ⊆ x, so x has a subset minimal in X – a contradiction!

20. Properties of finite families
Theorem
Let X be a finite family of sets. Then:
1. If X is non-empty then it has at least one minimal set.
2. If there is exactly one minimal set in X then it is the smallest set.
Family size Number of smallest sets and minimal sets
0 0/0
1 1/1
2 1/1 or 0/2
3 1/1 or 0/2 or 0/3
...
...
n 1/1 or 0/2 or 0/3 or ... or 0/n
...
...
infinity 0/0 or 0/1 or 1/1 or 0/2 or 0/3 or ... or 0/infinity

21. Properties of finite families, proof
Theorem
Let X be a finite family of sets. Then:
1. If X is non-empty then it has at least one minimal set.
2. If there is exactly one minimal set in X then it is the smallest set.
Proof
1. By the lemma above.
2. Take any finite family X of sets with a single minimal set x.
We will show that x is smallest in X.
Take any y ∈ X.
We need to show x ⊆ y.
By the previous lemma, y has a subset that is minimal in X.
As x is the only minimal set in X, we must have x ⊆ y.
Exercise: Show that the assumption of finiteness is essential for the theorem above.

22. Smallest sets vs. intersections
Proposition
Let X be a family of sets. Then:
1. x is smallest in X with respect to ⊆ iff x ∈ X and x = X.
2. x is greatest in X with respect to ⊆ iff x ∈ X and x = X.
Proof of 1.
→)
Assume that x is smallest in X.
Then, x ∈ X, and it remains to prove: x = X.
As x ∈ X, we have X ⊆ x, so it remains to prove: x ⊆ X.
Take any u ∈ x. Goal: u ∈ X.
Take any v ∈ X. Goal: u ∈ v.
As v ∈ X and x is smallest in X, we have x ⊆ v.
As u ∈ x and x ⊆ v, we have u ∈ v.

23. Proof, continued
We are proving:
x is smallest in X with respect to ⊆ iff x ∈ X and x = X.
←)
Assume that x ∈ X and x = X.
Goal: x is smallest in X.
Take any v ∈ X. Goal: x ⊆ v.
As v ∈ X, we have X ⊆ v.
As x = X and X ⊆ v, we have x ⊆ v.
Exercise
Disprove: Let X be a non-empty family of sets; x is smallest in X w.r.t. ⊆ iff x = X.
Disprove: Let X be a family of sets; x is greatest in X with respect to ⊆ iff x = X.