3.6 families ordered by inclusion

Introduction to set theory and to methodology and philosophy ofmathematics and computer programming

Families ordered by inclusion

An overview

by Jan Plaza

c©2017 Jan PlazaUse under the Creative Commons Attribution 4.0 International License

Version of March 10, 2017

A Hasse diagram of P({1, 2, 3}) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

{1, 2, 3}��

ZZZZ

{1, 2} {1, 3} {2, 3}

��

ZZZZ

��

ZZZZ

{1} {2} {3}

ZZZZ

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∅

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.

DefinitionLet X be a family of sets.

I 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 .

I 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 .

I x is minimal in X (with respect to ⊆) if x ∈ X and ∀y∈X y 6⊂ x

– x ∈ X , and no set in X is strictly “smaller” than x .

I x is maximal in X (with respect to ⊆) if x ∈ X and ∀y∈X x 6⊂ y

– x ∈ X , and no set in X is strictly “greater” than x .

ExampleP({1, 2, 3})− {∅}.

{1, 2, 3}��

ZZZZ

{1, 2} {1, 3} {2, 3}

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ZZZZ

��

ZZZZ

{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}.

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}.

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.

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 sets0 0/0

1 1/12 1/1 or 0/23 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

DefinitionSets 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

I Sets {1, 2} and {1, 2} are comparable.

I Sets {1} and {2} are incomparable.

I Sets {1, 2} and {1, 3} are incomparable.

ExerciseGive an example of four sets s.t. every two different sets are incomparable.

Two minimal setsProposition

1. ...

2. If in a family there are two or more different minimal sets,then there is no smallest set.


1 1/12 1/1 or 0/23 1/1 or 0/2 or 0/3...

...n 1/1 or 0/2 or 0/3 or ... or 0/n...

...


Two minimal sets, proofProposition

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 6⊂ x1 and s 6⊂ x2.So, s = x1 and s = x2. So, x1 = x2 – a contradiction.

Uniqueness of a smallest set

PropositionNo family of sets has two different smallest sets.(There can be only 1 or 0 smallest sets in a family.)


1 1/12 1/1 or 0/23 1/1 or 0/2 or 0/3...

...n 1/1 or 0/2 or 0/3 or ... or 0/n...

...


Uniqueness of a smallest set, proof

PropositionNo family of sets has two different smallest sets.(There can be only 1 or 0 smallest sets in a family.)

ProofAssume 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.

Smallest vs. minimalPropositionIf there exists a smallest set in a family,it is also a minimal set,and it is the only minimal set.


1 1/12 1/1 or 0/23 1/1 or 0/2 or 0/3...

...n 1/1 or 0/2 or 0/3 or ... or 0/n...

...


Smallest vs. minimal, proof

PropositionIf there exists a smallest set in a family,it is also a minimal set,and it is the only minimal set.

ProofFirst we will prove that if x is smallest then x is minimal. (Later we will still need toprove 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 6=x .As x is smallest, x ⊆ y .As x ⊆ y and y ⊆ x we obtain that x = y – contradiction.

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.

DefinitionLet 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 andif the two black dots belong to Y , then all the gray dots must belong to Y .

Minimal set in a downward closed subfamily

FactLet m ∈ Y ⊆ X , and let Y be downward closed in X .If m is minimal in Y then m is minimal in X .

Lemma

Let X be a finite family of sets.Then every member of X has a subset minimal in X .ProofConsider 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 familiesthere is a family X that has the smallest number of elements.X 6=∅, because the empty family satisfies (*).As X 6=∅, 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}.

Proof, continued

Notice that:

I Y ⊆ X ;

I Y ⊂ X , because x ∈ X − Y0;

I Y satisfies (*), because it has fewer elements than X ;

I y0 ∈ Y ;

I y0 has a subset m minimal in Y ;

I Y is downward closed in X ;

I m is minimal in X , by the fact above;

I m ⊆ y0 ⊆ x , so x has a subset minimal in X – a contradiction!

Properties of finite familiesTheoremLet 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.


1 1/12 1/1 or 0/23 1/1 or 0/2 or 0/3...

...n 1/1 or 0/2 or 0/3 or ... or 0/n...

...


Properties of finite families, proofTheoremLet 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.

Smallest sets vs. intersections

PropositionLet 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 .

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 .

ExerciseDisprove: 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 .

3.6 families ordered by inclusion

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