equivalence relations · bernd schroder¨ louisiana tech university, college of engineering and...

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Introduction Equivalence Relations Equivalence Classes Partitions

Equivalence Relations

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science


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Sometimes, We Want to Classify Objects ViaCertain Characteristics

1. We have not defined negative differences yet, but 3−5 and2−4 should be “the same”, even though they involvedifferent symbols. (But here we’ll actually think ofequivalence as equality, and we will see shortly that thatcan be problematic.)

2. Classification helps organize different entities. “food”,“clothing”, “pharmaceuticals”, etc. are in separate sectionsof a store. (Apples and steaks are both foods, so they areequivalent. They are certainly not equal.)

3. The elephant in the room: race. Take your presenter as anexample.



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1. We have not defined negative differences yet, but 3−5 and2−4 should be “the same”, even though they involvedifferent symbols.

(But here we’ll actually think ofequivalence as equality, and we will see shortly that thatcan be problematic.)





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2. Classification helps organize different entities. “food”,“clothing”, “pharmaceuticals”, etc. are in separate sectionsof a store.

(Apples and steaks are both foods, so they areequivalent. They are certainly not equal.)




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3. The elephant in the room: race.

Take your presenter as anexample.



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Relations Should Be The Right Tool

Typically we compare objects pairwise.1. When representing numbers, 3−5 is equivalent to 2−4,

because both represent the same number.2. When classifying items in a store, an apple is equivalent to

a steak in the sense that they are both foods.3. In terms of race, your presenter is equivalent to his wife.

The type of relation that we are looking for should retain keyproperties of equality without necessarily being equality.



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Relations Should Be The Right ToolTypically we compare objects pairwise.

1. When representing numbers, 3−5 is equivalent to 2−4,because both represent the same number.

2. When classifying items in a store, an apple is equivalent toa steak in the sense that they are both foods.

3. In terms of race, your presenter is equivalent to his wife.The type of relation that we are looking for should retain keyproperties of equality without necessarily being equality.



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3. In terms of race, your presenter is equivalent to his wife.

The type of relation that we are looking for should retain keyproperties of equality without necessarily being equality.



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3. In terms of race, your presenter is equivalent to his wife.The type of relation that we are looking for should retain keyproperties of equality without necessarily being equality.



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Definition.

Let X be a set. A relation ∼⊆ X×X is called anequivalence relation iff

1. ∼ is reflexive. That is, for all x ∈ X we have x∼ x.2. ∼ is symmetric. That is, for all x,y ∈ X we have x∼ y iff

y∼ x.3. ∼ is transitive. That is, for all x,y,z ∈ X we have that

x∼ y and y∼ z implies x∼ z.



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Definition. Let X be a set.

A relation ∼⊆ X×X is called anequivalence relation iff






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Definition. Let X be a set. A relation ∼⊆ X×X is called anequivalence relation iff






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1. ∼ is reflexive. That is, for all x ∈ X we have x∼ x.

2. ∼ is symmetric. That is, for all x,y ∈ X we have x∼ y iffy∼ x.

3. ∼ is transitive. That is, for all x,y,z ∈ X we have thatx∼ y and y∼ z implies x∼ z.



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y∼ x.

3. ∼ is transitive. That is, for all x,y,z ∈ X we have thatx∼ y and y∼ z implies x∼ z.



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Example.

The relation ∼⊆ N×N defined by m∼ n iff everypower of 2 that divides m also divides n and vice versa is anequivalence relation.

Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Example. The relation ∼⊆ N×N defined by m∼ n

iff everypower of 2 that divides m also divides n and vice versa is anequivalence relation.




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Example. The relation ∼⊆ N×N defined by m∼ n iff everypower of 2 that divides m also divides n and vice versa is anequivalence relation.




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Proof.

Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity.

Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N.

Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n

,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n.

So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.

Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry.

Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n.

Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa.

Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa.

Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.

Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity.

Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n.

Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m

and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n.

Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n.

The “vice versa” part is proved similarly, so k∼ n.



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Proof. Reflexivity. Let n ∈ N. Every power of 2 that divides n,divides n. So n∼ n.Symmetry. Let m,n ∈ N be so that m∼ n. Then every power of2 that divides m divides n and vice versa. Hence every power of2 that divides n divides m and vice versa. Therefore n∼ m.Transitivity. Let k,m,n ∈ N be so that k ∼ m and m∼ n. Thenevery power of 2 that divides k divides m and every power of 2that divides m divides n. Hence every power of 2 that divides kdivides n. The “vice versa” part is proved similarly

, so k∼ n.



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Definition.

Let ∼⊆ X×X be an equivalence relation on the setX. For each x ∈ X, the set [x] := {y ∈ X : y∼ x} is called theequivalence class of x.

Example. The equivalence classes of the relation ∼⊆ N×Ndefined by m∼ n iff every power of 2 that divides m also dividesn and vice versa are of the form

{2jk : k ∈ N,2 - k

}, where

j ∈ N0.



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Definition. Let ∼⊆ X×X be an equivalence relation on the setX.

For each x ∈ X, the set [x] := {y ∈ X : y∼ x} is called theequivalence class of x.


{2jk : k ∈ N,2 - k

}, where

j ∈ N0.



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Definition. Let ∼⊆ X×X be an equivalence relation on the setX. For each x ∈ X, the set [x] := {y ∈ X : y∼ x} is called theequivalence class of x.


{2jk : k ∈ N,2 - k

}, where

j ∈ N0.



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Example.

The equivalence classes of the relation ∼⊆ N×Ndefined by m∼ n iff every power of 2 that divides m also dividesn and vice versa are of the form

{2jk : k ∈ N,2 - k

}, where

j ∈ N0.



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Example. The equivalence classes of the relation ∼⊆ N×Ndefined by m∼ n iff every power of 2 that divides m also dividesn and vice versa

are of the form{

2jk : k ∈ N,2 - k}

, wherej ∈ N0.



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{2jk : k ∈ N,2 - k

}, where

j ∈ N0.



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Definition.

Let X be a set. Then a family {Xi}i∈I of subsets of Xis called a partition of X iff

1. All Xi are nonempty.2. The Xi are pairwise disjoint, that is, for i 6= j we have that

Xi∩Xj = /0.3.

⋃i∈I

Xi = X.

Proposition. Let X be a set. If ∼ is an equivalence relation onX, then the equivalence classes of ∼ form a partition of X.Conversely, if X is a set and {Xi}i∈I is a partition of X, thena∼ b iff there is an i ∈ I so that a,b ∈ Xi defines an equivalencerelation on X.



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Definition. Let X be a set.

Then a family {Xi}i∈I of subsets of Xis called a partition of X iff


Xi∩Xj = /0.3.

⋃i∈I

Xi = X.




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Definition. Let X be a set. Then a family {Xi}i∈I of subsets of Xis called a partition of X iff


Xi∩Xj = /0.3.

⋃i∈I

Xi = X.




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1. All Xi are nonempty.

2. The Xi are pairwise disjoint, that is, for i 6= j we have thatXi∩Xj = /0.

3.⋃i∈I

Xi = X.




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Xi∩Xj = /0.

3.⋃i∈I

Xi = X.




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Xi∩Xj = /0.3.

⋃i∈I

Xi = X.




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Xi∩Xj = /0.3.

⋃i∈I

Xi = X.

Proposition.

Let X be a set. If ∼ is an equivalence relation onX, then the equivalence classes of ∼ form a partition of X.Conversely, if X is a set and {Xi}i∈I is a partition of X, thena∼ b iff there is an i ∈ I so that a,b ∈ Xi defines an equivalencerelation on X.



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Xi∩Xj = /0.3.

⋃i∈I

Xi = X.

Proposition. Let X be a set.

If ∼ is an equivalence relation onX, then the equivalence classes of ∼ form a partition of X.Conversely, if X is a set and {Xi}i∈I is a partition of X, thena∼ b iff there is an i ∈ I so that a,b ∈ Xi defines an equivalencerelation on X.



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Xi∩Xj = /0.3.

⋃i∈I

Xi = X.

Proposition. Let X be a set. If ∼ is an equivalence relation onX, then the equivalence classes of ∼ form a partition of X.

Conversely, if X is a set and {Xi}i∈I is a partition of X, thena∼ b iff there is an i ∈ I so that a,b ∈ Xi defines an equivalencerelation on X.



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Xi∩Xj = /0.3.

⋃i∈I

Xi = X.




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Proof (equivalence relations induce partitions).

Let ∼ be anequivalence relation on X and let

{[x] : x ∈ X

}be the family of

equivalence classes of ∼. Then [x] 6= /0 for all x, because x ∈ [x].For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y]. Let [x], [y] be so that there is a z ∈ [x]∩ [y]. Then x∼ zand z∼ y, which implies x∼ y. Now let u ∈ [x]. Then u∼ x andx∼ y, so u∼ y and u ∈ [y]. Hence [x]⊆ [y], and we prove[y]⊆ [x] similarly. Thus the equivalence classes of ∼ arepairwise disjoint. Finally, for

⋃{[x] : x ∈ X

}= X first note that⋃{

[x] : x ∈ X}⊆ X is clear. For the reverse containment, let

x ∈ X. Then x ∈ [x] and [x]⊆⋃{

[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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Proof (equivalence relations induce partitions). Let ∼ be anequivalence relation on X and let

{[x] : x ∈ X

}be the family of

equivalence classes of ∼.

Then [x] 6= /0 for all x, because x ∈ [x].For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y]. Let [x], [y] be so that there is a z ∈ [x]∩ [y]. Then x∼ zand z∼ y, which implies x∼ y. Now let u ∈ [x]. Then u∼ x andx∼ y, so u∼ y and u ∈ [y]. Hence [x]⊆ [y], and we prove[y]⊆ [x] similarly. Thus the equivalence classes of ∼ arepairwise disjoint. Finally, for

⋃{[x] : x ∈ X




[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of

equivalence classes of ∼. Then [x] 6= /0 for all x, because x ∈ [x].

For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y]. Let [x], [y] be so that there is a z ∈ [x]∩ [y]. Then x∼ zand z∼ y, which implies x∼ y. Now let u ∈ [x]. Then u∼ x andx∼ y, so u∼ y and u ∈ [y]. Hence [x]⊆ [y], and we prove[y]⊆ [x] similarly. Thus the equivalence classes of ∼ arepairwise disjoint. Finally, for

⋃{[x] : x ∈ X




[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of

equivalence classes of ∼. Then [x] 6= /0 for all x, because x ∈ [x].For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y].

Let [x], [y] be so that there is a z ∈ [x]∩ [y]. Then x∼ zand z∼ y, which implies x∼ y. Now let u ∈ [x]. Then u∼ x andx∼ y, so u∼ y and u ∈ [y]. Hence [x]⊆ [y], and we prove[y]⊆ [x] similarly. Thus the equivalence classes of ∼ arepairwise disjoint. Finally, for

⋃{[x] : x ∈ X




[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of

equivalence classes of ∼. Then [x] 6= /0 for all x, because x ∈ [x].For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y]. Let [x], [y] be so that there is a z ∈ [x]∩ [y].

Then x∼ zand z∼ y, which implies x∼ y. Now let u ∈ [x]. Then u∼ x andx∼ y, so u∼ y and u ∈ [y]. Hence [x]⊆ [y], and we prove[y]⊆ [x] similarly. Thus the equivalence classes of ∼ arepairwise disjoint. Finally, for

⋃{[x] : x ∈ X




[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of

equivalence classes of ∼. Then [x] 6= /0 for all x, because x ∈ [x].For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y]. Let [x], [y] be so that there is a z ∈ [x]∩ [y]. Then x∼ zand z∼ y, which implies x∼ y.

Now let u ∈ [x]. Then u∼ x andx∼ y, so u∼ y and u ∈ [y]. Hence [x]⊆ [y], and we prove[y]⊆ [x] similarly. Thus the equivalence classes of ∼ arepairwise disjoint. Finally, for

⋃{[x] : x ∈ X




[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of

equivalence classes of ∼. Then [x] 6= /0 for all x, because x ∈ [x].For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y]. Let [x], [y] be so that there is a z ∈ [x]∩ [y]. Then x∼ zand z∼ y, which implies x∼ y. Now let u ∈ [x].

Then u∼ x andx∼ y, so u∼ y and u ∈ [y]. Hence [x]⊆ [y], and we prove[y]⊆ [x] similarly. Thus the equivalence classes of ∼ arepairwise disjoint. Finally, for

⋃{[x] : x ∈ X




[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of

equivalence classes of ∼. Then [x] 6= /0 for all x, because x ∈ [x].For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y]. Let [x], [y] be so that there is a z ∈ [x]∩ [y]. Then x∼ zand z∼ y, which implies x∼ y. Now let u ∈ [x]. Then u∼ x andx∼ y, so u∼ y and u ∈ [y].

Hence [x]⊆ [y], and we prove[y]⊆ [x] similarly. Thus the equivalence classes of ∼ arepairwise disjoint. Finally, for

⋃{[x] : x ∈ X




[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of

equivalence classes of ∼. Then [x] 6= /0 for all x, because x ∈ [x].For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y]. Let [x], [y] be so that there is a z ∈ [x]∩ [y]. Then x∼ zand z∼ y, which implies x∼ y. Now let u ∈ [x]. Then u∼ x andx∼ y, so u∼ y and u ∈ [y]. Hence [x]⊆ [y]

, and we prove[y]⊆ [x] similarly. Thus the equivalence classes of ∼ arepairwise disjoint. Finally, for

⋃{[x] : x ∈ X




[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of

equivalence classes of ∼. Then [x] 6= /0 for all x, because x ∈ [x].For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y]. Let [x], [y] be so that there is a z ∈ [x]∩ [y]. Then x∼ zand z∼ y, which implies x∼ y. Now let u ∈ [x]. Then u∼ x andx∼ y, so u∼ y and u ∈ [y]. Hence [x]⊆ [y], and we prove[y]⊆ [x] similarly.

Thus the equivalence classes of ∼ arepairwise disjoint. Finally, for

⋃{[x] : x ∈ X




[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of

equivalence classes of ∼. Then [x] 6= /0 for all x, because x ∈ [x].For pairwise disjointness, we prove that [x]∩ [y] 6= /0 implies[x] = [y]. Let [x], [y] be so that there is a z ∈ [x]∩ [y]. Then x∼ zand z∼ y, which implies x∼ y. Now let u ∈ [x]. Then u∼ x andx∼ y, so u∼ y and u ∈ [y]. Hence [x]⊆ [y], and we prove[y]⊆ [x] similarly. Thus the equivalence classes of ∼ arepairwise disjoint.

Finally, for⋃{

[x] : x ∈ X}

= X first note that⋃{[x] : x ∈ X

}⊆ X is clear. For the reverse containment, let


[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of


⋃{[x] : x ∈ X


[x] : x ∈ X}⊆ X is clear.

For the reverse containment, let


[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of


⋃{[x] : x ∈ X



x ∈ X.

Then x ∈ [x] and [x]⊆⋃{

[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of


⋃{[x] : x ∈ X



x ∈ X. Then x ∈ [x]

and [x]⊆⋃{

[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



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{[x] : x ∈ X

}be the family of


⋃{[x] : x ∈ X




[x] : x ∈ X},

so

X ⊆⋃{

[x] : x ∈ X}.



logo1



{[x] : x ∈ X

}be the family of


⋃{[x] : x ∈ X




[x] : x ∈ X}, so

X ⊆⋃{

[x] : x ∈ X}.



logo1


Proof (partitions induce equivalence relations).

Let {Xi}i∈Ibe a partition of X, and define a∼ b iff there is an i ∈ I so thata,b ∈ Xi. For every x ∈ X there is an Xi with x ∈ Xi and hencex∼ x. If x∼ y, then there is an Xi with x,y ∈ Xi and hencey∼ x. If x∼ y and y∼ z, then there are Xi and Xj so thatx,y ∈ Xi and y,z ∈ Xj. Therefore Xi∩Xj 3 y, which impliesXi = Xj. But then x,z ∈ Xi = Xj and hence x∼ z.



logo1


Proof (partitions induce equivalence relations). Let {Xi}i∈Ibe a partition of X, and define a∼ b iff there is an i ∈ I so thata,b ∈ Xi.

For every x ∈ X there is an Xi with x ∈ Xi and hencex∼ x. If x∼ y, then there is an Xi with x,y ∈ Xi and hencey∼ x. If x∼ y and y∼ z, then there are Xi and Xj so thatx,y ∈ Xi and y,z ∈ Xj. Therefore Xi∩Xj 3 y, which impliesXi = Xj. But then x,z ∈ Xi = Xj and hence x∼ z.



logo1


Proof (partitions induce equivalence relations). Let {Xi}i∈Ibe a partition of X, and define a∼ b iff there is an i ∈ I so thata,b ∈ Xi. For every x ∈ X there is an Xi with x ∈ Xi and hencex∼ x.

If x∼ y, then there is an Xi with x,y ∈ Xi and hencey∼ x. If x∼ y and y∼ z, then there are Xi and Xj so thatx,y ∈ Xi and y,z ∈ Xj. Therefore Xi∩Xj 3 y, which impliesXi = Xj. But then x,z ∈ Xi = Xj and hence x∼ z.



logo1


Proof (partitions induce equivalence relations). Let {Xi}i∈Ibe a partition of X, and define a∼ b iff there is an i ∈ I so thata,b ∈ Xi. For every x ∈ X there is an Xi with x ∈ Xi and hencex∼ x. If x∼ y, then there is an Xi with x,y ∈ Xi and hencey∼ x.

If x∼ y and y∼ z, then there are Xi and Xj so thatx,y ∈ Xi and y,z ∈ Xj. Therefore Xi∩Xj 3 y, which impliesXi = Xj. But then x,z ∈ Xi = Xj and hence x∼ z.



logo1


Proof (partitions induce equivalence relations). Let {Xi}i∈Ibe a partition of X, and define a∼ b iff there is an i ∈ I so thata,b ∈ Xi. For every x ∈ X there is an Xi with x ∈ Xi and hencex∼ x. If x∼ y, then there is an Xi with x,y ∈ Xi and hencey∼ x. If x∼ y and y∼ z, then there are Xi and Xj so thatx,y ∈ Xi and y,z ∈ Xj.

Therefore Xi∩Xj 3 y, which impliesXi = Xj. But then x,z ∈ Xi = Xj and hence x∼ z.



logo1


Proof (partitions induce equivalence relations). Let {Xi}i∈Ibe a partition of X, and define a∼ b iff there is an i ∈ I so thata,b ∈ Xi. For every x ∈ X there is an Xi with x ∈ Xi and hencex∼ x. If x∼ y, then there is an Xi with x,y ∈ Xi and hencey∼ x. If x∼ y and y∼ z, then there are Xi and Xj so thatx,y ∈ Xi and y,z ∈ Xj. Therefore Xi∩Xj 3 y, which impliesXi = Xj.

But then x,z ∈ Xi = Xj and hence x∼ z.



logo1


Proof (partitions induce equivalence relations). Let {Xi}i∈Ibe a partition of X, and define a∼ b iff there is an i ∈ I so thata,b ∈ Xi. For every x ∈ X there is an Xi with x ∈ Xi and hencex∼ x. If x∼ y, then there is an Xi with x,y ∈ Xi and hencey∼ x. If x∼ y and y∼ z, then there are Xi and Xj so thatx,y ∈ Xi and y,z ∈ Xj. Therefore Xi∩Xj 3 y, which impliesXi = Xj. But then x,z ∈ Xi = Xj and hence x∼ z.



equivalence relations · bernd schroder¨ louisiana tech university, college of engineering and...

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