Theorem about equivalence relations.

Question

Is this theorem true ?

Theorem: Assume R is a relation on a set A.The relation R partitions the elements of A into classes if and only if R is an equivalence relation.

My counter example is:

Let A = {1,2} and R = {(1,2),(2,1)} The class generated by 1 is {2} and the class generated by 2 is {1}. And {1} and {2} completely partition A but R is not reflexive.

Presumably the context surrounding the Theorem says something to the effect that the classes defined by a relation consist of elements which are all related to each other (and that the elements in two different classes are not related). In your example, there would be no classes. — lulu, Aug 04 '23 at 13:35
What does "The relation R partitions the elements of A into classes" means? You can talk about classes only if there is already an equivalence relation. — jjagmath, Aug 04 '23 at 13:36
https://proofwiki.org/wiki/Fundamental_Theorem_on_Equivalence_Relations — CyclotomicField, Aug 04 '23 at 13:39
Here is the definition of a class in my book: "If R is a relation on A and a∈A define the class generated by a to be the set {b∈A: a R b}" — gokalp gokalp, Aug 04 '23 at 13:48
You have misunderstood. An equivalence relation is reflexive, symmetric and transitive. Your relation is only symmetric. — John Douma, Aug 04 '23 at 13:49
@JohnDouma That's the point of the OP, the relations is not a equivalence relation. — jjagmath, Aug 04 '23 at 13:53
@JohnDouma As stated, the theorem asks you to prove that a relation is an equivalence relation (given the partition). The OP's has proposed a counterexample. — lulu, Aug 04 '23 at 13:54
Ok, that's an unusual definition of "class". But with that definition of class the relation does generate a partition of $A$, so your counterexample is correct. — jjagmath, Aug 04 '23 at 13:54
@gokalpgokalp If nothing else, that ought to be called a right-class (or a left-class, depending on conventions) as you might get different classes if you used ${b\in A:,bRa}$, for instance. But if that's really the definition they want, then your counterexample is valid. — lulu, Aug 04 '23 at 13:55
@jjagmath The equivalence relation generated by the partition ${1,2}$ contains $(1,1)$ and $(2,2)$. — John Douma, Aug 04 '23 at 13:59
I'll just use another resource.I think that author doesn't mean the "class" that i understood.Thanks. — gokalp gokalp, Aug 04 '23 at 14:01
@JohnDouma First: ${1,2}$ is not a partition. That would be ${{1},{2}}$. Second. The OP is working with a general relation (not necessarily an equivalence relation) and giving its definition of class generated by that relation. — jjagmath, Aug 04 '23 at 14:04
I don't have that text at hand. If I did, I would want to check the exact wording of the definition and theorem. Based on what you wrote, I agree with lulu. — Matthew Leingang, Aug 04 '23 at 14:44

John Douma · Accepted Answer · 2023-08-04T14:33:18.327

I see the problem. Your definition of a class is the set of all things equivalent to a given element. Your class generates the partition $\{1\}, \{2\}$

Note that your relation, although it generates the class, is not an equivalence relation. The equivalence relation for your partition is $a$ is equivalent to $b$ if and only if $a=b$.

The theorem states that a partition is the same as an equivalence relation. It does not say that if any relation generates a partition then that relation is an equivalence relation. In fact there are two separate theorems:

If $R$ is an equivalence relation on a non-empty set $A$ then the set of equivalence classes forms a partition of $A$.
Given a partition $P=\{A_1,\dots, A_n\}$ then the relation induced by the partition, $x$ is equivalent to $y$ if and only if $x$ and $y$ are in the same $A_i$, is an equivalence relation.

Therefore, the theorem that you wrote is incorrect.

See this for more information. – John Douma Aug 04 '23 at 14:36 — John Douma, Aug 04 '23 at 14:36

Theorem about equivalence relations.

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