Proof involving partitions and equivalence class

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SUMMARY

This discussion focuses on the relationship between partitions of a set X and equivalence relations. It establishes that every partition of X, denoted as Ai for i in an enumeration set I, naturally defines an equivalence relation R on X, where xRy if and only if there exists an i in I such that both x and y belong to Ai. The properties of reflexivity, symmetry, and transitivity confirm that R is indeed an equivalence relation. The main challenge discussed is formalizing the proof that the equivalence classes correspond to the subsets in the partition.

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  • Understanding of set theory and partitions
  • Familiarity with equivalence relations and their properties
  • Knowledge of mathematical proof techniques
  • Basic notation in mathematical logic
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  • Study the properties of equivalence relations in detail
  • Explore examples of partitions and their corresponding equivalence classes
  • Learn about the role of enumeration sets in defining partitions
  • Practice formalizing proofs in set theory
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Students of mathematics, particularly those studying abstract algebra or set theory, as well as educators looking to enhance their understanding of equivalence relations and partitions.

Korisnik
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Homework Statement


Show that every partition of X naturally determines an equivalence relation whose equivalence classes match the subsets from the partition.

Homework Equations


( 1 ) we know that equivalence sets on X can either be disjoint or equal

The Attempt at a Solution


Let Ai be a partition of X, i ∈ I, where I is an enumeration set, R is a relation on X. Let x,y∈X, xRy ⇔ ∃i∈I : x, y∈ Ai.

Since X is partitioned into sets, we choose Ai by taking an arbitrary number i∈I, Ai isn't empty.

Let x∈Ai. Thus xRx (reflexivity).
Let x, y∈Ai. x, y∈Ai ⇒ y, x ∈Ai ⇔ xRy ⇒ yRx (symmetry).
Let x, y∈Ai and y, z∈Ai then x, z∈Ai ⇔ xRy ∧ yRz ⇒ xRz (transitivity) (1).

Thus R is an equivalence relation.

The part so far is pretty clear. What I actually have to prove is the bolded part, and I'm not sure what is it that I have to prove. So far I've done this:

By definition of an equivalence class, each such equivalence relation (such that every pair of two elements of the same subset are in that relation) determines an equivalence class on each different subset.

Which essentially means: it's obvious from the definition... and I'm not satisfied with it. What should I prove, i.e. how do you formalize “equivalence classes are equal to subsets in a partition set”? How to start?
 
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Korisnik said:

Homework Statement


Show that every partition of X naturally determines an equivalence relation whose equivalence classes match the subsets from the partition.

Homework Equations


( 1 ) we know that equivalence sets on X can either be disjoint or equal

The Attempt at a Solution


Let Ai be a partition of X, i ∈ I, where I is an enumeration set, R is a relation on X. Let x,y∈X, xRy ⇔ ∃i∈I : x, y∈ Ai.

Since X is partitioned into sets, we choose Ai by taking an arbitrary number i∈I, Ai isn't empty.

Let x∈Ai. Thus xRx (reflexivity).
Let x, y∈Ai. x, y∈Ai ⇒ y, x ∈Ai ⇔ xRy ⇒ yRx (symmetry).
Let x, y∈Ai and y, z∈Ai then x, z∈Ai ⇔ xRy ∧ yRz ⇒ xRz (transitivity) (1).

Thus R is an equivalence relation.

The part so far is pretty clear. What I actually have to prove is the bolded part, and I'm not sure what is it that I have to prove. So far I've done this:

By definition of an equivalence class, each such equivalence relation (such that every pair of two elements of the same subset are in that relation) determines an equivalence class on each different subset.

Which essentially means: it's obvious from the definition... and I'm not satisfied with it. What should I prove, i.e. how do you formalize “equivalence classes are equal to subsets in a partition set”? How to start?
You don't have to start, you are already done. :oldsmile:
"Let x,y∈X, xRy ⇔ ∃i∈I : x, y∈ Ai" is the bolded text in symbols, and you have proved that this defines an equivalence relation.
 
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Okay, I guess you're right. I should leave it alone.
 

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