partition is equivalent to an equivalence relation


Proposition 1.

There is a one-to-one correspondence between the set Part⁡(S) of partitions of S and the set Equiv⁡(S) of equivalence relationsMathworldPlanetmath on S.

Proof.

Let S be a set.

Suppose P={Pi∣i∈I} is a partition of S. Form E=⋃{Pi×Pi∣i∈I}. Given any a∈S, a∈Pi for some i∈I. Then (a,a)∈Pi×Pi∈E. If (a,b)∈E, then (a,b)∈Pi×Pi for some i∈I, so a,b∈Pi, whence (b,a)∈Pi×Pi⊆E. Finally, suppose (a,b),(b,c)∈E. Then (a,b)∈Pi×Pi and (b,c)∈Pj×Pj, or a,b∈Pi and b,c∈Pj. Since b∈Pi∩Pj, Pi=Pj since P is a partition. So a,c∈Pi=Pj, or (a,c)∈Pi×Pi⊆E.

Conversely, suppose E is an equivalence relation on S. For each a∈S, define

[a]={b∈S∣(a,b)∈E}.

Then each a∈[a] since E is reflexiveMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath. If b∈[a], then (a,b)∈E or (b,a)∈E as E is symmetricPlanetmathPlanetmathPlanetmath. So a∈[b] as well. Next, pick any c∈[b]. Then (b,c)∈E. But (a,b)∈E. So (a,c)∈E since E is transitiveMathworldPlanetmathPlanetmathPlanetmathPlanetmath. Therefore c∈[a], which implies [b]⊆[a]. Collect all the information we have so far, this implies [a]=[b]. Therefore P={[a]∣a∈S} forms a partitition of S (P is being interpreted as a set, not a multiset). In fact, E=⋃{[a]×[a]∣a∈S}. ∎

From the above proof, we also have

Corollary 1.

For every partition P={Pi∣i∈I} of S, E=⋃{Pi2∣i∈I} is an equivalence relation. Conversely, every equivalence relation on S can be formed this way.

Title partition is equivalentMathworldPlanetmathPlanetmathPlanetmath to an equivalence relation
Canonical name PartitionIsEquivalentToAnEquivalenceRelation
Date of creation 2013-03-22 16:44:55
Last modified on 2013-03-22 16:44:55
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 7
Author CWoo (3771)
Entry type DerivationPlanetmathPlanetmath
Classification msc 03-00