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 relations 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 reflexive. If b∈[a], then (a,b)∈E or (b,a)∈E as E is symmetric
. 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 transitive
. 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=⋃{P2i∣i∈I} is an equivalence relation. Conversely, every equivalence relation on S can be formed this way.
Title | partition is equivalent![]() |
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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 | Derivation |
Classification | msc 03-00 |