Difference between revisions of "Equivalence classes are either equal or disjoint"

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Latest revision as of 14:36, 13 November 2016

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Statement

Let [ilmath]X[/ilmath] be a set, let [ilmath]\sim\subseteq X\times X[/ilmath] be an equivalence relation on [ilmath]X[/ilmath], let [ilmath]\frac{X}{\sim} [/ilmath] denote the quotient of [ilmath]X[/ilmath] by [ilmath]\sim[/ilmath][Note 1] and lastly let [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] given by [ilmath]\pi:x\mapsto[x][/ilmath] be the canonical projection of the equivalence relation. Then:

  • We claim that [ilmath]\frac{X}{\sim} [/ilmath] is a partition of [ilmath]X[/ilmath]. That is:
    1. [ilmath]\forall x\in X\exists y\in\frac{X}{\sim}[x\in y][/ilmath] - all elements of [ilmath]x[/ilmath] belong to an element of the partition
    2. [ilmath]\forall u,v\in\frac{X}{\sim}[u\cap v\neq\emptyset\implies u\eq v][/ilmath] - if [ilmath]u[/ilmath] and [ilmath]v[/ilmath] are not disjoint, they are equal
      • Equivalently (by contrapositive[Note 2]): [ilmath]\forall u,v\in\frac{X}{\sim}[u\neq v\implies u\cap v\eq\emptyset][/ilmath]

Proof

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This is an easy and routine proof. First year friendly and all. However:
  • Caution:I must make sure to prove the requirements e.g. that [ilmath][x][/ilmath], the equivalence class containing [ilmath]x[/ilmath] makes sense. As if I use a property like:
    • [ilmath]y\in [x]\iff y\sim x[/ilmath]
    I'm sort of indirectly using part 2 of this theorem if I ever use the transitive property of equivalence relations involving [ilmath]x[/ilmath] and [ilmath]y[/ilmath].
Basically - be careful. It's easy to be circular here as the results are so engrained into my every day mathematical work

This proof has been marked as an page requiring an easy proof

Notes

  1. ↑ In other words: [ilmath]\frac{X}{\sim} [/ilmath] is the set of equivalence classes of [ilmath]\sim[/ilmath]
  2. ↑ The contrapositive of [ilmath]A\implies B[/ilmath] is [ilmath](\neg B)\implies(\neg A)[/ilmath]. That is to say:
    • [ilmath]\big(A\implies B\big)\iff\big((\neg B)\implies(\neg A)\big)[/ilmath]

References

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Do not worry I know this content really well, this result is true. I promise
  • I need to find a book that deems this theorem worthy of making explicit!


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