Equivalence classes are either equal or disjoint

From Maths
Revision as of 14:36, 13 November 2016 by Alec (Talk | contribs) (Created page with "{{Stub page|grade=B|msg=I'm sure I've already done this SOMEWHERE - find it!}} __TOC__ ==Statement== Let {{M|X}} be a set, let {{M|\sim\subseteq X\times X}} be an equiva...")

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search
Stub grade: B
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
I'm sure I've already done this SOMEWHERE - find it!

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

Grade: B
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
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

Grade: D
This page requires references, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:
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!