Equivalence classes are either equal or disjoint
From Maths
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!
Contents
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:
- [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
- [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 proof has been marked as an page requiring an easy proof
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].
This proof has been marked as an page requiring an easy proof
Notes
- ↑ In other words: [ilmath]\frac{X}{\sim} [/ilmath] is the set of equivalence classes of [ilmath]\sim[/ilmath]
- ↑ 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:
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!
Categories:
- Stub pages
- Pages requiring proofs: Easy proofs
- Pages requiring proofs
- Pages requiring references
- Theorems
- Theorems, lemmas and corollaries
- Abstract Algebra Theorems
- Abstract Algebra Theorems, lemmas and corollaries
- Abstract Algebra
- Elementary Set Theory Theorems
- Elementary Set Theory Theorems, lemmas and corollaries
- Elementary Set Theory
- Set Theory Theorems
- Set Theory Theorems, lemmas and corollaries
- Set Theory