Describing Equivalence Classes

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SUMMARY

This discussion focuses on the concept of equivalence classes in the context of a mapping function f: A-->B. The user successfully demonstrates that the relation defined by x~y iff f(x) = f(y) is an equivalence relation by proving reflexivity, symmetry, and transitivity. When f is one-to-one (1-1), the equivalence classes consist of singleton sets, meaning each element in set A maps uniquely to a single element in set B. If there is only one equivalence class, it indicates that f is a constant function, where all elements in A map to the same value in B.

PREREQUISITES
  • Understanding of equivalence relations in mathematics
  • Familiarity with functions and mappings, specifically one-to-one functions
  • Basic knowledge of set theory and notation
  • Ability to prove mathematical properties such as reflexivity, symmetry, and transitivity
NEXT STEPS
  • Study the properties of equivalence relations in more depth
  • Learn about functions and their classifications, focusing on injective (one-to-one) and surjective (onto) functions
  • Explore examples of equivalence classes in different mathematical contexts
  • Investigate the implications of constant functions in functional analysis
USEFUL FOR

Students studying abstract algebra, mathematicians interested in set theory, and educators teaching concepts related to functions and equivalence relations.

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Hey guys!
So I am having trouble understanding equivalence classes. How are they determined??
Anyways here is my problem!

Homework Statement


Let A and B be two sets, and f: A-->B a mapping. A relation on A is defined by: x~y iff f(x) = f(y)
a) Show ~ is an equivalence relation
b) Describe the equivalence classes when f is 1-1
c) What can be said about f if ~ has only one equivalence class?


Homework Equations





The Attempt at a Solution



a) I've already done this and understand it:
reflexive: f(x)=f(x)
symmetric: f(x) = f(y), f(y) = f(x)
transitive f(x) = f(y) and f(y) = f(z) then f(x) = f(z)

b) Okay here is where I am having trouble
so if f is 1-1, it means f(x) = f(y) --> x = y
Then would the equivalence class be something like all x that are in A which get mapped to f(x)?
So [x] = { x ϵ A | f(x) = f(y) } = {x ϵ A | x = y } = {x}

:confused:

c) I don't know get the above question so I don't understand this one either...
 
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Yes, for b, the equivalence classes are all "singleton sets"- each set contains only one member of A.

For (c) it is exactly the opposite- all members of A are in the same equivalence class so for all x and y, f(x)= f(y)- f is a "constant function".
 

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