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

c) I don't know get the above question so I don't understand this one either...
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}

c) I don't know get the above question so I don't understand this one either...