# Thread: Equivalence relation and Functions

1. ## Equivalence relation and Functions

The problem is:
"Suppose that A is a nonempty set and R is an equivalence relation on A. Show that there is a function f with A as its domain such that (x,y) are elements of R if and only if f(x)=f(y)"

I don't understand how to connect a relation with a function

thank you

2. Originally Posted by mremwo
The problem is:
"Suppose that A is a nonempty set and R is an equivalence relation on A. Show that there is a function f with A as its domain such that (x,y) are elements of R if and only if f(x)=f(y)"

I don't understand how to connect a relation with a function

thank you
A function is actually a relation, but that's just not how many people think about it. Consider that a function is really just it's graph $\displaystyle \Gamma_f=\left\{(x,f(x)):x\in X\right\}$. This $\displaystyle \Gamma_f$ is the relation.

3. Try finding a function $\displaystyle f:A\rightarrow \mathcal{P}(A)$ (the power set of $\displaystyle A$) such that two elements are in the same equivalence class under $\displaystyle R$ if and only if they have the same image under $\displaystyle f$. Can you think of a natural choice for the image under $\displaystyle f$ of an element from $\displaystyle A$?

4. Originally Posted by topspin1617
Try finding a function $\displaystyle f:A\rightarrow \mathcal{P}(A)$ (the power set of $\displaystyle A$) such that two elements are in the same equivalence class under $\displaystyle R$ if and only if they have the same image under $\displaystyle f$. Can you think of a natural choice for the image under $\displaystyle f$ of an element from $\displaystyle A$?
What does this have to do with anything? Unless you really meant $\displaystyle 2^{A\times A}$.

5. No. The function $\displaystyle f:A\rightarrow \mathcal{P}(A)$ given by $\displaystyle x \mapsto [x]$ does exactly what the question asks.

6. Originally Posted by topspin1617
No. The function $\displaystyle f:A\rightarrow \mathcal{P}(A)$ given by $\displaystyle x \mapsto [x]$ does exactly what the question asks.
Touche sir, touche. I misread the question.