Originally Posted by

**ElectroSpecter** It makes sense as a statement to me, I guess I just meant it didn't seem to sufficiently convey what I was trying to show.

But, I think I understand now! So maybe this is a better proof?:

**Proof**: Let $\displaystyle f : X \to Y$ be a function. Define a relation on $\displaystyle X$ via $\displaystyle x_1 \sim x_2$ iff $\displaystyle f(x_1) = f(x_2)$.

Reflexive: Let $\displaystyle x \in X$. We see that $\displaystyle f(x)=f(x)$. So $\displaystyle x \sim x$.

Symmetric: Let $\displaystyle x_1, x_2 \in X$ where $\displaystyle f(x_1) = f(x_2)$. We see that $\displaystyle f(x_1) = f(x_2)$ and similarly $\displaystyle f(x_2)=(x_1)$. So $\displaystyle x_1 \sim x_2$.

Transitive: Let $\displaystyle x_1, x_2, x_3 \in X$ where $\displaystyle f(x_1) = f(x_2)$ and $\displaystyle f(x_2) = f(x_3)$. Transitively, we see that $\displaystyle f(x_1) =f(x_3)$. So $\displaystyle x_1 \sim x_3$.

Therfore, by definition, $\displaystyle \sim$ is an equivalence relation.