So I read in wikipedia that "It follows from the properties of an equivalence relation that $x \sim y$ $⟺$ $[x] = [y]"$, but there seems to be no further elaboration on why $x \sim y$ $⟺$ $[x] = [y]$
I believe it is the transitivity and symmetric property of the equivalence relation that leads to this if-and-only-if relationship, but I'm not sure.
Here's my attempt at a proof:
For any $x$ belongs to$ X$, the equivalence class of $x$ is {$y$ belongs to $X$ such that $x \sim y$}. Now because of the transitivity property of the equivalence relation, which is that $x \sim y$, $y \sim z$ $\Rightarrow$ $x \sim z$, this means that everything which is related to y is also related to x, which means that everything in the equivalence class of $y$ is also in the equivalence class of $x$. Also, because equivalence relations are symmetric this means $y \sim x$, $x \sim z $$\Rightarrow$$y \sim z$, therefore everything in the equivalence class of $x$ is in the equivalence class of $y$.
Is this proof correct?
Let $x\sim y$
Then $z\in[x]\Rightarrow z\sim x.$But we know that $x\sim y$.So by transitivity $z\sim y$.Thus $z\in [y]$.
And $z\in[y]\Rightarrow z\sim y.$But we know that $x\sim y$.By symmetry we have $y\sim x$.So by transitivity $z\sim x$.Thus $z\in [x]$.
So $[x]=[y]$
Conversely assume $[x]=[y]$
As $\sim$ is reflexive we have $x\sim x$ and thus $x\in[x]=[y]$.So $x\in [y]\Rightarrow x\sim y$