# Disjoint Equivalence

Why do equivalence classes, on a particular set, have to be disjoint? What's the intuition behind it? I'd appreciate your help

Thank you!

-

To explain the intuition, like you asked, think about what would happen if two non-equal equivalence classes were not disjoint: let $X$ be a set and let $\sim$ be an equivalence relation on $X$; now choose two elements $a$ and $b$ that have different equivalence classes. This gives two sets; $A=\{ x\in X \mid x \sim a \}$ and $B=\{ x\in X \mid x \sim b \}$. We assumed $A$ and $B$ aren't equal, but suppose they aren't disjoint, meaning there is some element $c\in A\cap B$. Then we have that $c\sim a$ and $c \sim b$, but by definition, equivalence relations are transitive, so this would imply that $a \sim b$, thus they are in the same equivalence class. So for all $x \in X$ such that $x \sim a$, it is also true that $x\sim b$. From definition, now, $A=B$, which we assumed was false.

In short, if some element is equivalent to both $a$ and $b$, then $a$ and $b$ are in the same equivalence class since the equivalence relation $\sim$ is transitive.

-

Suppose that $R$ is an equivalence relation, and $X$ and $Y$ are two $R$-equivalence classes. Suppose $c$ is an element of $X$. Then $X$ consists of just those elements which are $R$-related to $c$ (do you understand why?). Likewise suppose $c$ is an element of $Y$. Then now $Y$ consists of just those elements which are $R$-related to $c$.

Suppose $X$ and $Y$ are not disjoint, so have a common element $c$. Then $X$ and $Y$ both consist of just those elements which are $R$-related to $c$, i.e. are the same set.

So, contraposing, if $X$ and $Y$ are not one and the very same equivalence class, they have no common element.

-
With equivalence relation, if so, we allow two things to be "almost equal" (namely equal where it count, and we don't care about their other distinctive properties). So the equivalence class of an object is the class of things which are "almost equal" to it. Clearly if $x$ and $y$ are almost equal they have to have the same class of almost equal objects; and similarly if they are not almost equal then it is impossible to have an object almost equal to both.