Prove that $M/{\sim}$ forms a partition of $M$.

Question

My attempt:

Suppose $\sim$ is an equivalence relation on $M$. If $a \in M$, let $\bar{a}=\{m \in M \,|\, m\sim a\}$. Since each element $a$ of $M$ is in its own equivalence class $\bar{a}$, the union of all the $\bar{a}$'s equals $M$.

Questions:

1. Do I need to mention any property of equivalence relations to claim that: "Since each element $a$ of $M$ is in its own equivalence class $\bar{a}$, the union of all the $\bar{a}$'s equals $M$."? I feel that I need to bring up the reflexive property of the equivalence relations to say that since $\sim$ is a reflexive relation on $\bar{a}$, $a\sim a$.

2. By definition of partition, I need to show that all distinct equivalence classes that make up $M$ are disjoint? Any hints as to how to do this?

Answer 1

you do need the properties of equivalence relations since if $a \sim a$ is not true then you cannot be sure that the union indeed covers the whole set. also to get that the classes are disjoint, you need the symmetry and transitivity property otherwise for e.g suppose $a \sim b$ but $b \nsim a$ then $\bar{b}$ contains $a$ but $\bar{a}$ does not contain $b$ but contains $a$ which shows they are neither disjoint nor equal.

symmetry and transitivity show that classes are disjoint. suppose $\bar{b} \cap\bar{a}\neq \varnothing$ then say $c \in \bar{b} \cap\bar{a}$ so $c \sim b$ and $c \sim a$ hence $a \sim b$, i.e $\bar{b}=\bar{a}$

Answer 2

1. I think your argument is fine.
2. Go by contradiction -- assume it is not the case. Say, let $c \in \bar{a}$ and $c \in \bar{b}$ with $\bar{a} \ne \bar{b}$. Argue by symmetry and transitivity of equivalence relations that if that were the case, $\bar{a}$ and $\bar{b}$ would in fact be the same equivalence class.