Let $\sim$ be an equivalence relation on a set $A$ and let $a, b, c \in A$. Prove that if $a \in [ b ]$ and $c \notin [ a ]$, then $c \notin [b]$.

Question

Let $a,b\in A$, and assume $a \sim b$. Let $c\in [a]$. This means $c\sim a$. Since $c\sim a$ and $a\sim b$, so therefore $c\sim b$, which means $c\in [b]$.

Would this be the correct way to show a contradiction?

Answer 1

Your "proof" is incorrect. What you have shown is instead the converse, which is not equivalent to the original statement.

To prove by contradiction, we assume that the conclusion is false, i.e. $c \in [b]$. Since $a \in [b]$ and $\sim$ is an equivalence relation, we have that $[a] = [b]$, so $c \in [a]$. This contradicts our hypothesis that $c \notin [a]$.

Answer 2

Suppose that $a \in [b]$ and $c \not \in [a]$. Further, suppose for contradiction that $c \in [b]$. Then since $a \sim b$ and $c \sim b$, we have by transitivity that $c \sim a$, a contradiction, since $c \not \in [a]$. Therefore, we have that $c \not \in [b]$, as desired.