Prove the theorem of equivalence class

Question

Let $R$ be equivalence relation on $A$, and let $A_x$ be equivalence of $R$ which contain $x$, $$A_x=\{a\in A\mid (a,x)\in R\}.$$ Prove if $y\in A_x$ then $A_x=A_y$.

This is my effort:

Given $y\in A_x$ so $y\in A$ such that $(y,x)\in R$. Because of $R$ is equivalence relation, $$(x,y)\in R.$$ We have $x\in A$ such that $(x,y)\in R$. So, $x\in A_y$.

I'm confused and I can't prove this theorem. How we can get $A_x=A_y$?

Answer 1

It is enough to prove that $A_y \subseteq A_x$ and $A_x \subseteq A_y$.

• $A_y \subseteq A_x$: Let $z \in A_y$. Then $z \sim y$. Since $y \sim x$, we get $z \sim x$ by transitivity. Thus, $z \in A_x$.

• $A_x \subseteq A_y$: Let $z \in A_x$. Then $z \sim x$. Since $y \sim x$, we get $x \sim y$ by symmetry and so $z \sim y$ by transitivity. Thus, $z \in A_y$.

Answer 2

Hint:

Use symmetry to prove that, if $y\in A_x$, then $x\in A_y$.

Use transitivity to prove that $A_y\subseteq A_x$.

Answer 3

If $y \in A_x$, then to show $A_x = A_y$ try showing that $A_x \subseteq A_y$ and $A_y \subseteq A_x$.

For the first inclusion, if $a \in A_x$, then $(a,x) \in R$. But since $y \in A_x$, $(y,x) \in R$. Transitivity means $(a,y) \in R$ so $a \in A_y$.

I'll leave the other inclusion for you to try.