# Prove the theorem of equivalence class

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 $$A_x=A_y$$?

• How to prove equality of two sets? What is the recipe? Well, you need to prove that A_x is contained in A_z and vice versa. What is the recipe to prove containment? You need to take an arbitrary element z from A_x and prove that z belongs also to A_y. Try it! – mzg147 Dec 13 '19 at 12:59

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, $$z \in A_y$$.

Hint:

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

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

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.