2
$\begingroup$

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 $A_x=A_y$?

$\endgroup$
  • 2
    $\begingroup$ 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! $\endgroup$ – mzg147 Dec 13 '19 at 12:59
2
$\begingroup$

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, $z \in A_y$.

$\endgroup$
5
$\begingroup$

Hint:

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.