Prove that a relation is ordering relation

Question

$\beta\subset A^2$ is reflexive and transitive relation.$\alpha$ is an equivalence relation defined on A by $x\alpha y \iff x\beta y$ and $y\beta x$. A factor set (no idea what the english term is for this) $A/_\alpha$ is defined such that it contains all the classes of set A. Prove that the relation $\le$ defined by $x^\alpha \le y^\alpha \iff x\beta y$ is ordering on $A/_\alpha$.

Here is my solution so far. Obviously we need to prove that $\le$ is antisymmetric,reflexive and transitive.

Reflexive :

Since $\beta$ is reflexive if $x\in\beta$ then $x\beta x \implies x^\alpha\le x^\alpha$

Antisymmetry:

Here is where my problem occurs. We have no information whether $\beta$ is symmetric or not. Perhaps I don't even need it.

Transitivity:

Since $\beta$ is transitive then there exist $x,y,z\in A$ such that $x\beta y$ and $y\beta z \implies x\beta z$. By the definition of $\le$ from $x\beta y \implies x^\alpha \le y^\alpha $, $y\beta z \implies y^\alpha \le z^\alpha $ and , $x\beta z \implies x^\alpha \le z^\alpha$.

Are my proofs about reflexivity and transitivity enough. How would you approach proving antisymmetry?

Answer 1

What you need to prove is not asymmetry (i.e., the fact that a relation is not symmetric) but antisymmetry.

You need to prove that if $x^\alpha\leq y^\alpha$ and $y^\alpha\leq x^\alpha$, then $x^\alpha=y^\alpha$.

To do that:

1. Assume that $x^\alpha\leq y^\alpha$ and $y^\alpha\leq x^\alpha$
2. Therefore, because $x^\alpha\leq y^\alpha$, we know that $x\beta y$.
3. Also, because $y^\alpha\leq x^\alpha$, we know that $y\beta x$.

Can you conclude, from (2) and (3), that $x^\alpha=y^\alpha$?

(Remember $x^\alpha=y^\alpha$ if and only if $x\alpha y$, so you need to prove that).

Answer 2

To prove antisymmety, you must assume $X\le Y$ and $Y \le X$ and then find a way to conclude $X=Y$.

Let's set $X=x^\alpha$ and $Y=y^\alpha$. What we know is then that $x\mathrel{\beta}y$ and $y\mathrel{\beta} x$. But that means exactly that $x\mathrel{\alpha} y$.

Can you see that this implies that $x^\alpha=y^\alpha$?