# Isn't x/E = y/ E ⇔ x E y deduced, not just x/E = y/ E ⇒ x E y from (a) x/ E≠ Ø, (b) x/E ∩ y/ E ≠ Ø ⇔xEy?

"Theorem 3. Let $$\mathscr E$$ be an equivalence relation on a nonempty set $$X$$. Then
(a) Each $$x/\mathscr E$$ is a nonempty subset of $$X$$.
(b) $$x/\mathscr E \cap y/\mathscr E \neq \emptyset$$ if and only if $$x\mathscr Ey$$.
(c) $$x\mathscr Ey$$ if and only if $$x/\mathscr E = y/\mathscr E$$"

In the following proof of (c) I don't think it have to be as lengthy as like that

"Proof of Theorem 3

(c) It follows immediately from (a) and (b) above that x/$$\mathscr E$$ = y/$$\mathscr E \Rightarrow x \mathscr E$$ y

We need to prove that x$$\mathscr E$$y $$\Rightarrow x/\mathscr E = y/\mathscr E$$

Let x$$\mathscr E$$y. Then
$$z\in x/\mathscr E \Rightarrow z\mathscr E x$$ Def. 6
$$z\mathscr Ex$$ $$\land$$ $$x\mathscr E y$$ $$\Rightarrow$$ $$z\mathscr Ey$$ $$\space\space\space\space\space\mathscr E$$ is transitive.
$$\Rightarrow z\in y/\mathscr E$$ Def. 6
Since $$z$$ is arbitrary, it follows that $$x/\mathscr E \subseteq y/\mathscr E$$. A similar argument gives $$y/\mathscr E \subseteq x/\mathscr E$$; hence $$x/\mathscr E \subset x/\mathscr E$$ "
Source: Set Theory by You-Feng Lin, Shwu-Yeng T.Lin

The proof of Theorem 3 (c) can be shortened to the following:

$$x/\mathscr E = y/\mathscr E$$ $$\Leftrightarrow$$ $$x/\mathscr E \cap y/\mathscr E = x/\mathscr E \neq \emptyset$$ by Idemp, (a)
$$\space\space\space\space\space\space\space\Leftrightarrow$$ $$x\mathscr E y$$ by (b)

So there's no need to prove that $$x\mathscr E y$$ $$\Rightarrow$$ $$x/\mathscr E = y/\mathscr E$$, isn't it?

FYI

Idempotency law of set (Idemp.): P$$\bigcap P\Leftrightarrow P$$, P$$\bigcup P\Leftrightarrow P$$

Definition of equivalence class of $$x$$ in $$X$$ in symbols, $$x/\mathscr E=\{y∈X| y \mathscr E x\} = \{y\in X | (y, x) \in\mathscr E\}$$

"

Definition 6. Let $$\mathscr E$$ be an equivalence relation on a nonempty set X . For each $$x\in X$$, we define

X/​$$\mathscr E$$={y∈X∣y$$\mathscr E$$x}

which is called the equivalence class determined by the element x.

The set of all such equivalence classes on X is denoted by X/$$\mathscr E$$; that is, X/$$\mathscr E$$={x/$$\mathscr E$$∣x∈X} X/ɛ={x/$$\mathscr E$$$$x\in X$$}.

The symbol X/ɛ is read "X modulo $$\mathscr E$$," or simply "X mod $$\mathscr E$$".

• Related post: math.stackexchange.com/questions/238940/… Jan 25, 2016 at 13:51
• I think that Michael Hardy's comment on your previous post applies to the formatting of this post, too. Jan 25, 2016 at 13:53
• The right-to-left implications in your attempt at a shorter proof of (c) reads $x E y\to x/E\cap y/E=x/E\ne \phi \to x/E=y/E.$ PROBLEMS: From (a) and (b) you have $x E y \to x/E\cap y/E \ne \phi,$ but how does $x/E \cap y/E \ne \phi$ imply that $x/E\cap y/E=x/E$ unless you already know what you are trying to prove (which is $xEy \to x/E=y/E$ )? Jan 25, 2016 at 22:23
• @user254665 Aha. The right to left implication, doesn't suppose x/E=y/E, right? That cleared up my problem! Now I understand the proof Jan 26, 2016 at 5:16
• Exactly. You will find that partitions and equivalence relations are a widely used, very useful tool in many subjects. Jan 26, 2016 at 20:56

Comment

From Def. 6 we have that: if $x \mathrel{\mathscr{E}}y$, then $x \in y/\mathscr E$.

From Th. 3(a) we have that: $x \in x/\mathscr{E}$

Thus, we have that:

if $x \mathrel{\mathscr{E}}y$, then $x \in x/\mathscr{E}\cap y/\mathscr{E} \neq \emptyset$.

Thus, how to conclude from: $x/\mathscr{E}\cap y/\mathscr{E}\neq \emptyset$ that $x/\mathscr{E} = y/\mathscr{E}$ ?

Consider the sets $A = \{ c \}$ and $B = \{ b, c \}$; we have two non-empty sets whose intersection is: $A \cap B = \{ c \} = A \ne \emptyset$.

But $A \ne B$.

$P \cap P = P$;
• Did you check the Theorem (3) and my reason? $$"x/E = y/E ⇔ [x/E⋂ y/E = x/E] ≠∅ ⇔ xEb"$$ Jan 25, 2016 at 10:54
• @buzzee - but from Theorem 4 [page 42] (b) The idempotency laws: $A U A = A$ and $A ∩ A = A$, it does not follows that : "if $A ∩ B = A$, then $A=B$. Jan 25, 2016 at 11:06