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?

Question

"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 $\mathscr E$".

Answer 1

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 $A \ne B$.

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Please, note that your version of "Idempotency" is wrong; we have:

$P \cap P = P$;

it is equality between sets, and not equivalence between sentences regarding sets.