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

$x/ \mathscr E=\{y \in X \mid y\mathscr Ex\}$

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, $X/\mathscr E=\{x/\mathscr E \mid x \in X\}$. The symbol $X/\mathscr E$ is read "$X$ modulo $\mathscr E$," or simply "$X$ mod $\mathscr E$".

"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) $x/\mathscr E \bigcap y/\mathscr E \neq \emptyset$ if and only if $x\mathscr Ey$.

(c) $x\mathscr E y$ if and only if $x/\mathscr E = y/\mathscr E$."

"Theorem 4 Let $\mathscr E$ be an equivalence relation on a nonempty set X. Then X/$\mathscr E$ is a partition of X.

[Proof] By Theorem 3(a) and Definition 6, X/​$\mathscr E$ ={x/$\mathscr E$ | $x \in $X} is a family of nonempty subsets of X. We next show that

x/$\mathscr E \neq$ y/$\mathscr E$ ⇒ x/$\mathscr E \bigcap$ y/$\mathscr E$ = $\emptyset$

by showing its contrapositive : x/$\mathscr E \bigcap y$/$\mathscr E \neq \emptyset \Rightarrow$ x/$\mathscr E$=y/$\mathscr E$.

The last assertion is a direct consequence of Theorem 3(b) and (c). Finally, we have to show that $\bigcup\limits_{x\in X} x$/ $\mathscr E$ = $X$. This is also trivial, since each x in X belongs to x/$\mathscr E$."

I don't understand the last paragraph above. If each x in X belongs to x/$\mathscr E$ doesn't it mean $\bigcap \limits_{x\in X} x$/ $\mathscr E$ = $X$?

The reasong for my thought is like the following:

"for any x", "for all x", "for any x" are universal quantifiers denoted by $\forall x$. On the other hand, "there exists x", "there is at least one x", "for some x" are existential quantifiers denoted by "$\exists x$".

"each x in X" means "for all x in X" so "each x in X" would be denoted by "$\forall x \in X$" in symbols, then when I consider the definition 2.6.6 and 2.6.7 below, universal quantifier is translated to union of sets, while existential quantifier is translated to intersection of sets.

FYI

"Definition 2.6.6 Let F be an arbitrary family of sets. The union of the sets in F, denoted by $\bigcup\mathscr F$, is the set of all elements that are in A for some $A\in\mathscr F$.​

$\bigcup\limits_{A \in \mathscr F}A$={$x\in U$|$x \in A$ for some $A\in \mathscr F$}"

"Definition 2.6.7 Let F be an arbitrary family of sets. The intersection of sets in F, denoted by $\bigcap\limits_{A\in\mathscr F}A$ or $\bigcap\mathscr F$, is the set of all elements that are in A for all $A \in\mathscr F$.​ " $\bigcap\limits_{A\in\mathscr F}A$={$x\in U$| x$\in$A for all $A\in \mathscr F$}

Source: Set Theory by You-Feng Lin, Shwu-Yeng T. Lin.

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Your statement "universal quantifier is translated to union of sets, while existential quantifier is translated to intersection of sets." is incorrect, in fact it is other way around.

For all B, x in B means that x is in every B in other words in their intersection Exists B , x in B means that x is in some B, or that it is in their union.

Usually when the quantifiers refer to elements (such as numbers) it implies some sets. Say: for every epsilon greater than zero x is smaller than epsilon means x is in every set (-infinity, epsilon), hence in the intersection (-infinity 0] or that x is smaller or equal to zero.

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  • $\begingroup$ So for "each x in X" is in every each X, in other words in their intersection? $\endgroup$ – buzzee Jan 13 '16 at 11:44
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Each $x\in X$ belongs to its own equivalence class $x/\mathscr{E}$, but not to the others. For example, consider the equivalence relation $\mathscr{E}$, "same parity", on $\mathbb{Z}$: two integers are equivalent if they are both even or both odd. There are two equivalence classes: $0/\mathscr{E}$ (evens) and $1/\mathscr{E}$ (odds). Clearly their intersection is empty - there is no number which is both even and odd.

EDIT: Perhaps another way to see this is to remember that the intersection is defined as $$\bigcap_{x\in X} A_x=\{y: \forall x\in X, y\in A_x\}.$$ Note the use of a different variable here - since "$x$" is already used as an indexing variable in the expression for the intersection, we have to use a different variable here.


Note the difference between the (true) statement $$\forall x\quad x\in x/\mathscr{E}$$ with the (false) statement $$\forall x, y\quad x\in y/\mathscr{E}.$$

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  • $\begingroup$ Isn't "each $x \in X$" a universal quantifier? $\endgroup$ – buzzee Jan 13 '16 at 11:40
  • $\begingroup$ Yes, it is - but so what? Do you understand the example I gave? $\endgroup$ – Noah Schweber Jan 13 '16 at 12:00
  • $\begingroup$ Then since that is a universal quantifier "each x in X belongs to x/$\mathscr E$" means $\bigcap\limits_{x \in X}$ x/ $\mathscr E$ = X, but in the paragraph in my book, it's written "Finally, we have to show that $\bigcup\limits_{x\in X} x$/ $\mathscr E$ = $X$. This is also trivial, since each x in X belongs to x/$\mathscr E$.". Is my book wrong? Apart from that, about your answer, where are the equivalence class $0/\mathscr{E}$(evens)? 0 is neither even nor odd. $\endgroup$ – buzzee Jan 13 '16 at 12:45
  • $\begingroup$ @buzzee First of all, 0 is indeed even (see math.stackexchange.com/questions/15556/is-zero-odd-or-even). Second, and more importantly, your book is right and you are wrong - in the expression "$\bigcap_{x\in X}x/\mathscr{E}$", the set $x/\mathscr{E}$ is dependent on $x$ - there is not a single $x/\mathscr{E}$ to which every element of $X$ belongs. (cont'd) $\endgroup$ – Noah Schweber Jan 13 '16 at 13:15
  • $\begingroup$ A better way to see this might be to represent "$\bigcap_{x\in X}x/\mathscr{E}$" as "$x_1/\mathscr{E}\cap x_2/\mathscr{E}\cap x_3/\mathscr{E}\cap . . .$" (as $x_i$ vary over the elements of $X$); each $x_i$ is in $x_i/\mathscr{E}$, but there's no reason for instance for $x_1\in x_2/\mathscr{E}$. But really, if you first understand a concrete case - like the evens/odds above - you'll get better intuition for the whole than if you just try to attack the abstract notation. $\endgroup$ – Noah Schweber Jan 13 '16 at 13:16

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