# Ordered pairs and set -builder as function class

I'm trying to read "Set Theory" by Thomas Jech and i'm confused by statement in page 14 (1.14) It states:

The separation Axioms follow from the Replacement Schema.
(1) Given $\varphi$, let $F = \{(x,x):\varphi(x)\}$.
(2) Then $\{x\in X:\varphi(x)\}=F(X)$, for every $X$.

What I don't understand is how (2) follows form (1), specifically:
if we have set $S = \{e_1, e_2, e_3 \dots \}$ and if all elements are not set themselves then
form (1) and $F\cap S= F(S)$ assuming all elements of $S$ satisfy $\varphi$ we get $\{\{\{e_1\}\},\{\{e_2\}\},\{\{e_3\}\} \dots\}$, set of singleton sets. This is from definition of ordered pair and singleton set:
$(a,b) = \{\{a\},\{a,b\}\}$ then
$(a,a) = \{\{a\},\{a,a\}\} = \{\{a\},\{a\}\} = \{\{a\}\}$
but from (2) we clearly get set of original elements, or set $S$.
What do i miss here?

Edit1:
Is nontation $F= \{(x,y):\varphi(x,y,p)\}$ shorthand or convention for something? And if it is for what exactly?

I think there's some confusion over the notation "$F(X)$": by definition, $F(X)=\{y: \exists x\in X((x, y)\in F)\}$. It should then be clear that $F(X)=\{x: \varphi(x)\}$: we have $$x\in F(X)\iff \exists y\in X((x, y)\in F)\iff \exists y\in X(y=x\wedge \varphi(x))\iff x\in X\wedge \varphi(x).$$
• In other words, $F$ is the restriction of the identity to the class $\{\,x:\phi(x)\,\}$. – Hagen von Eitzen Jul 21 '15 at 6:19