Help with the proof of Theorem 1 (Chapter 2) of Suppes' "Axiomatic Set Theory"

Question

See the proof of

Theorem 1 : $x \notin 0$

page 21 and page 22.

Anybody can help me with the first step of the proof.

I don't understand why the author uses in this step "x belongs to empty set" and "x not equal to x".

Thank you.

Answer 1

The theorem to be proved is :

$x \notin \emptyset$.

The author uses the Axiom schema of separation ( top page 21, slightly modifying the letters used ) :

$(\exists A)(\forall x)(x \in A \leftrightarrow x \in B \land \varphi(x))$.

The author consider the instance of the axiom schema with $\emptyset$ as $B$ and where formula $\varphi(x)$ is $(x \ne x)$.

The result is the formula :

$(\exists A)(\forall x)(x \in A \leftrightarrow x \in \emptyset \land (x \ne x))$.

Thus, using the Separation axiom we have proved that a set $A$ such that ... exists.

Now, consider what happens for an $x$ whatever such that $x \in A$.

From : $x \in A \leftrightarrow x \in \emptyset \land (x \ne x)$ we have : $x \in \emptyset \land (x \ne x)$ and then in turn :

$(x \ne x)$.

But this is impossible (because $(\forall x)(x=x)$ is an equlity axiom) and thus we have to conclude that our assumption : $x \in A$, leads to a contradiction.

This means that we are forced to conclude with :

$x \notin A$,

and from the fact that $x$ is an object whatever, that :

$(\forall x)(x \notin A)$.

Now we have to use Definition 1 (page 19) stating that an object

$y \text { is a set iff } (\exists x) (x \in y) \lor (y = \emptyset)$

and apply it to set $A$.

$(\forall x)(x \notin A)$ is equivalent to $\lnot (\exists x) (x \in A)$, and thus, by Disjunctive Syllogism :

$A = \emptyset$.