Axiom of Specification as in Halmos' Naive set theory

Question

As I understand, to apply the Axiom of Specification in ZF set theory, we have to specify a set $A$ to apply it to. However, I don't quite understand why Halmos says that

In case $S(x)$ is $(x\notin x)$ or $(x=x)$, the specified $x$'s do not constitute a set.

by the end of Section 3 in his book Naive Set Theory. I want to verify what kind of set he is using to apply the Axiom of Specification. Besides, in the previous section he already showed that we can produce a set from an arbitrary set with the sentence $x \notin x$. So how come that ”the specified x’s do not constitute a set”? I would appreciate if someone could help explain this to me.

Answer 1

See page 7 :

To specify a set, it is not enough to pronounce some magic words (which may form a sentence such as "$x \notin x$"); it is necessary also to have at hand a set to whose elements the magic words apply.

We have to review the text of the Axiom of Specification :

To every set $A$ and to every condition $S(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $S(x)$ holds.

This means that we can use the formula $x \notin x$ as "condition $S(x)$" only in conjunction with and already existent $A$ to "carve out" the subset $B$ of $A$ of all and only those elements $x \in A$ that satisfy the condition.

In symbols :

$B = \{x : x \in A \text { and } S(x) \}$.

Having said that :

what kind of set he is using to apply the Axiom of Specification ?

A set $A$ whatever, provided that we already know that it exists. This is the gist of the axiom : it does not produce new sets out of nothing, but allows use to manufacture them only as subsets of alredy existsing sets.

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Regarding Halmos' explanation :

in this notation, the role of $S(x)$ is now played by $x \notin x$.

It follows that, whatever the set $A$ may be, if $B = \{x : x \in A \text { and } x \notin' x \}$, then, for all $y$,

$y \in B \text { if and only if } (y \in A \text { and } y \notin y)$.

we have only to note that the "condition" $S(x)$ is expressed by a formula whatever of the language of set theory, with a free variable $x$ (a "parameter").

how come that ”the specified x’s do not constitute a set”?

Because in ordet to specify a set we have to consider an already existsing set $A$ and a "specifying condition" $S(x)$.