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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.

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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.


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" $S(x)$.

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  • $\begingroup$ This I understand. However, when Halmos uses notations like ${x:S(x)}$, he's not explicitly stating the set $A$. Does he only tacitly do this when there's no confusion that possibly arise (such as when $S(x)$ stands for $x\neq x$)? $\endgroup$ – Macrophage Jan 18 at 10:48
  • $\begingroup$ Thank you for adding extra explanations about formulas. However, I'm not sure if that address my problem. With $A(x)=x \notin x$ we can still find a set B for any set A...so how does Halmos say what I quoted. still didn't get it. $\endgroup$ – Macrophage Jan 18 at 14:33
  • $\begingroup$ But what I read is that in the section for the Axiom of Specification he used that sentence on any arbitrary set A to prove there's a set B that does not belong to A? So from any arbitrary set A, this B should be an existent set that can be induced with $S(x)=x \notin x$? $\endgroup$ – Macrophage Jan 18 at 14:45
  • $\begingroup$ I'm not sure what is the set he specifies for the Axiom in the section for unordered pairs where I quoted the sentence in my question, though. $\endgroup$ – Macrophage Jan 18 at 14:51
  • $\begingroup$ @Macrophage - now the issue is about Axiom of Pairing ? This is a new one: it asserts that, given two sets $a$ and $b$, it exists a third set - call it $A$ - that contain them. Again, the following comment is a little bit subtle... Given Pairing and using Spec, Halmos proves that exists a set with only $a$ and $b$. Pairing asserts that $A$ contains $a$ and $b$ but not asserts that it contains only them... $\endgroup$ – Mauro ALLEGRANZA Jan 18 at 14:59

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