# Axiom of Specification as in Halmos' Naive set theory

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.

## 1 Answer

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" $$S(x)$$.

• 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 $x\neq x$)? – Macrophage Jan 18 at 10:48
• 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. – Macrophage Jan 18 at 14:33
• 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 $S(x)=x \notin x$? – Macrophage Jan 18 at 14:45
• 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. – Macrophage Jan 18 at 14:51
• @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... – Mauro ALLEGRANZA Jan 18 at 14:59