# How can I write the Axiom of Specification as a sentence?

I began reading Paul Halmos' "Naive Set Theory", and encountered 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.

Earlier in the same section, I learned that statements in set theory should be "sentences". A sentence was defined by

There are two basic types of sentences, namely, assertions of belonging, $x \in A$, and assertions of equality, $A = B$; all other sentences are obtained from such atomic sentences by repeated applications of the usual logical operators...

A more complete definition of a sentence follows, which can be read on Google Books here: http://goo.gl/XvK2B

I tried to translate the axioms and theorems in the book into sentences, but it seems like the Axiom of Specification is not a sentence. It refers to "every condition", but I have no way to build a sentence that refers to "every condition" because the atomic sentences only refer to sets.

Is the Axiom of Specification a sentence? If not, does that mean that statements about set theory do not need to be sentences? What other sorts of statements are allowed? (I'm using "statement" colloquially since I don't know the technical term.)

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The axiom of specification is not a sentence. It's an "axiom scheme", which is to say that it is a family of sentences. This is one of the things that becomes more clear when you move to axiomatic set theory, instead of naive set theory.

For each sentence $S(x)$ that does not mention $B$, the axiom of specification includes the axiom $$\forall A \exists B \forall x ( x \in B \Leftrightarrow x \in A \land S(x)).$$

What that axiom says, informally, is that given a set $A$ and a definition $S$ of a subset of $A$, that subset actually exists. The scheme is slightly more general than my previous formula, because the scheme allows sentences with "parameters".

The restriction that $S$ does not mention $B$ is to avoid paradoxes. Otherwise we would have as an axiom (letting $A = \{0\}$ and letting $S$ be "$x \not \in B$") $$\exists B \forall x ( x \in B \Leftrightarrow x \in \{0\} \land x \not \in B).$$ That set is paradoxical - it contains 0 if and only if it doesn't contain $0$.

The reason that we cannot quantify over sentences is that set theory is formalized using the logical system of "first order logic". That system is not able to quantify over sentences. This isn't an arbitrary choice; the inability to quantify over sentences is a necessary result of certain logical properties of first-order logic that are desirable. There are other logics in which one can quantify over sentences, but these logics do not have nice properties (and some have argued these logics themselves include set theory).

All of this is explained, in great detail, in books on axiomatic set theory. One reasonable book is Levy's Basic set theory. The standard graduate textbook is Kunen's Set theory: an introduction to independence proofs, and it can be used to learn axiomatic set theory, but it is somewhat terse at the beginning and is better as a second book on axiomatic set theory in my opinion.

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Thank you, Carl. That and your comments below clear things up to the point where I realize it would take a lot of work learning axiomatic set theory to make things totally clear! I think I have the basic idea, though. – Mark Eichenlaub Nov 9 '10 at 20:17

$$\forall A, \forall S, \exists B ; x \in A, x \in S \Rightarrow x \in B \text{ and } x \in B \Rightarrow x \in A, x \in S$$

Read it as: "For all $A$, for all $S$, there exists $B$ such that, if $x$ is in $A$ and $x$ is in $S$, then $x$ is also in $B$ and, if $x$ is in $B$, then $x$ is in $A$ and $x$ is in $S$."

(Saying that a property $S(x)$ is true is usually translated as $x \in S$).

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Ah, thank you. I didn't realize that conditions actually are sets. – Mark Eichenlaub Nov 9 '10 at 8:36
@Mark Eichenlaub : If I remember correctly, this is explained later in the book. Note that this axiom could be interpreted as the axiom of intersection, since basically it describes the intersection of 2 sets. You can see in the book that the Union needs an axiom, but the intersection doesn't, since it would be equivalent to specification. – Djaian Nov 9 '10 at 9:06
This isn't right; conditions are not sets. The point of the axiom is to guarantee that definable subsets exist. – Carl Mummert Nov 9 '10 at 11:51
@Carl Mummert : you can write $S(x)$ instead of $x \in S$ if you want, but in set theory, everything is set. You are right when you say that the point of the axiom is to garantuee existence of subsets defined by conditions, however in reality these subsets are just the intersection of your beginning set and the set that describes the property / condition. – Djaian Nov 9 '10 at 12:04
There is no way to know that the set defining the property exists, if all you start with is a (compound) sentence that defines the property. It is true that every object is a set, but there are also syntactic formulas that are used to define properties of sets. This syntactic element of set theory is one of the things that is (intentionally) obscured in naive set theory. – Carl Mummert Nov 9 '10 at 12:12