I am studying Zermelo-Frankel set theory from Jech's Set Theory book. I understood it like functions are sets but the book uses the phrase "If a class F is a function" in Axiom Schema of Replacement. Why does it call it a class F instead of a set?
Are all functions sets?
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2 Answers
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You're right that technically that's a bad phrasing. It's informal language to help motivate the axiom, which is a bit technical.
More precisely, Replacement says:
If $F$ is a class such that $(i)$ each element of $F$ is an ordered pair, and $(ii)$ for each $(x, y), (x, y')\in F$ we have $y=y'$, then [the rest of the axiom].
A class satisfying $(i)$ and $(ii)$ is called a class function; basically, Replacement says "any class function whose domain is a set, is a set" (actually it says that the range is a set, but it's easy to see that these statements are equivalent).
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I only have Jech's first edition. He is clear about the notion of class as a proxy for working with formulas because Zermelo-Fraenkel set theory has no proper classes. Fortunately, this unfortunate simplification in his presentation of replacement is accompanied by the actual form of the schema members:
$$ \begin{array}{c} \forall x \forall y \forall z [ \phi ( x, y, p ) \wedge \phi ( x, z, p ) \rightarrow y = z]\\ \rightarrow \\ \forall X \exists Y \forall y [ y \in Y \leftrightarrow (\exists x \in X) \phi( x, y, p )], \\ \end{array} $$
where $\phi ( x, y, p )$ is a given formula.
There are no proper classes in Zermelo-Fraenkel set theory. So, all functions are sets. Axiom schemas relate elements of the language to the objects of the domain. If one accepts the cardinality of a language as an epistemological limitation, schemas delimit the domain. This is why the question of which functions exist in a given model determine numerous properties for a model. A countable language can describe no more than a countable collection of functions.