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The distinction between predicative and impredicative definitions is important in mathematics. As first approximation, impredicativity means circularity. Let me give you an example of an impredicative definition.

Let $V$ be a vector-space over a field $K$, and $S \subseteq V$ a set of vectors. The ${span}$ of $S$ is the intersection of all sub-vector-spaces $V'$ of $V$ that also contain $S$.

$$ \operatorname{span}(S) = \bigcap \{V'\ |\ S \subseteq V', V'\ \text{is a sub-vector-space of}\ V\} $$

In a set theory like ZF(C), this definition is impredicative because $\operatorname{span}(S)$ is itself a member of $\{V'\ |\ S \subseteq V',\ \text{is a sub-vector-space of}\ V\}$. In some sense this definition is circular. In this particular case, we can easily get around this impredicativity, for example by defining

$$ \operatorname{span}(S) = \{\Sigma _{i=1}^{n} k_i.v_i\ |\ n \geq 0, k_i \in K, v_i \in S\} $$

but it's not always that easy. For example in ZF(C) the natural numbers are often defined as follows.

$$ \mathbb{N} = \bigcap \{S \ |\ S\ \text{is an inductive set}\} $$

where $S$ set is inductive if it contains $0$ and is closed under successor. Clearly, $\mathbb{N}$ is itself inductive.

Such circularities are not considered problematic in classical mathematics, in the sense that no contradictions have ever been derived from such impredicative definitions. Nevertheless, impredicate definitions don't always sit well with constructive mathematics. This leads to my question: is it always easy to see if a definition is (im)predicative? More precisely:

Is it decidable if a formula $F$ is predicative in a theory $T$?

I'm most interested in the case where the theory $T$ is some set theory.

Note that I have not formally defined (im)predicativity. Such a definition itself appears to be difficult, but I would be happy to hear about answers to my question for any of the extant formal or informal notions of (im)predicativity.

Martin Berger
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  • In predicative second-order arithmetic (ACA), all sets that can be constructed have a normal form, namely "${ n : φ(n) }$" where $φ$ is in normal form, these being recursively defined as closed under boolean connectives and quantification over sets in normal form. So if one defines a predicative definition as one in normal form, it is indeed decidable. Depending on your choice of formal system, you can also extend this a little, such as allowing quantification over only previously defined objects. Applied to ZFC, this forbids von Neumann ordinals. – user21820 Nov 14 '16 at 07:03
  • @user21820 What does that do to the usual impredicative definition of $\mathbb{N}$? – Martin Berger Nov 14 '16 at 08:02
  • One cannot use the "intersection of all inductive sets" because that boils down to quantification over all sets. But one can first construct one inductive set $I$ (given by the axiom of infinity), and then define the natural numbers as the intersection of all inductive subsets of $I$. The question then reduces to whether or not you accept the various axioms of ZFC as being sufficiently predicative or not. In this case the most severely contentious one is the power-set axiom, which you need so as to quantify over all subsets of $I$. If you accept that notion as well-defined, then it's fine. – user21820 Nov 14 '16 at 08:23
  • And I should clarify that I meant that enforcing predicativity in ZFC forbids definition of von Neumann ordinals, and does not forbid their existence. In particular one can easily construct small von Neumann ordinals and prove that they are indeed transitive well-orderings under $\in$. – user21820 Nov 14 '16 at 08:26
  • Why can we intersect over all subsets of $I$ and get $\mathtt{N}$ in ACA, but not in ZF(C)? – Martin Berger Nov 14 '16 at 11:13
  • I don't know what you're talking about. ACA is a 2-sorted theory with intended domain $\mathbb{N}$ for individuals. If you're not familiar with it read https://www.sas.upenn.edu/~htowsner/prooftheory/ReverseMath.pdf. – user21820 Nov 14 '16 at 13:54
  • Thanks. I wasn't aware that ACA, like many type-theories, takes $\mathtt{N}$ as given. That neatly sidesteps the problem. – Martin Berger Nov 14 '16 at 15:06
  • Well I go further and say that it's not really side-stepping the problem but addressing it directly. The reason is that any formal system whatsoever is based on string manipulation, and (in any reasonable meta-system) we already assume various properties (such as string length and extracting head symbol and tail substring). The fact is that all these properties are essentially equivalent to PA, so we already are sort of taking as given a structure satisfying PA. It is thus reasonable to take N as given, as you say many type theories do. See http://math.stackexchange.com/a/1808558 for more. – user21820 Nov 14 '16 at 16:19
  • More concretely, no logician doubts the fact that PA has a concrete approximation in terms of finite strings in some physical medium such as (extendable) electronic storage. The same cannot be said for set theories in general. So a logician with a predicative bent is somewhat justified in refusing to accept anything beyond higher-order arithmetic, on ontological or epistemological grounds. However, ZFC as a formal system is certainly an object that can be studied in predicative setting. – user21820 Nov 14 '16 at 16:27

1 Answers1

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1] The usual characterization of an impredicative definition is that it defines some object (property, relation, function, etc.) by means of a quantification over some domain which, if the definition succeeds, contains that object (property, relation, function, etc.). Here quantifications will be taken to include Russellian definite descriptions, as when we define an object as the unique object such that ...

In the context of a formalized theory with typed variables, we typically implement a ban on impredicative definitions by banning definitions of things of type $t$ via formulas that involve quantifiers of type $t$. Thus, predicative second-order arithmetic is characterised by only allowing instances of the comprehension schema defining a numerical property [or set of numbers] to contain first-order quantifiers over numbers (and not second-order quantifiers over properties [sets of numbers]). And in such a context it is readily decidable by inspection whether a formula is predicative and can feature in the comprehension scheme or other kind of definition.

2] But, as the OP hints, it can be interesting to ask when an impredicative definition has a co-extensive predicative counterpart. There's surely not usually going to be a decidable routine to determine that even within a fixed formalized theory (if only because we can't usually effectively determine co-extensionality).

3] As a footnote, pace Russell, it isn't particularly helpful to think of impredicativity as a species of circularity. To use Ramsey's example, if I pick out Jane as the tallest woman in the room, that's impredicative (I pick her out by a quantification over a totality including her); but in what sense is that circular?

Peter Smith
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  • Thanks for your informative answer. Maybe self-referential is a better term in this context than circular. BTW I am not sure that Ramsey's example is impredicative. Informally, Jane is not freshly constructed, she's existing already. What we define is a predicate $tallestInRoom$ (that happens to hold of Jane). And the predicate is not defined in a self-referential way. This is quite different with the usual impredicative definition of eg. the natural numbers in set theory, where you bring about the set only by the impredicative definition. – Martin Berger May 30 '13 at 08:07
  • Another question: would you happen to have a (natural) example of a definition in ZF(C) where it's unclear if it's impredicative or not? – Martin Berger May 30 '13 at 08:08
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    Ah, but that's exactly the issue (as Gödel pointed out): should we think of e.g. a set of numbers as "brought about" by its definition or as existing there already, waiting to be picked out by us [like Jane]. Should we be constructivists or realists about e.g. sets of numbers? If you are a constructivist about $X$s, you'll care about avoiding impredicativity about $X$s; if you are a realist about $X$s then you won't. – Peter Smith May 30 '13 at 08:18
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    So the "Jane" example is an example of an unworrying impredicative way of picking something out, as we are naively realistic about Jane [unless we have been corrupted by philosophy!] But it is technically impredicative. – Peter Smith May 30 '13 at 08:22
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    I think the real reason people (used to) worry about impredicativity isn't to do with realism or otherwise, is was more because there was a lingering fear that impredicative definitions would eventually turn out to be contradictory. Time has assuaged this worry. That said, I don't particularly care about such issues. I'm interested in the syntactic shape of (im)predicative definitions. I have an intuitive idea why e.g. the impredicative definition of the natural numbers in ZF is unproblematic, and it's a syntactic criterion. – Martin Berger May 30 '13 at 08:44
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    Regarding Jane. It depends on what exactly you mean by picking out Jane. If it's something like this: $ tallest = {x \in R|\forall y \in R. (x \neq y \rightarrow h(y) < h(x))} $ where $R$ is the set of people in the room, and $h$ a function returning the height of an individual, then I would imagine that the predicate $tallest$ is not defined in an impredicative way. – Martin Berger May 30 '13 at 08:45
  • "$j$ is the $\varphi$" counts as an impredicative specification of $j$ if $\varphi$ embeds a quantifier running over a domain including $j$. Which fits the Jane example. – Peter Smith May 30 '13 at 14:03
  • Thanks, what is the set-theoretic semantics of "$j$ is the $\phi$"? – Martin Berger May 31 '13 at 07:06
  • I do not think your definition of "impredicative definition" as "a definition with quantification with range including the object to be defined" works. In ACA for example we can easily add (conservatively) a new $1$-input function-symbol $min$ such that $min(X)$ is the least element of $X$ if $X$ is non-empty and $0$ otherwise. This is because ACA satisfies induction and hence $\forall X\ ( \exists n\ ( n \in X ) \to \exists! m\ ( m \in X \land \forall n\ ( n \in X \to m \le m ) ) )$. But $min$ is certainly predicatively defined. – user21820 Nov 14 '16 at 06:50
  • And $min(X)$ too, which would be a member of $X$. If you argue that the definition of $min(X)$ (as I have in fact written) does not need to quantify over $X$, then neither does your definition of $tallest$. – user21820 Nov 14 '16 at 06:54