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I wonder if there are some objects that one could define in a predicative way and also in an impredicative way. I mean, two defenitions for the same object.

Thanks.

Edit: although I did get an answer for my question from Derek Elkins. I am editing for future googlers because I recieved some comments about the topic not being cleared. Examples for objects:

  • R = {S | S ∉ S} (Russell's paradox) - Impredicative, because R is a set (in the naive sense) and S is ranging over all possible sets
  • Let x be the smallest element in S” - Impredicative, because x is a member of S
  • Let π be the ratio between the circumference and diameter of a circle” - predicative

I wanted to find a predicative defenition and an impredicative defenition for the same object.

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    What do you mean by "defining a statement"? It would be helpful if you provided some background to your question. – 2017-01-27
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    I'm NotSure anyone knows what you are asking. What is "statment"? What is "predicative" and "impredicative"? And what do you mean by "two defenitions"? There is no universally accepted definition of "predicative", and it is largely a philosophical matter unless you can give a precise **mathematical definition** for us to work with. – 2017-01-28
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    Edited my question inorder for it to be clearer. – 2017-01-28

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As briefly covered on the nLab for Grothendieck toposes, it turns out you can't prove that a Grothendieck topos (by one definition) is cocomplete in a finitist, predicative foundation. Given either W-types, including natural numbers, or impredicativity you can. The core problem is the notion of transitive closure. In a finitist, predicative mathematics, you can't define the transitive closure of an arbitrary relation. In more traditional mathematics, there are two equivalent ways of defining the transitive closure of a relation.

The first is to say the transitive closure of a relation, $R$, is the smallest transitive relation that contains $R$. The definition doesn't require any infinite sets to exist, but it does require that you can quantify over "all transitive relations containing $R$" which is impredicative in general.

The second is to say the transitive closure of a relation, $R \subseteq X\times X$, is the relation $R^*$ where $R^*(x,y)$ if and only if there exists a list of elements of $X$, $(x_i)_{i=0}^n$ such that $x = x_0$ and $x_n = y$ and $R(x_{i-1},x_i)$ for $i \in \{1,\dots, n\}$. This definition is predicative but not finitist since it's based on lists.

The nLab page on pretoposes gives a brief but decent break-down of what kinds of math you get as you assume different structure. (Of course, you don't need to start at a pretopos.) This provides one way of interpreting what "finitist" or "predicative" mathematics is.