2
$\begingroup$

Working in Quine's $\sf NFU$, with urelements being at least as many as sets. Formally the latter is: $|Ur| \geq |Set|$.

Where $Ur$ is the set of all urelements and $Set$ is the set of all sets. We shall reserve $Set^2$ for the set of all sets of sets.

Lets call a set small, if and only if, it is a set of sets and it is strictly smaller in cardinality than the set of all double singletons of urelements.

$\operatorname {small}(x) \iff x \in Set^2 \land |x| < |\{ \{\{y\}\} \mid y \in Ur\}| $

Is it consistent with $\sf NFU$ for the Category of those small sets to be Cartesian Closed? And thus constitute a topos.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
11
$\begingroup$

No. The same obstruction to cartesian closedness exists as in the case of the category of all sets and functions in NFU. In brief, the problem is that the set of functions from a one element set to A is not the same size as A in NFU, and this remains true with the bound on size you adopt: the bound must force all sets involved to be cantorian, at least.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Cartesian closedness implements currying, and currying is inherently unstratified and does not work in general in NFU. $\endgroup$
    – Randall Holmes
    Commented Oct 25 at 16:45
  • 1
    $\begingroup$ The point being that the collection of small sets that you describe contains sets which are not cantorian and so satisfy the unpleasant condition "A is not the same size as 1 -> A", which breaks cartesian closedness. $\endgroup$
    – Randall Holmes
    Commented Oct 25 at 16:50
5
$\begingroup$

Not an answer, but it is legitimately everything that the literature knows on the topic. Check out:

Category Theory with Stratified Set Theory, by Forster, Lewicki, and Vidrine.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .