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Let us work in ZFC set theory. 1: We name "Very very weak universe (VVWU)" a set u such that if a and b are two member sets of u, then every function between a and b is also a member set of u; The empty set, and the singleton of the empty set are such sets. Do we know a description of VVWU's ? 2: We name "Very weak univers (VWU)" a set u that is VVWU and moreover such that if a and b are two set members of u, so is the pair {a,b}. Do we know a description of VWU ? 3: What is the least VWU having the empty set as a member set ? Gérard Lang

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    $\begingroup$ Your universe concept will depend on how you represent functions in ZFC and in particular on the ordered-pair encoding you adopt. $\endgroup$ – Joel David Hamkins Jan 10 '17 at 10:49
  • $\begingroup$ I would preferably encode functions using Kuratowski's definition of ordered pairs $\endgroup$ – Gérard Lang Jan 10 '17 at 15:49
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Unfortunately, I think that this universe concept will not be very useful, since although you add all functions from $a$ to $b$, you haven't ensured that the elements of $a$ or $b$ are in the universe, and so you won't be able to evaluate those functions while living inside that universe. That is, if you try to use this universe as a model of set theory of some kind, you won't necessarily be able to tell different functions apart; you won't be able to evaluate the functions; there will be automorphisms of the universe that fix a set $a$ but move function from $a$ to $a$ around.

Even your VWU concept suffers from this. For example, take any set $a$, an infinite set, say, and then, starting with $\{a\}$, close under the pairing operation and functions, in $\omega$ many steps. This will produce a VWU. But since pairs and functions always have larger von Nuemann rank (if one uses the usual Kuratowski pairing function), you will never add any elements of $a$. So from the point of view of the resulting univese $u$, the set $a$ has no elements at all, even though it has many different functions from $a$ to $a$. None of those functions can be evaluated inside $u$, since $u$ has no elements from the domain. So $u$ will fail to satisfy the axiom of extensionality and all kinds of other things will go wrong, if one wants to treat it as a set-theoretic universe.

To fix this kind of issue, you probably want to insist that $u$ is a transitive set, so that every $a\in u$ has $a\subseteq u$.

Incidentally, you can find some weak universe concepts in my answer to the question: What interesting results in algebraic geometry require universes?. My universe concepts, however, are considerably stronger than what you mention here.

Regarding your particular questions, however, let me try to answer.

You've defined your universe concepts by being closed under certain operations, and I think it will be harder to improve upon that to find a simpler characterization in general. Any set can be placed into such a VVWU or VWU simply by closing under those operations in $\omega$ many steps. Note, however, that all the sets added during this closure process are unordered pairs or functions. So any set that is not like that will not be added.

In the case of $\emptyset$, you inquire what is the smallest VWU $u$ with $\emptyset\in u$. As I mentioned, this depends on what you mean by function and what you mean by ordered-pair. But let us take the usual set-theoretic definitions: a function is a set of ordered-pairs with the functional property, and we'll use the Kuratowski definition of ordered pair.

In this case, it is easy to see that the class HF of hereditary finite sets is definitely a VWU containing $\emptyset$, since it is closed under pairing and any function from an HF set to another is still HF. So HF is an upper bound.

But the smallest $u$ will be considerably smaller than HF. In fact, let me argue that $u$ will not even have all the natural numbers in it. I claim it won't have the von Neumann number $3$, which is $3=\{0,1,2\}=\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}$ should be in $u$. We can get $1=\{0\}$ by an instance of pairing, and then we can get $2=\{0,1\}$ by another instance of pairing. But how are we to get $\{0,1,2\}$? It can never be added, since it is not a pair and it is not a function. So it will never be added. But your minimal $u$ will be infinite, since it will have sets of every finite rank, because in particular you add the singleton of every set you have.

Meanwhile, however, the smallest VWU $u$ containing $\emptyset$ as an element will be a transitive set, which you can prove by induction through the closing process. Whenever you add a pair, you already had the elements, and whenever you add a function from $a$ to $b$, then since you already had all the elements of $a$ and $b$, you will also add all the Kuratowski pairs, and so you will have all elements of the function. So it will be transitive.

Incidentally, some of the problematic issues could be fixed if you close your universes under the operation known as adjunction, which is $(a,b)\mapsto a\cup\{b\}$, adding an element to a set. If you start with $\{\emptyset\}$ and close under adjunction, you get the entire HF, which is a model of $\text{ZFC}_{\neg\infty}$, a very attractive theory that is bi-interpretable with Peano arithmetic.

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  • $\begingroup$ Dear JD Hamkins, thank you very much. Do I understand that my questions become more untrusting if i add the condition that u must be a transitive set ? $\endgroup$ – Gérard Lang Jan 10 '17 at 15:49
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    $\begingroup$ Yes, I think your questions become more interesting in that case. My final remarks about the minimal VWU having $\emptyset$, however, are not affected by this, since as I mentioned, it will be transitive anyway. $\endgroup$ – Joel David Hamkins Jan 10 '17 at 15:53
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    $\begingroup$ With Kuratowski pairs, the smallest VWU containing $\varnothing$ is exactly $H_3$, consisting of the sets heriditarily of cardinality at most 2. $\endgroup$ – Henning Makholm Jan 10 '17 at 19:23
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    $\begingroup$ Henning, I think you should post your comment as an answer. $\endgroup$ – Joel David Hamkins Jan 10 '17 at 22:09
  • $\begingroup$ I see only now this nice answer of Henning Makholm ! $\endgroup$ – Gérard Lang Jan 14 '17 at 22:20

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