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A set $\mathscr{U}$ is a universe if the following conditions are met:

  • For any $x \in \mathscr{U}$ we have $x \subseteq \mathscr{U}$

  • For any $x,y \in \mathscr{U}$ we have $\{x,y\} \in \mathscr{U}$,

  • For any $x \in \mathscr{U}$ we have $\mathcal{P}(x) \in \mathscr{U}$,

  • For any family $(x_i)_{i \in I}$ of elements $x_i \in \mathscr{U}$ indexed by an element $I \in \mathscr{U}$ we have $\bigcup_{i \in I} x_i \in \mathscr{U}$.

Grothendieck introduced an addition axiom $\mathscr{U}$A which says that every set $x$ is contained in some universe $\mathscr{U}$.

I've seen some authors use the concept of the successor universe $\mathscr{U}^+$ of a given universe $\mathscr{U}$. It is the smallest universe which contains $\mathscr{U}$. However, I'm not sure how to prove that such a thing exists in $\mathsf{ZFC}$ (provided that $\mathscr{U}$ exists in the first place). If we knew that for any two universes $\mathscr{U}$ and $\mathscr{V}$ we have either $\mathscr{U} \in \mathscr{V}$ or $\mathscr{V} \in \mathscr{U}$, it would be easy. But I'm not sure if we can prove that latter without showing first that universes are equivalent to $V_\kappa$ for inaccessible cardinals $\kappa$.

Edit. The question, as evident from the accepted answer is turned out to be quite trivial. I apologize for that.

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The intersection of a nonempty set of universes is a universe. Now, let $U$ be a universe, and suppose that there exists a universe $V$ with $U\in V$. Then, the set of all universes $W$ with $U\in W\subseteq V$ exists and is nonempty. Its intersection is a universe, and this is readily checked to be the smallest universe containing $U$.

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  • $\begingroup$ Oh, I forgot we could take intersections of proper classes in $\mathsf{ZFC}$ provided they are nonempty. Indeed, if $\phi(x,u_1,...,u_n)$ is a formula in the language of $\mathsf{ZFC}$ for some fixed $u_1,...,u_n$ and if $A$ is a set satisfying $\phi(A,u_1,...,u_n)$, then there is the intersection of the "class" of such sets, namely $\{ x \in A \mid \forall B, \phi(B,u_1,...,u_n) \Longrightarrow x \in B \}$. Thanks for reminding that. My question now seems too trivial for MO. I apologize. $\endgroup$ – Jxt921 Nov 18 '18 at 21:51
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    $\begingroup$ @Jxt921 I guess you mean one can take the intersection of all sets in a proper class, not that intersections $C\cap C'$ are possible, for $C,C'$ proper classes :-) And since the class if proper, it can't be empty, or it would be a set. $\endgroup$ – David Roberts Nov 18 '18 at 22:44
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    $\begingroup$ No need for classes - see my edit. $\endgroup$ – Fred Rohrer Nov 19 '18 at 10:10
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ZFC doesn't even prove that universes exist in the first place. However:

  • ZFC proves that universes are exactly sets of the form $V_\kappa$ for $\kappa$ inaccessible. In particular, "any two universes are $\in$-comparable" is provable in ZFC alone; what's not provable is that there are many universes, or indeed any.

  • The successor universe of $\mathcal{U}$ is therefore a well-defined concept, and every universe has a successor universe assuming enough universes exist in the first place: ZFC proves that there is exactly one universe of height $\kappa$ for each inaccessible cardinal $\kappa$, so to get the successor universe we just "go up to the next inaccessible." ZFC can't prove that there always is a next inaccessible, but it does prove that "every set is contained in a universe" is equivalent to "there is a proper class of inaccessibles," and that each of these implies "every universe has a successor universe."

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    $\begingroup$ The successor universe is a well-defined concept without assuming enough universes exist in the first place. Fear not the void! $\endgroup$ – Andrej Bauer Nov 18 '18 at 20:45
  • $\begingroup$ @AndrejBauer Whoops, I was writing too quickly. Fixed! $\endgroup$ – Noah Schweber Nov 18 '18 at 21:19
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    $\begingroup$ I think it's a psychological thing. Our brains don't like to concern themselves about the empty set of predators, although they do seem to be worried about the empty set of wives or husbands, as the case may be. $\endgroup$ – Andrej Bauer Nov 19 '18 at 7:22
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The question confounds definition of an object and its existence. These are two different notions.

Definition: The universe $\mathcal{V}$ is the successor of the universe $\mathcal{U}$ iff $\mathcal{U} \in \mathcal{V}$, and for all universes $\mathcal{W}$, if $\mathcal{U} \in \mathcal{W}$ then $\mathcal{V} \subseteq \mathcal{W}$.

There is no mystery in the definition. I take it that the question is really asking about existence of successor universes, in which case the title of the question should be modified.

As is well-known, in ZFC a set $\mathcal{U}$ is a universe if, and only if, $\mathcal{U} = V_{\kappa}$ for some (unique) inaccessible cardinal $\kappa$. Here $V_\kappa$ is the $\kappa$-th level of the cummulative hierachy.

Because in ZFC the the cardinals are well-ordered, it follows that a universe $V_\kappa$ has a successor if, and only if, there exists an inaccessible larger than $\kappa$, in which case the successor universe is $V_\lambda$, for $\lambda$ the least inaccessible above $\kappa$.

Nowhere in the answer did we have to assume that inaccessibles or universes exist. We observed that existence of universes is equivalent to existence of inaccessible cardinals. ZFC does not prove that inacessible cardinals exist.

The void is psychologically scary but is mathematically quite tame.

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  • $\begingroup$ You are right. I apologize for my poor wording. I edited the title and the question. First of all, I wanted to write "construct" instead of "define". Second, I wanted to prove not the existence of a universe in $\mathsf{ZFC}$, but the implication "$\mathscr{U}$ exists $\Longrightarrow \mathscr{U}^+$ exists". $\endgroup$ – Jxt921 Nov 18 '18 at 21:55
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    $\begingroup$ @Jxt921 That implication isn't provable in ZFC either - what is provable is that every universe has at most one successor, and that "every universe has a successor" is equivalent to "there is no largest inaccessible." $\endgroup$ – Noah Schweber Nov 18 '18 at 23:57

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