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What axioms need to be added to ZFC2 (second-order ZFC) before the theory has a unique model (up to isomorphism)?

I was thinking: adjoin the generalized continuum hypothesis (GCH) and a statement asserting that no inaccessible cardinals exist.

goblin GONE
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2 Answers2

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We already know that the models of $\sf ZFC_2$ are exactly the sets $V_\kappa$ for $\kappa$ inaccessible, so we really just want an axiom $\varphi$ such that $\sf ZFC_2+\varphi$ would only have one inaccessible cardinal which satisfies it.

For example $\varphi$ could be "There are no inaccessible cardinals" (satisfied by the first inaccessible); or "There is exactly one inaccessible cardinal" (satisfied by the second inaccessible); and so on.

Namely if $K$ is the set of inaccessible cardinals then $\varphi$ should tell us what is $K\cap V_\kappa$ in a unique way. This is not necessarily possible when that set is complicated enough (or when $\kappa$ is large enough that all sort of crazy reflections are going on), but if we can characterize $K\cap V_\kappa$ uniquely then the axiom guarantees that at most one $V_\kappa$ can satisfy that axiom.

Some words on the issue of $\sf GCH$. Suppose that the universe of set theory $V$ satisfies $\sf GCH$, in this case it is clear that any $V_\kappa$ satisfies $\sf GCH$ as well, and in particular for inaccessible $\kappa$, which means that every model of $\sf ZFC_2$ is a model of $\sf GCH$.

If, on the other hand, $V$ does not satisfy $\sf GCH$ and the example for the failure is below the first inaccessible cardinal of $V$ then for sufficiently big $\kappa$ (which is still much smaller than the inaccessible cardinals) $V_\kappa$ fails to satisfy $\sf GCH$. In particular all the models of $\sf ZFC_2$ would.

Think of something worse, suppose there are exactly two inaccessible cardinals, $\kappa_1<\kappa_2$. $\sf GCH$ holds below $\kappa_1$ but fails at $\kappa_1$. It follows that $V_{\kappa_1}\models\sf ZFC_2+GCH$ and $V_{\kappa_2}\models\sf ZFC_2+\lnot GCH$. And $V_{\kappa_1}$ is the unique model of $\sf ZFC_2$ in which there are no inaccessible cardinals, whereas $V_{\kappa_2}$ is the unique model of $\sf ZFC_2$ in which there is exactly one inaccessible cardinal.

The reason for this is that full semantics second-order logic does not have the completeness theorem. The fact there is just one model up to isomorphism does not mean that the theory can prove or disprove any proposition in the language.

Asaf Karagila
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  • Does this answer apply to "models" in the conventional sense, or is it specifically about "inner models"? And why is the GCH not needed to remove the possibility of "weird intermediate cardinals"? – goblin GONE Mar 01 '13 at 11:11
  • @user18921: Inner models are inner models. We already know that $(M,E)\models\sf ZFC_2$ if and only if there exists an inaccessible $\kappa$ such that $(M,E)\cong(V_\kappa,\in)$. So I mean really models. This removes the need for GCH, because we really don't care how the cardinals behave, they are still "locked" underneath the inaccessible. – Asaf Karagila Mar 01 '13 at 11:14
  • Forgive me if this is a silly question, but does mean that for any such $\phi$, it holds that $\mathrm{ZFC}_2 + \phi$ proves GCH? – goblin GONE Mar 01 '13 at 11:28
  • Of course not. But [full semantics] second-order logic does not have completeness and compactness theorems. – Asaf Karagila Mar 01 '13 at 11:40
  • Just reread this answer and its really very clear. Good job! – goblin GONE Jul 14 '13 at 10:11
  • Actually, one question. When you say that the models of $\mathrm{ZFC}2$ are exactly the sets $V\kappa$ for $\kappa$ inaccessible, we're talking specifically about transitive (and therefore standard) models, right? – goblin GONE Jul 14 '13 at 10:17
  • @user18921: Yes, I just reread it as well, and it quite clear! Thanks. – Asaf Karagila Jul 14 '13 at 10:19
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    @user18921: It's not very hard to show that any model of $\sf ZFC_2$ is well-founded, so we can collapse it. Hint: If there is a decreasing sequence then we can find a subset of the universe which enumerates this sequence, by second-order replacement the sequence is in the model. – Asaf Karagila Jul 14 '13 at 10:28
  • Gotcha.$,,,!!!$. – goblin GONE Jul 14 '13 at 11:16
  • Why does given $\phi=$first inaccessible mean that $ZFC_2+\phi$ state that there are none - it seems on the face of it that there is at least one? Have I misinterpreted your notation? – Mozibur Ullah Feb 17 '14 at 09:20
  • @Mozibur: We know that all the models of $\sf ZFC_2$ are isomorphic to $V_\kappa$ for an inaccessible $\kappa$. So $\varphi$ is a statement that we can prove is satisfied by at most one inaccessible cardinal, and then $\sf ZFC_2+\varphi$ can only have a unique model up to isomorphism. Such $\varphi$ are, for example "There are no inaccessible cardinals" or "There are exactly $42$ inaccessible cardinals" and so on. – Asaf Karagila Feb 17 '14 at 09:24
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My suggestion is essentially the same as Asaf's, but phrased a little differently to avoid directly mentioning inaccessible cardinals etc.

A well-known way of making an axiomatisation of a structure categorical is to add a suitable second-order induction axiom – for instance, second-order arithmetic. So the obvious thing to try in set theory is second-order $\in$-induction:

If $U \subseteq \mathbf{V}$, and $(\forall x \in \mathbf{V} . \, x \in^{\mathbf{V}} y \to x \in U) \to y \in U$, then $U = \mathbf{V}$.

Unfortunately this doesn't actually help here: it just asserts that $\in^\mathbf{V}$ is well-founded, and so if the "real" universe has lots of inaccessibles, then there will be lots of non-isomorphic models satisfying this axiom.

The problem, as far as I can tell, is that this axiom doesn't imply that $\mathbf{V}$ can be constructed inductively from a finite set of operations. In words, it simply says:

If $U$ is a subset of $\mathbf{V}$, such that $\emptyset^\mathbf{V}$ is a member of $U$, and for all elements $y$ of $\mathbf{V}$, if $y$ is a subset of $U$ (in the appropriate sense), then $y$ is a member of $U$ as well.

To put it in type-theoretic terms, the difficulty comes from the fact that the constructors for elements of $\mathbf{V}$ are themselves parametrised by $\mathbf{V}$. So let's think about the fundamental ways of constructing sets:

  1. There are two constants: $\emptyset^\mathbf{V}$ and $\omega^\mathbf{V}$.
  2. If $x$ and $y$ are elements of $\mathbf{V}$, then so is $\{ x, y \}^{\mathbf{V}}$.
  3. If $x$ is an element of $\mathbf{V}$, then so is $\bigcup^{\mathbf{V}} x$.
  4. If $x$ is an element of $\mathbf{V}$, then so is $\mathscr{P}^{\mathbf{V}}(x)$.
  5. If $x$ is an element of $\mathbf{V}$ and $C$ is a subset of $\mathbf{V}$, then $\{ y : y \in^\mathbf{V} x \land y \in C \}^\mathbf{V}$ (or $\mathsf{sep}^\mathbf{V}(C, x)$ for short) is also an element of $\mathbf{V}$.
  6. If $x$ is an element of $\mathbf{V}$, and $F : \mathbf{V} \rightharpoonup \mathbf{V}$ is a partial function, then so is $\{ F (y) : y \in^\mathbf{V} x \land y \in \operatorname{dom} F \}^\mathbf{V}$ (or $\mathsf{repl}^\mathbf{V}(F, x)$ for short).

This suggests the following induction principle:

If $U$ is a subset of $\mathbf{V}$ and $U$ is closed under the above-mentioned constructors, then $U = \mathbf{V}$.

Informally, we are declaring that everything in $\mathbf{V}$ must be constructed using one of the above rules, and this definitely precludes the possibility of (uncountable) inaccessible cardinals in $\mathbf{V}$. Thus, by Mostowski collapse, any model of second-order ZF satisfying the above induction principle must be isomorphic to $(V_\kappa, \in)$, where $\kappa$ is the least inaccessible cardinal.

One also notes that we can economise a little and drop the $\mathsf{sep}^\mathbf{V}$ constructor without changing anything; what remains is more-or-less the definition of Grothendieck universe, at least once we erase all the $\mathbf{V}$ superscripts and add the requirement of transitivity.

Zhen Lin
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