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I'm reading through Joel David Hamkins' set theory lecture notes. On page 14, on the subject of inaccessible cardinals and submodels of ZFC in $V$, he defines a universe cardinal to be a cardinal $\gamma$ such that $V_\gamma$ models ZFC. For example, if an inaccessible cardinal $\kappa$ exists, then $\kappa$ is a universe cardinal. But furthermore, by the Löwenheim-Skolem theorem, there are then lots of models of ZFC inside $V_\kappa$, and Hamkins says that this proves that $\kappa$ is the $\kappa$th universe cardinal. Specifically, $\{\gamma < \kappa: V_\gamma \prec V_\kappa\}$ is evidently closed in $\kappa$, and is "unbounded by Löwenheim-Skolem".

  1. This is confusing to me. The Löwenheim-Skolem theorem says that $V_\kappa$ contains submodels of arbitrary cardinality. But in order to have any universe cardinals at all, you need models of ZFC of the form $V_\gamma$, which don't seem to come from Löwenheim-Skolem. Why should these exist? Is there something you can do to a model to put it in the form $V_\gamma$?

  2. Relatedly, it's also true (this is in Kanamori's Higher Infinite) that the only models for second-order ZFC are $V_\kappa$ for $\kappa$ inaccessible. Can someone explain the difference between these two concepts? I'm finding the distinction between first-order and second-order Replacement a little hair-splitting.

I hope I've phrased this the right way -- I'm a set theory rookie. Thanks in advance!

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    $\begingroup$ First-order replacement schema is handle the definable class function, while second-order replacement axiom can handle arbitrary class function (even the given class function is not definable.) $\endgroup$
    – Hanul Jeon
    Commented May 22, 2014 at 15:13
  • $\begingroup$ @tetori: thanks. So the idea is that if $V_\kappa$ models second-order ZFC, then every chain of ordinals of length less than $\kappa$ is the image of a function, so $\kappa$ must be regular. Whereas if $V_\kappa$ just models first-order ZFC, you could evade regularity of $\kappa$ by somehow blocking all these functions from being definable in $V_\kappa$. Correct? $\endgroup$ Commented May 22, 2014 at 16:25
  • $\begingroup$ @PaulVonKaoghnett I don't think about this part so I don't assure that your arguement is correct. But by Asaf's answer says that the front part of your argument is right. $\endgroup$
    – Hanul Jeon
    Commented May 23, 2014 at 6:29
  • $\begingroup$ It is known that minimal $\kappa$ which $V_\kappa$ makes the model of ZFC is singular; see related post in Cantor's attic $\endgroup$
    – Hanul Jeon
    Commented May 23, 2014 at 6:30

2 Answers 2

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Yes, you do need a tiny additional argument to get universe cardinals below an inaccessible $\kappa$. The additional step is an elementary chain argument, which tends to occur relatively frequently.

Starting from an inaccessible $\kappa$, Löwenheim-Skolem gives a small elementary submodel $M_0\prec V_\kappa$. We can then put $M_0$ inside $V_{\alpha_0}$ for some cardinal $\alpha_0<\kappa$. Applying Löwenheim-Skolem again gives a new elementary submodel $V_{\alpha_0}\subseteq M_1\prec V_\kappa$. We then put $M_1$ into some $V_{\alpha_1}$ again. After repeating this $\omega$ many times we get intertwined chains of models $M_n$ of ZFC and initial segments $V_{\alpha_n}$ of the universe which both union up to $V_{\sup \alpha_n}$, which is elementary in $V_{\kappa}$. This implies that $\sup\alpha_n<\kappa$ is a universe cardinal.

By the way, the terminology has shifted slightly since Joel wrote those notes and the term now used for these cardinals is worldly cardinals.

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  • $\begingroup$ Thanks, this answers my question! By the way, can you take arbitrary (longer than $\omega$) unions of chains of submodels? $\endgroup$ Commented May 22, 2014 at 16:18
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    $\begingroup$ @PaulVanKoughnett Indeed. The argument is flexible enough to give worldly cardinals of arbitrary cofinality (below $\kappa$). $\endgroup$ Commented May 22, 2014 at 16:32
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    $\begingroup$ What's blocking the application of this argument to show (the false statement) that if $\kappa$ is a $\beth$-fixed point of uncountable cofinality, then there's some $\gamma$ such that $V_\gamma\prec V_\kappa$? $\endgroup$ Commented Sep 8, 2020 at 22:45
  • $\begingroup$ @JasonZeshengChen Why is that statement false? $\endgroup$ Commented Sep 9, 2020 at 0:10
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    $\begingroup$ @JasonZeshengChen Alright, this is a great question, and it took me a while to figure out. Here is where the argument fails: with the notation of the answer, the model $M_1$ is a small subset of $V_\kappa$. Even thought $V_\kappa=H_\kappa$, you can't conclude that $M_1\in V_\kappa$ if $\kappa$ isn't regular (the transitive closure of $M_1$ is a small union of small sets, but if $\kappa$ is singular it might end up having size $\kappa$ and not appearing in $V_\kappa$). And if $M_1\notin V_\kappa$, the rest of the argument breaks down. $\endgroup$ Commented Sep 9, 2020 at 18:22
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Miha gave a very good answer to the first question. Let me add a bit on the second question.

The difference between first-order and second-order replacement is subtle. First-order replacement is a schema, and it says that if $f$ is an internally definable function with a set domain, then its range is a set; whereas the second-order replacement axiom is a single axiom, and it states that whenever $f$ is a subset of $V_\kappa$ which is a function, and its domain is a set of $V_\kappa$, then its range is an element of $V_\kappa$ as well.

If $V_\kappa$ satisfies the second-order axiom, then it $\kappa$ to be regular. To see this note that:

  1. If $A\subseteq\kappa$ is an unbounded set, then $A\notin V_\kappa$.
  2. Every unbounded set of ordinals is the range of a function which enumerates it.

Now if $\kappa$ is singular, there is some $\gamma<\kappa$, and a function $f\colon\gamma\to\kappa$ which enumerates an unbounded set. This function $f$ is a subset of $V_\kappa$, and its domain is an element of $V_\kappa$, but its range is not.

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