Let us work in the standard ZFC universe $S$. Within it there also exist nonstandard models of ZFC. In our standard universe we have (standard) $\mathbb{N}$ with countable cardinality. Within the nonstandard model $L$ of ZFC we can have a nonstandard model $K$ of Peano Arithmetic that has some uncountable cardinality by the theorem of Lowenheim and Skolem.

But because the nonstandard ZFC model is within our standard model of ZFC, we can find an injective function $f$ that maps the standard set $\mathbb{N}$ to $K$. What cardinality does the image $I$ of $f$ have with respect to the nonstandard ZFC model $L$?

From what I have been able to find [Theorem 2 and following text, page 663, here], $I$ is not actually regarded as a valid set in $K$. Why is this -- is $\in^L$ different from $\in^S$? Is $\in^L$ some complex operator or is it merely $\in^S$ with added stipulations?

Or, does $I$ have some nonstandard finite cardinality in $K$, thus appearing finite with respect to $L$? What does it even mean to have a nonstandard cardinality? For example, if $\mathbb{N}$ has nonstandard cardinality $H$, then what about $H-1$? Something must have cardinality $H-1$, but then this would be a set that is not, from the perspective of the standard universe, finite; but it has cardinality less than $\mathbb{N}$.

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    $\begingroup$ "$I$ is not actually regarded as a valid set in $K$. Why is this?" This just means that the set $I$ is not an element of the nonstandard model $L$. $\endgroup$ Aug 23, 2017 at 15:41
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    $\begingroup$ what is "the standard ZFC universe" ? $\endgroup$
    – mercio
    Aug 23, 2017 at 15:50
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    $\begingroup$ @mercio $V{{}}$. $\endgroup$ Aug 23, 2017 at 15:52
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    $\begingroup$ @Zeno What does it mean for $I$ to "be a set with respect to $K$"? $K$ is a model of PA, and models of PA don't know anything about sets. $\endgroup$ Aug 23, 2017 at 16:50
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    $\begingroup$ I know you're just throwing letters, but L and K have special meaning in set theory. Especially when talking about "models of set theory". $\endgroup$
    – Asaf Karagila
    Aug 24, 2017 at 7:40

1 Answer 1


The injection $f$ is an element of $V$, not $L$ (incidentally, you really shouldn't use "$L$" to denote a nonstandard model of ZFC). There is no reason to believe that its image is in $L$ at all. Its image will be a subset of $L$, but not all subsets of $L$ correspond to elements of $L$. (Brute force way to see this: in $V$, $L$ has $2^{\vert L\vert}$-many subsets, but clearly only $\vert L\vert$-many of them can correspond to elements of $L$.)

Basically: whenever you have some object defined by referencing $V$, there is no reason to expect that it or anything defined in terms of it should live inside your other model.

To be absolutely precise: there is no reason to believe that there is some $x\in L$ such that $$\{y\in L: L\models y\in x\}=im(f).$$ (Incidentally, if you prefer we can write "$\{y\in L: L\models y\in x\}$" as "$\{y\in^V L: y\in^L x\}$.")

  • $\begingroup$ Thanks. Does your argument that not all subsets of $L$ correspond to elements of $L$ work for $V$ as well, so we have shown there exist sets that don't exist in $V$? -- I'm guessing the thing is we can't talk about $V$ as a set of things, thus can't talk about its power set. We can do it for $L$ because it is a set within $V$. Is this so? $\endgroup$
    – user474153
    Aug 23, 2017 at 20:16
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    $\begingroup$ @Zeno Yes, the issue is that $V$ isn't a set so this argument doesn't apply. (The set/class distinction here is similar to the finite/infinite distinction elsewhere: e.g. for any finite $n$ (ok fine $n>1$), there are more natural numbers $<n$ than there are even natural numbers $<n$, but this fails for $\omega$.) $\endgroup$ Aug 23, 2017 at 23:10
  • $\begingroup$ Makes sense, thanks. $\endgroup$
    – user474153
    Aug 24, 2017 at 1:16

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