The Von Neumann universe $V$ satisfies ZFC, and there are other models within $V$ that are non-standard and satisfy ZFC. If we look at one of these non-standard models $M$ with a non-standard model of PA $P$, then internally to $M$ we have a notion of 'numbers' in $P$.

Consider a non-standard number $H$ in $P$. I want to understand what a set $A$ (in the non-standard model of ZFC $M$) of cardinality $H$ looks like. What will such a set's external (i.e., w.r.t. $V$) cardinality be? What about a set $B$ of cardinality $H+1$? It seems like you could just take the union of $A$ with some element of $P$ not in $A$ to get a set of cardinality $H+1$.

I am seeing a few ways to resolve this:

  1. Sets of nonstandard cardinalities don't exist within $M$.

  2. $A$ and $B$ exist in $M$; they have different cardinalities within $M$. However, with respect to $V$, they are both infinite and of the same cardinality. The only reason they have different cardinalities in $M$ is that the bijection between $A$ and $B$ is not within $M$.

  3. $A$ and $B$ are infinite and have different external cardinalities, but you can't take the union of $A$ and some element of $P$ to get something of cardinality $H+1$ for some reason.

  4. $A$ and $B$ are infinite and have different external cardinalities, but taking the union of $A$ and some element of $P$ is somehow externally not just adding one thing to $A$, but enough things to change its cardinality.

Which of these, if any, is the right answer?

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    $\begingroup$ Let $\mathbb M=(M,E)$ be a non-standard model that satisfies $(ZFC,\in). $....If $h,x \in M$ then, because $\mathbb M$ satisfies Extensionality and Union , we have $(\{x\})^{\mathbb M}=\{x\}$ and $(h\cup \{x\})^{\mathbb M}=h\cup \{x\}.$.... But if $h$ is a non-standard positive integer of $\mathbb M$ (or of $P$) then $h$ is an infinite set so the cardinal of $h\cup \{x\}$ equals the cardinal of $h$. However if $\neg (xEh).$... (i.e. $\mathbb M\Vdash (x\not \in h$)...then there does not exist $f\in M$ such that $\mathbb M\Vdash$ "$f$ is a bijection from $h$ to $h\cup \{x\}$". $\endgroup$ – DanielWainfleet Aug 31 '17 at 3:07

Option 2 is correct: $M$ doesn't see the bijection that exists externally (= in $V$).

Specifically, let's take $P$ to be $M$'s actual version of $\mathbb{N}$ (I think this is what you had in mind anyways) and make our $A$ and $B$ really simple: let $a$ be some nonstandard natural number in $M$, and set $A=\{n\in\mathbb{N}^M: M\models n<a\}$ and $B=\{n\in\mathbb{N}^M: M\models n\le a\}=A\cup\{a\}$. Then:

  • In $M$, both $A$ and $B$ are finite and $A\subsetneq B$; so in $M$ there is no bijection between $A$ and $B$.

  • In $V$, both $A$ and $B$ are infinite, and $B$ is $A$ plus one element, so there is a bijection $f:A\rightarrow B$ in the true model $V$.

Another example of this phenomenon, which doesn't involve arithmetic, is: suppose $M$ is a countable transitive submodel of $V$ (ok fine, we need a weak hypothesis to guarantee such an $M$ exists, but ignore that for now). Then $\omega_1^M$ is an ordinal which in $V$ is countable; the reason $M$ doesn't see a bijection between $\omega_1^M$ and $\omega^M$ (exercise: $\omega^M=\omega$, since $M$ is transitive and hence well-founded) is because the countability of $\omega_1^M$ is only seen outside of $M$.

Incidentally, the following is a great exercise for becoming comfortable with internal vs. external reasoning.

Suppose that $M$ is a countable elementary (not transitive) submodel of $V$. Show that (i) $\omega_1^M=\omega_1$ and $\omega^M=\omega$, (ii) $Ord^M$ is not closed downwards in $V$ (and so we can conclude that there are no countable elementary transitive submodels of $V$), but (iii) $\omega_1\cap M$ is closed downwards.

Part (i) is immediate from elementarity and part (ii) follows from part (i); part (iii), however, takes some thought.

This isn't an exercise about nonstandard models, of course - everything we're talking about is well-founded - but it is all about being careful about what model believes what statement.

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    $\begingroup$ Do you really need choice for the bijection in $V$? Can't you just map $0$ to $a$ and shift the standard naturals (which $V$ can see) down by $1$? $\endgroup$ – Alex Kruckman Aug 23 '17 at 23:34
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    $\begingroup$ Also, in the example (not the exercise), you didn't mean to say "elementary", right? (Such a model doesn't exist, by the exercise!) $\endgroup$ – Alex Kruckman Aug 23 '17 at 23:37
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    $\begingroup$ @AlexKruckman Yes, on both points. For the former, though, what I was getting at was the statement "An infinite set + one element is in bijection with the original infinite set" is not ZF-provable; but you're right that the particular application here is. $\endgroup$ – Noah Schweber Aug 24 '17 at 1:04

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