9

The incompleteness theorem prevents PA from pinning down $\Bbb N$ uniquely. However, each model of each set theory brings its own $\Bbb N$, i.e. each model of e.g. ZFC "knows" (up to isomorphism) what the natural numbers are without ambiguity (for example by some second-order definition). But the incompleteness strikes for ZFC too. So this does not really help. Different models of $\Bbb N$ inside different models of ZFC could look wildely different. Or couldn't they?

Could it be that any two representations (e.g. second-order PA models) of $\Bbb N$ inside different models of ZFC (or another reasonable set theory) are isomorphic?

Does this question make sense? Can one even talk about isomorphism between structures defined inside other models, where also the functions, relations etc. making up these (inner) models are interpretations w.r.t. to these (outer) models?

The other way around:

Are there two models of ZFC (or another reasonably set theory) whichs representation of $\Bbb N$ are non-isomorphic?

M. Winter
  • 29,928
  • If two sets $A$ and $B$ both represent $\Bbb N$, but are not isomorphic, then which one actually is $\Bbb N$? Doesn't any difference that make them not isomorphic make one not the natural numbers? –  May 15 '17 at 18:12
  • Note that ZFC is a first-order theory, so asking about a "second-order PA model" inside models of ZFC needs more exposition. – hardmath May 15 '17 at 18:14
  • @hardmath Second-order logic needs an underlying set theory anyway. And some set theory up this chain must be defined in first-order terms. Not right? – M. Winter May 15 '17 at 18:15
  • @vrugtehagel Thats the big crisis for me, see here. – M. Winter May 15 '17 at 18:18
  • I deleted my answer because I now understand how I misunderstood the question. – Stella Biderman May 15 '17 at 19:27
  • @StellaBiderman If you see any way to make this question clearer, please suggest! – M. Winter May 15 '17 at 19:29
  • The discussion of PA-unique vs set theoretically unique was helpful for my comprehension. – Stella Biderman May 15 '17 at 19:30

1 Answers1

6

There is a way in which two models of ZFC can have different natural numbers. For what follows, I will assume that the "standard" representation of $\mathbb{N}$ inside a model of ZFC is the model's first infinite ordinal.

Indeed, assume first of all that ZFC does indeed have a model (and therefore multiple ones). Let $M_1$ be a countable model (to get one, you can simply apply the Löwenheim-Skolem theorem) of ZFC.

Now consider an uncountable cardinal $\kappa$ and a set of distinct symbols $\{c_i\mid i \in\kappa\}$.

Consider then the theory $T=$ ZFC+$\{c_i \epsilon \omega \mid i \in \kappa\}$ which is written in the usual language of set theory (where $\epsilon$ is the relational symbol used to denote membership - here it's important to differentiate between both, because there might be some confusion) + the constant symbols $c_i$; where $z\epsilon \omega$ is an abbreviation for "$z$ belongs to the smallest infinite ordinal" (which is perfectly definable, by the axioms of ZFC).

By assumption, ZFC is consistent. Let $M$ be a model of ZFC. Let $T_0$ be a finite subset of $T$, and thus it's included in a set of the form ZFC+$\{c_i\epsilon \omega \mid i\in I_0\}$ where $I_0$ is a finite subset of $\kappa$, say of cardinality $n$. Let $z\in M$ be such that $M\models z=\omega$ (where $z=\omega$ is the abbreviation for an obvious formula). Then of course there are enough elements $a_1,...a_n$ in $M$ such that for each $k\leq n$, $M\models a_k \epsilon \omega$.

Interpret the $c_i$'s accordingly in $M$ to get a model of $T_0$.

Thus $T$ is finitely consistent, and by compactness it is consistent.

Let $M_2\models T$

Then the set $\{x \in M_1 \mid M_1\models x\epsilon \omega \}$ is countable, and the set $\{x \in M_2 \mid M_2\models x\epsilon \omega\}$ is uncountable, so the two representations of $\mathbb{N}$ can't really be isomorphic.

To get an even more striking picture, you can find models (of course, assuming ZFC is consistent) where there are uncountably long decreasing sequences of natural numbers (but the model doesn't know it- I advise you to try and show it if you know a bit of model theory), which shows that natural numbers aren't "unique" in any way (at least w.r.t. ZFC models - a philosophical approach would be to say that all these models of ZFC aren't "good" ones, but this would be leaving maths)

Maxime Ramzi
  • 43,598
  • 3
  • 29
  • 104
  • Some questions:
    1. Can we prove that the smallest infinite ordinal satisfies PA?
    2. How does $M\vDash z=\omega$ make sense, because $z$ is no symbol of the language, hence $z=\omega$ no valid formula, right?
    3. My question only makes sense for inner models of ZFC, right? Otherwise we cannot talk about isomorphism between their respective representations of $\Bbb N$.
    4. Is $T$ a model in which there are uncountably many "finite" sets? Uncountably w.r.t. the meta theory, and finite w.r.t. $T$.
     – M. Winter May 16 '17 at 10:44
  • Yes, it's quite easy actually you should try it. 2. If you'd rather, $M\models (v_1 = \omega)[z]$ which means that the formula $v_1=\omega$ is satisfied when $v_1$ is interpreted by $z$. It was simply an abbreviation. 3. Well the isomorphism we would be talking about wouldn't sit indisde the model. But nonetheless, my proof still works because you can simply pick $M_1$ a countable inner model of $M_2$, rather than an arbitrary countable model. You can even have $M_1$ elementarily equivalent with $M_2$. 4. $M_2$ is a model in which there are uncountably many finite ordinals. $T$ is a theory
    • – Maxime Ramzi May 16 '17 at 12:12