Is there a finite ordinal in a model of ZF that is not of the form $\mathbf{n}$?

Question

Let $\mathbf{U}$ be a model of ZF. The recursively defined set $\mathbf{n + 1} = \mathbf{n} \cup \{ \mathbf{n} \}$ with $\mathbf{0} = \emptyset$ is a finite ordinal with $n + 1$ elements.

The exercise asks me now to show the existence of a ZF-model $\mathbf{U}$, which contains a finite ordinal $\alpha$ not being of the form $\mathbf{n}$ for any $n \in \mathbf{N}$.

I can't quite get my head around that. Any finite totally ordered set is a chain, as can be shown by induction. So how can a finite ordinal not be isomorphic to $\mathbf{n}$?

Answer 1

Hint : assume ZF is consistent. Take $L= \{\in, =\}\cup\{c_i, i\in \Bbb{N}\}$ to be the language of set theory and some distinct constant symbols.

Is the theory ZF + $\{ c_{i+1}\in c_i \land c_i\in \omega, i\in \Bbb{N}\}$ finitely consistent ?

Answer 2

Max's approach works - it will yield a model $M$ of $\operatorname{ZF}$ such that $\mathbb N ^M$ contains a copy of $\mathbb Z$. Hence $\mathbb N^M$ can not be embedded into $\mathbb N$ and hence has to contain nonstandard natural numbers.

With a similar approach you can obtain a model $N$ of $\operatorname{ZF}$ that has natural numbers contradicting the Archimedian property:

Add a single constant symbol $c$ to your language and consider the theory

$$ \operatorname{ZF} \cup \{ \lceil c \text{ is a natural number} \rceil \} \cup \bigcup_{ n \in \mathbb N } \{ \lceil n < c \rceil \}. $$

(The formulas in $\lceil . \rceil$ are informal but can easily be translated into proper first order $\mathcal L_{\in}$ statements.)

This theory is finitely consistent (witnessed by interpreting $c$ as a large enough natural number in some $V_{\kappa}$, $\kappa > \omega$ regular). And if $N$ is a model of this theory, $c^N$ cannot be a standard natural number. In fact $$ \{ n \in N \mid N \models n < c^N \} $$ already contains a copy of $\mathbb N$.