$\mathbb R$ vs. $\omega+\omega$ 
  
*
  
*Show that there is a subset of $\mathbb R$ which is order-isomorphic to $\omega+\omega$.
  
*Show that all axioms of ZF except the scheme of Replacement hold in $V_{\omega+\omega}$.
  



*

*How about $\{n/(n+1)\mid n\in\mathbb N\} \cup \mathbb N$? My idea is to associate $\{1/2, 2/3, \ldots\}$ with $\{\emptyset, \emptyset^+, \ldots\}$ and $\{1,2,\ldots\}$ with $\{\omega, \omega+1, \ldots\}$, does that work?

*I'm confused, how can any ZF axiom not hold for any $V_\alpha$? I thought the von-Neumann hierarchy described the universe of ZF sets.

 A: *

*Your proposed example works.  Remember that an order isomorphism is just a bijection that preserves order, i.e. a strictly increasing bijection.

*Certainly the axioms hold in $\bigcup_\alpha V_\alpha$ (taking the union over all ordinals).  Some axioms assert the existence of something.  If the thing whose existence is asserted is not in $V_\alpha$ for some $\alpha$, then $V_\alpha$ does not satisfy the axioms.  For example, in $V_{\omega+1}$ the ordinal $\omega$ has no successor.
A: Regarding 1. we have the following fact:
For every countable ordinal $\alpha$ there is a subset $X \subseteq \mathbb Q$ such that $(\alpha, \in)$ and $(X, <)$ are isomorphic, where $<$ is the usual strict order on $\mathbb Q$ restricted to $X$.
Proof. Consider $Y = \alpha \times \mathbb Q$ and let $\prec$ be the strict lexicographical order on $Y$. So, for any $(\beta, q), (\gamma, r) \in Y$ we have $(\beta, q) \prec (\gamma, r)$ iff $\beta \in \gamma$ or [$\beta = \gamma$ and $q < r$]. Now $Y$ is a countable, dense linear order without endpoints and thus there is an isomporphism $\pi \colon (Y, \prec) \to (\mathbb Q, <)$. Let $\rho \colon (\alpha, \in) \to (Y, \prec), \beta \mapsto (\beta,0)$ be the "natural embedding". Then $\pi \circ \rho \colon (\alpha, \in) \to \pi \circ \rho " \alpha$ is an isomorphism from $(\alpha, \in)$ to a suborder of $(\mathbb Q, <)$.
Usually the above result is proved by induction on $\alpha$, but recently it occurred to me, that Cantor's characterization of countable, dense linear orders without endpoints allowed for a shorter proof (provided that Cantor's result is known by the reader).
A: In general $V_{\alpha}$ does not satisfy ZF, and if we could prove from ZF that some $V_{\alpha}$ satisfies ZF then by Godel's First Incompleteness Theorem  we could prove $1=0$ from ZF.
For brevity let $a=\omega +\omega$  and $X=V_a.$ Assume $X$ satisfies ZF. Then we can prove (1).$a=a^X\in X.$... (2). $B(b)=(B(b))^X$ for  all $b\in a,$ where $B$ is the Beth function.... (3). $V_b=(V_b)^X$ for  all $b\in a.$ But then $X$ satisfies $\;\forall b\in a\;\exists ! c\; (c=V_b),\;$ which is an instance of Replacement, so $X$ satisfies $\;\exists d \;\forall b\in a \;(V_b\in d).$ But then $X$ satisfies$\;\exists d\; (\cup d\supset V) .$
Remark:Whether any of (1),(2),(3) actually does hold in ZF is not needed to show that $X$ does not satisfy ZF. 
In order for any $V_a$ to satisfy all of ZF it is sufficient that $a$ is a strongly inaccessible cardinal. If ZF is consistent then neither ZF nor ZFC can prove that such a cardinal exists.
