# Ordinals that are not Von Neumann?

I am having a disagreement with a teacher of mine over a question. The problem was asking whether $V_{\kappa}$ models ZF if $\kappa$ is weakly inaccessible. After a discussion on this site I came out with the conclusion that this condition on $\kappa$ is not sufficient to say whether $V_{\kappa}\models$ 'replacement', which my teacher disagrees with.

I argued that the function sending each $\alpha\in V_{\kappa}$ to the cardinal $2^{\alpha}$ might fail replacement, should there exist $\lambda$ with $\kappa\leq 2^{\lambda}$, because the ordinals in $V_{\kappa}$ are those $<\kappa$.

• Their answer was that the von Neumann ordinals in $V_{\kappa}$ are those $<\kappa$, so this example does not work. What is going on here? I was under the impression the ordinals were the von Neumann ordinals.
• I am told $V_{\kappa}\not\models$ 'replacement' if $|\kappa|<|V_{\kappa}|$. I know that these cardinals are equal if $\kappa$ is strongly inaccessible; is it consistent that this inequality holds if $\kappa$ is weak? (if so, why?)
• What is $2^\alpha$ for $V_\kappa$ if it does not exist in $V_\kappa$? I suppose what your teacher wanted to tell you is that as the ordinals in $V_\kappa$ are only those $<\kappa$, you cannot even define $2^\alpha$ inside $V_\kappa$ and you cannot define the function $\alpha \mapsto 2^\alpha$ and so you cannot test replacement on it. – Jonathan Jan 28 '17 at 16:55
• @Jonathan I can define the function taking ordinals to the cardinal number of their power set, no? – user409172 Jan 28 '17 at 19:27
• In $V_\kappa$ there is no ordinal bijective to $\mathcal{P}(\alpha)$ if $2^\alpha \geq \kappa$. – Jonathan Jan 29 '17 at 19:06

So for example, in $V_{\omega+\omega}$ there are uncountable well-ordered sets, so there are "uncountable ordinals" there.
The von Neumann ordinal assignment is matching each well-ordered $(A,<_A)$ set a unique transitive set $\alpha$ such that $(A,<_A)\cong(\alpha,\in)$. Indeed in $V_\kappa$, the von Neumann ordinals are just $\kappa=\{\alpha\mid\alpha<\kappa\}$.
If you go over the proof of Hartogs' theorem, the proof actually doable without Replacement: Given a set $X$, we can find a well-ordered set $A$ such that $|A|\nless|X|$. So now if $\kappa$ is not a strong limit cardinal, then there is some $\lambda<\kappa$ such that $\kappa\leq2^\lambda$. This means that $\mathcal P(\lambda)\in V_\kappa$, but we can construct inside $V_\kappa$ a well-ordered which is not isomorphic to any von Neumann ordinal.
To get that $|V_\kappa|=\kappa$, you essentially need that $\kappa=\beth_\kappa$, since $|V_{\omega+\alpha}|=\beth_\alpha$. So for a sufficiently large ordinal, $\omega+\alpha=\alpha$. In particular, if $\kappa>\omega$ is a cardinal. These cardinals, $\beth$-fixed points, make a closed and unbounded proper class of ordinals. But if there are no inaccessible cardinals, then there are no regular cardinals satisfying this.