# How can it be seen that the Von Neumann universe $V_{\omega+\omega}$ does not model the Fraenkel axiom

Consider the Von Neumann universe $V_{\omega+\omega}$. As mentioned on the Wikipedia page on Von Neumann universes, $(V_{\omega+\omega},\in)$ is a model for $\rm Z$, but not for the Fraenkel axiom of replacement.

How can it be seen this is not a model of this axiom? I have already followed the note on the Wiki page, but was not succesful in retrieving a proof of this fact.

• Look at the function $f : \omega \to \omega + \omega$ defined by $f(n) = \omega + n$. – Paul McKenney Jan 20 '16 at 16:41
• @Paul: That should have been an answer! – Asaf Karagila Jan 20 '16 at 17:37
• – Noah Schweber Jan 20 '16 at 21:53

Look at the function $f : \omega \to \omega + \omega$ defined by $f(n) = \omega + n$. This function is not a member of $V_{\omega + \omega}$, but it's definable using parameters from $V_{\omega + \omega}$, and its domain is in $V_{\omega + \omega}$. Its range is not a member of $V_{\omega + \omega}$.
You can use this argument in more generality to show that, if $V_\alpha$ satisfies replacement, and $\alpha$ is a limit ordinal, then $\alpha$ must be a regular cardinal.
• The last line is not true. The least $\alpha$ such that $V_\alpha\models\sf ZFC$ has to be of countable cofinality. It's just that there is no $\varphi$ which defines a cofinal sequence inside $V_\alpha$. (See also Worldly cardinals.) What is true, however, is that if $V_\alpha$ satisfies the second-order variant of replacement (now a single axiom) then it is necessarily the case that $\alpha$ is a strongly inaccessible cardinal. – Asaf Karagila Jan 20 '16 at 21:32