Two exercises on set theory about Cantor's equation and the von Neumann hierarchy

Question

Good evening to all. I have two exercises I tried to resolve without a rigorous success:

1. Is it true or false that if $\kappa$ is a non-numerable cardinal number then $\omega^\kappa = \kappa$, where the exponentiation is the ordinal exponentiation?

   I found out that the smallest solution for this Cantor's Equation is $\epsilon_0$ but it is numerable. I was thinking about $\omega_1$ but i don't know how to calculate something like $\omega^{\omega_1}$ and see if it is equal or not to $\omega_1$.

2. Exists an ordinal number $\alpha > \omega$ that verify $\alpha \times \alpha \subseteq V_\alpha$ , where $V_\alpha$ is the von Neumann hierarchy?

Thanks in advance

Answer 1

1) is true and the answer to 2) is yes.

For 1): Show that if $\alpha$ is countable, then $\omega^\alpha$ is countable. For this, use this characterization of ordinal exponentiation. Also, show that ordinal exponentiation is continuous at limits: $$\omega^\tau=\sup_{\alpha<\tau}\omega^\alpha $$ for $\tau$ a limit ordinal. This gives that $\omega^{\omega_1}=\omega_1$. Similarly, if $\rho$ has infinite size $\kappa$, then $\omega^\rho$ also has size $\kappa$. This shows that for any uncountable infinite cardinal $\kappa$ (whether successor or limit), if $\mu<\kappa$ then $\omega^\mu$ has size less than $\kappa$, and therefore continuity gives us that $\omega^\kappa=\kappa$.

For 2): This is a closure argument: Given $\alpha$, let $\beta_0\ge\alpha$ be least such that $\alpha\times\alpha\subset V_{\beta_0}$. In general, given $\beta_n$, define $\beta_{n+1}$ to be least such that $\beta_n\times\beta_n\subset V_{\beta_{n+1}}$. Let $\beta_\omega$ be the supremum of the $\beta_n$. Then, by construction, $\beta_\omega\times\beta_\omega\subset\bigcup_n V_{\beta_n}=V_{\beta_\omega}$.

Answer 2

You really just have to apply the definitions, in both cases, and see what happens.

For the first one, recall that the definition in the case of exponentiation is as follows: $$\alpha^0=1;\ \alpha^{\beta+1}=\alpha^\beta\cdot\alpha;\ \alpha^\delta=\sup\{\alpha^\gamma\mid\gamma<\delta\}.$$

Since $\kappa$ is a limit ordinal then we have that $\omega^\kappa=\sup\{\omega^\delta\mid\delta<\kappa\}$. It is a nontrivial claim that one should prove first, but $|\omega^\delta|=|\delta|$ for every infinite ordinal $\delta$. Now it's simple, since $\kappa$ is a cardinal, we have that $|\omega^\delta|=|\delta|<\kappa$ for $\delta<\kappa$, and so the supremum in the definition has to be $\kappa$.

As for the second problem, note that $\alpha\times\alpha=\{\{\{x\},\{x,y\}\}\mid x,y\in\alpha\}$. So you just have to verify what properties $\alpha$ has that $\{\{x\},\{x,y\}\}$ has rank smaller than $\alpha$. (Hint: what is the rank of $\{x\}$ and $\{x,y\}$?)