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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

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mcdos
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2 Answers2

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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, $\beta_\omega\times\beta_\omega\subset\bigcup_n V_{\beta_n}=V_{\beta_\omega}$.

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Andrés E. Caicedo
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  • I blame the annoying keyboard on the tablet dock. Had I been at home, there's no chance you'd beat me to it by seven seconds! :-) – Asaf Karagila Sep 17 '13 at 17:56
  • Probably not: I spent a bit of time wondering whether to post an answer or a comment. – Andrés E. Caicedo Sep 17 '13 at 17:58
  • (I had to reboot my tablet before posting the answer, because there's something wrong with the WiFi driver.) – Asaf Karagila Sep 17 '13 at 17:59
  • Do you really need a closure argument for (2)? – Brian M. Scott Sep 17 '13 at 18:03
  • @BrianM.Scott No, of course not: For all limit ordinals $\beta$ we have $\beta\times\beta\subset V_\beta$, and this containment only holds at $0$ and limit ordinals. It just seemed fastest to argue that way, as the argument applies in many more situations. Looking at what I wrote, one simply notices that the $\beta_n$ only increase by finite amounts each time. – Andrés E. Caicedo Sep 17 '13 at 18:08
  • @Andres: Phew! You had me doubting myself for a minute. :-) Hard to decide which argument to offer here; closure arguments are awfully important, but on the other hand it’s good to understand how the rank behaves here. (By the way, do you prefer Andrés?) – Brian M. Scott Sep 17 '13 at 18:16
  • @BrianM.Scott: Thanks, I do, actually (but my keyboard is not accents friendly). – Andrés E. Caicedo Sep 17 '13 at 18:20
  • If you use Windows, you might look into Unichars; it’s what I use. – Brian M. Scott Sep 17 '13 at 18:22
  • Thanks very much for yours answers! – mcdos Sep 17 '13 at 19:01
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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 $\{x,y\}$?)

Asaf Karagila
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  • Thanks Asaf! I'm interested in proving that $|\omega^\delta| = |\delta|$ for every infinite ordinal $\delta$ but i don't figure out how to construct a bijection between them (in particular to prove the $\leq$ inequality) – mcdos Sep 17 '13 at 19:08
  • @mcdos: Assume this is true for every $\gamma<\delta$; if $\gamma+1=\delta$ then $\omega^\delta=\omega^\gamma\cdot\omega$ which is an order on the set $\omega^\gamma\times\omega$ which has size $|\omega^\gamma|\cdot|\omega|=\max{|\omega^\gamma|,\omega}=|\omega^\gamma|=|\gamma|=|\gamma+1|=|\delta|$. For the limit case note that $|\delta|\leq|\omega^\delta|$ (try and find an injection) and $|\omega^\delta|$ is at most the cardinality of the disjoint union of the sets $\omega^\gamma$ for $\gamma<\delta$, this disjoint union has $\delta$ terms, each of size $\leq\delta$, so it has size $\delta$. – Asaf Karagila Sep 17 '13 at 19:16