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I'm trying to understand ordinal arithmetic. If one had an ordered list of the some subset of countable ordinal numbers, what order would the following 6 countably infinite ordinals be in? If the following order is not correct, what is correct order and why is that the correct order?

$$\omega\;<\; \omega^2 \;<\; 2^\omega \;<\; \omega^\omega \;<\; {^\omega}2 \;<\;{^\omega} \omega$$

I know $\epsilon_0 = {^\omega} \omega$ is the largest, but is still countable, but I'm not sure where the powers of $2$ fit in versus the powers of $\omega$.

I understand why $\omega^2$ or $\omega^n$ for any finite value of $n$ needs to be countable. But, why does $\omega^\omega$ need to be countable? For cardinal numbers, $2^{\aleph_0}$ is uncountably infinite. Presumably, there would be some contradiction in mathematics if any finite ordinal arithmetic equation involving $\omega$ generated an uncountable infinity.

Sheldon L
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    What does the notation ${^\omega} 2$ mean? Never saw that (and it's hard to see what to type into Google to find it...) – David C. Ullrich Jul 13 '15 at 19:12
  • Also http://math.stackexchange.com/questions/154985/do-omega-omega-2-aleph-0-aleph-1 and http://math.stackexchange.com/questions/197863/how-to-define-countability-of-omega-omega-and-omega-1-in-set-theory and http://math.stackexchange.com/questions/598224/what-is-the-cardinality-of-omega-omega and http://math.stackexchange.com/questions/318827/ordinal-exponentiation-2-omega-omega – Asaf Karagila Jul 13 '15 at 19:13
  • The defining sequence for $^{\omega} 2$ is $2;;2^2;;2^{2^2};;2^{2^{2^2}};;2^{2^{2^{2^2}}}....$ – Sheldon L Jul 13 '15 at 19:15
  • So then in fact ${^\omega}2=\omega$. – David C. Ullrich Jul 13 '15 at 19:17
  • ok, presumably that is consistent, but it seems a bit odd, that such a fast growing sequence is regarded as smaller than $\omega^2$ – Sheldon L Jul 13 '15 at 19:18
  • I don't think this should be marked as duplicate; I would like someone to verify that size ordering is apparently: $${^\omega} 2 \qquad 2^\omega = \omega,\qquad \omega^2,\qquad, \omega^\omega, \qquad, {^\omega} \omega$$ – Sheldon L Jul 13 '15 at 19:21
  • Sheldon, the answer to your question appears there. Moreover, if you know that $\varepsilon_0$ is the largest and it is countable, how can you not know that the others are countable too? – Asaf Karagila Jul 13 '15 at 19:23
  • @AsafKaragila I wanted to know the order of the sizes – Sheldon L Jul 13 '15 at 19:23
  • Size is cardinality. They are all equal, because all of the same cardinality. If you're asking about their order as ordinals, this is indeed a different question. But in that case, you should probably edit your question to clarify that. – Asaf Karagila Jul 13 '15 at 19:25
  • @AsafKaragila They are definitely NOT all equal. My question was clearly talking about ordinal arithmetic and the relative sizes or ordinal numbers – Sheldon L Jul 13 '15 at 19:26
  • I never said they are equal. Read my comment again. – Asaf Karagila Jul 13 '15 at 19:29
  • @AsafKaragila I tried rewriting to make my question clearer. – Sheldon L Jul 13 '15 at 19:45

1 Answers1

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First note that $$ ^\omega2 := \sup \{ \underbrace{2^{2^{2^\ldots}}}_{n\text{-times}} \mid n < \omega \} = \omega. $$

[..]but it seems a bit odd, that such a fast growing sequence is regarded as smaller than $\omega^2$.

Well, that's the thing when dealing with infinities. Sometimes our intuition fails us. While $(\underbrace{2^{2^{2^\ldots}}}_{n\text{-times}})_{n < \omega}$ could be regarded as "a fast growing sequence", each of its elements is finite and therefore $\omega$ is an upper bound. As $\omega$ certainly is the least upper bound, we get $^\omega2 = \omega$.

By an analogous argument we get that $$2^\omega := \sup \{2^n \mid n < \omega \} = \omega$$

So we are left with ordering $\omega, \omega^2, \omega^\omega$ and $^\omega \omega$.

We have $$\begin{align}\omega^2 &:= \omega \cdot \omega \\ &= \sup \{ w \cdot n \mid n < \omega\} \\ &\ge \omega \cdot 2 \\ &> \omega \end{align}$$

and

$$\begin{align}\omega^\omega &= \sup \{ \omega ^n \mid n < \omega \} \\ &\ge \omega^3 \\ &= \sup \{(\omega^2) \cdot n \mid n < \omega \} \\ &\ge \omega^2 \cdot 2 \\ &> \omega^2. \end{align}$$

Finally $$\begin{align}^\omega \omega &:= \sup \{ \underbrace{\omega^{\omega^{\omega^ \ldots}}}_{n\text{-times}} \mid n < \omega \} \\ &\ge \omega^{\omega^\omega} \\ &= \sup\{\left( \omega^\omega \right)^n \mid n < \omega \} \\ &\ge \left(\omega^\omega \right)^2 \\ &= \omega^\omega \cdot \omega^\omega \\ &= \sup \{\omega^\omega \cdot \alpha \mid \alpha < \omega^\omega \} \\ &\ge \omega^\omega \cdot 2 \\ &> \omega^\omega. \end{align}$$

Combining these calculations we get the desired order: $$ \omega = 2^\omega = {^\omega} 2 < \omega^2 < \omega^\omega < ^\omega\omega $$

In general, ordinal and cardinal arithmetic are very different beasts and every ordinal arithmetic expression using only ordinals $\le \omega$ is countable.

proof (sketch)

Take a countable transitive model $M$ of a large enough fracture of $ZFC$. Every ordinal expression using only ordinals $\le \omega$ can be computed correctly inside $M$ (<- this requires some work). As $M$ only contains countable ordinals (as it is transitive), the result follows.

Stefan Mesken
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