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.


1 Answer 1


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


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

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.

  • 1
    $\begingroup$ That is a horrible notation, ${}^\omega2$. $\endgroup$
    – Asaf Karagila
    Jul 14, 2015 at 6:34
  • $\begingroup$ Thanks for the proof; much to learn. $\endgroup$
    – Sheldon L
    Jul 14, 2015 at 8:39

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