Tetration of Alephs

Question

Does this even mean anything?

$\underbrace{x^{x^{x^{...^{x^x}}}}}_n$

Where $n = \aleph_0$?

Because I "know" it converges when (say) $x = .5$ and $n \to \infty$

Answer 1

This is meaningful for ordinal numbers only. This operation has no meaning in cardinal arithmetic (explained below). We can define an operation $\uparrow \uparrow$ using transfinite recursion as follows:

$$\alpha \uparrow \uparrow 0 = 1$$

$$\alpha \uparrow \uparrow \beta + 1 = \alpha^{ \alpha \uparrow \uparrow \beta} $$

$$\alpha \uparrow \uparrow \gamma = \sup \{ \alpha \uparrow \uparrow \beta : \beta < \gamma \} \text{ where $\gamma$ is a limit ordinal}$$

EDIT: This is only meaningful if we are talking about this as an operation with ordinal numbers. This has no meaning for cardinal numbers. However, we can talk about this operation with the ordinal number $\omega$, for example.

The problem is that we want our hypothetical operation $2\uparrow \uparrow \omega$ or $2 \uparrow \uparrow \aleph_0$ to be a fixed point so that $2^{ 2 \uparrow \uparrow \omega } = 2 \uparrow \uparrow \omega$. But if $\aleph_\kappa$ is an infinite cardinal, then $2^{\aleph_\kappa} \ne \aleph_\kappa$ so our operation would have no fixed points.

Answer 2

Since $2^{\aleph_0} = \aleph_0^{\aleph_0}$, we can generalize and have that $\wp(κ) = κ ↑↑ 2$. Technically, $\aleph_0 ↑↑ \aleph_0 = \beth_{\omega}$ (I think someone else on this site noted this elsewhere, but I can't find the posting). You can also use Wolfram Alpha to iterate self-exponentiation of $\aleph_0$ and get the $\beth_n$ from this.

Tetration also shows up in accounts of ultrafilters, via the expression "2^2κ." See e.g. page 4 of this article.

As for 2 ↑↑ $\aleph_0$, this would depend on what $2^{\aleph_0}$ (as 2 × 2 × ...) is (note that 2 + 2 + ... as 2 × $\aleph_0$ = $\aleph_0$). Presumably, it would be some $\kappa$ greater than $\aleph_0$, but how much so? We wouldn't know...