## FANDOM

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The pressing question: what does $$\alpha \uparrow\uparrow \beta$$ mean for ordinals $$\alpha$$ and $$\beta$$? An ideal solution must suit the following criteria:

• Be total and closed over $$\text{On}$$
• Be compatible with finite arrow notation
• Provide fundamental sequences for all suitable ordinals
• Be as internally consistent as necessary

We can assume the following things:

• $$\alpha \uparrow\uparrow 1 = \alpha$$
• $$\alpha \uparrow\uparrow (\beta + 1) = \alpha \uparrow (\alpha \uparrow\uparrow \beta)$$ for $$\beta < \omega$$
• $$\alpha \uparrow\uparrow \beta = \sup\{\gamma < \beta: \alpha \uparrow\uparrow \gamma\}$$
• $$(\alpha \uparrow\uparrow \beta)[n] = \alpha \uparrow\uparrow \beta[n]$$

The second and third "axioms" give us $$\omega \uparrow\uparrow \omega = \varepsilon_0$$ as we'd expect. But what's $$\omega \uparrow\uparrow (\omega + 1)$$? If we blindly apply the second rule ignoring the condition that $$\beta < \omega$$, we get $$\omega \uparrow\uparrow (\omega + 1) = \omega \uparrow\uparrow \omega$$, which is clearly not what we want! The hole in our definition is what $$\alpha \uparrow\uparrow (\beta + 1)$$ is for infinite $$\beta$$.

Let's look at what $$\varepsilon_1$$ should be. We could deal with either of the fundamental sequences $$\varepsilon_0 + 1, \omega^{\varepsilon_0 + 1}, \omega^{\omega^{\varepsilon_0 + 1}}, \ldots$$ (A) or $$\varepsilon_0, \varepsilon_0^{\varepsilon_0}, \varepsilon_0^{\varepsilon_0^{\varepsilon_0}}, \ldots$$ (B) Ideally we would want to equate these with $$\omega \uparrow\uparrow (\omega + n)$$ for finite $$n$$.

Fundamental sequence A has a simple pattern, but it starts with $$\varepsilon_0 + 1$$ and not $$\varepsilon_0$$. Fixing this breaks the pattern in an ugly way:

• $$\alpha \uparrow\uparrow (\beta + 2) = \alpha \uparrow (\alpha \uparrow\uparrow (\beta + 1))$$
• $$\alpha \uparrow\uparrow (\beta + 1) = \alpha \uparrow ((\alpha \uparrow\uparrow \beta) + 1)$$ for limit ordinals $$\beta$$

So yeah, that sucks. Sequence B looks nicer, being directly reliant on exponentiation. But unfortunately, its definition isn't directly inductive. How do we compute $$\varepsilon_0^{\varepsilon_0^{\varepsilon_0}}$$ from $$\varepsilon_0^{\varepsilon_0}$$? The process of fixing this is also a bit cumbersome:

• $$\alpha \uparrow\uparrow (\beta + n) = (\alpha \uparrow\uparrow \beta) \uparrow\uparrow (n + 1)$$ for $$\beta \geq \omega$$, $$n < \omega$$

Nevertheless, it's far better than the above.

We can jump ahead a little and apply this definition to arrow notation in general:

• $$\alpha \uparrow^k 1 = \alpha$$
• $$\alpha \uparrow^{k + 1} (\beta + 1) = \alpha \uparrow^k (\alpha \uparrow^{k + 1} \beta)$$ for $$\beta < \omega$$
• $$\alpha \uparrow^{k + 1} (\beta + n) = (\alpha \uparrow^{k + 1} \beta) \uparrow^{k + 1} (n + 1)$$ for $$\beta \geq \omega$$, $$n < \omega$$
• $$\alpha \uparrow^k \beta = \sup\{\gamma < \beta: \alpha \uparrow^k \gamma\}$$
• $$(\alpha \uparrow^k \beta)[n] = \alpha \uparrow^k \beta[n]$$

lending to the following (annoying) Infinite Catastrophic Rule:

• Infinite Catastrophic Rule. If there is no limiter and $$p > \omega$$:
• Let $$\alpha + n = p$$, where $$\alpha$$ is a limit ordinal and $$n < \omega$$.
• Define $$B$$ as $$A$$, but with $$B(1) = \alpha$$.
• Define $$A'$$ as $$A$$, but with $$A'(0) = v(B)$$ and $$A'(1) = n + 1$$.
• $$v(A) = v(A')$$.
In the interest of full disclosure, an earlier version of this post had me claiming that $$\omega \uparrow\uparrow (\omega + 1) = \varepsilon_1$$. I've removed it not because I'm embarrassed (although I am), but because I'd like to make room for this new content.