It's pretty cool that you worked this out. Do you think this gives an enumeration of $\varepsilon_0$? I thought a bit about how you would enumerate $\varepsilon_0$. You could use the usual snaking diagonals proof of the countability of the rationals, applied over and over to enumerate our way up to $\varepsilon_0$.I think this is what he WANTS to do since what he wrote makes little sense. But he writes things very differently and I think he makes several mistakes in his post.

$\varepsilon_0 = \bigcup_{\mathbb N} {^{n}} \omega$ where $^{n} \omega$ is an $n$-fold exponential tower of $\omega$'s.

It suffices to enumerate each of the $^{n} \omega$, then enumerate the enumerations, and apply the snaking diagonals.

It's obvious how to enumerate $\omega$. Now $\omega^2$ is just $\omega$ copies of $\omega$, and you can stack them vertically and snake the diagonals.

Likewise $\omega^3 = \omega^2 \omega$, which is $\omega$ copies of $\omega^2$ and you can "stack and snake" those. And likewise for $\omega^n$. Stacking and snaking the union we have enumerated our way to $\omega^\omega = \ ^{2} \omega$. We can then convince ourselves this will work for any finite power tower, and we attain an enumeration $\varepsilon_0$.

That's the best I can conceptualize enumerating $\varepsilon_0$. It would be a lot of work to come up with a formula for it, but in principle it's along the lines of the Cantor pairing function.

Last edited: