Pardon my ignorance, but isn't TREE(3) a finite number?

Question

Pardon my ignorance, but isn't TREE(3) a finite number? -Dylan Thurston

It is my understanding as well that TREE(3) is finite (Proof that TREE(n) where n >= 3 is finite?).

However, I have seen statements such as:

TREE(3) is known to exceed the $\Gamma_0$-level, which is much higher than the $\epsilon_0$-level. -Source

It is also my understanding that $\omega < \epsilon_0 < \Gamma_0$ (where $\omega$, $\epsilon_0$ & $\Gamma_0$ are transfinite).

If $\Gamma_0<TREE(3)$, wouldn't that imply TREE(3) is transfinite?

Answer 1

This is indeed a confusing bit of language. $TREE(3)$ is indeed finite. I think the later comment by Peter clears things up:

"For example, Graham's number is approximately $f_{\omega+1}(64)$, so we say that Graham's number is at level $f_{\omega+1}$, or [for] short at the $\omega+1$-level."

Basically, we want to say that a number is "at level $\kappa$" if it is $f_\kappa(s)$ for some "small" $s$ (e.g. here $64$ is considered small). Note that this is a subjective distinction - what's the least non-small number? :P - but it still gives a sense of the size of the number involved. The point, roughly, is to say: $TREE(3)$ is so huge that, even with a special symbol for $f_{\Gamma_0}$, it is still infeasible to express $TREE(3)$.

There are various precise versions of this - e.g. the statement "$f_{\Gamma_0}(3)<TREE(3)$" (per Andres) is a perfectly precise statement, and (I believe) is known to be true - but I think it's better to view this as a general piece of descriptive language