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Reading online, it generally seems accepted that TREE(n) where n >= 3 is a finite number, but large enough to be incomputable and only has extremely loose lower bounds today. TREE(n) is the function defined by Harvey Friedman, based on Joseph Kruskal's tree theorem. A simplified definition:

"TREE(n) = the maximum length of a set of rooted trees Tm with n possible vertex labels, where Tm has m or fewer vertices, and subsequent trees do not homeomorphically embed a preceding tree."

We can trivially show that TREE(1) = 1 and TREE(2) = 3.

My question is, how can it be known that for values of n greater than 2, the series of possible trees is finite? If we can't compute the solution, how do we know that it ends? If there is an existing write up online describing proof that these numbers are finite, please link me to it, I'm finding limited reading material on the TREE function. Thanks.

Simply Beautiful Art
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Ivan G.
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    "If we can't compute the solution" We can in principle, it's just that it's so huge that the computation is impossible in the real world. – Noah Schweber Nov 12 '17 at 21:20
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    Incidentally, I strongly disagree with the downvote - it is not at all trivial to show that $TREE(3)$ is finite. – Noah Schweber Nov 12 '17 at 21:30

1 Answers1

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The finiteness of $TREE(n)$ for each $n$ is a consequence of Kruskal's tree theorem. There are various proofs of Kruskal's theorem, including a particularly short one by Nash-Williams.

Ultimately, the proof - similarly to the proof of Goodstein's theorem - amounts to showing that an appropriate partial order is well-founded by assigning invariants to elements of the partial order; the proof then concludes by showing that these invariants are in fact ordinals, and so the theorem follows from an appropriate transfinite induction principle (e.g. Goodstein's theorem follows from, and is in fact equivalent to in an appropriate sense, the well-foundedness of $\epsilon_0$). There is also a connection between such arguments and consistency proofs, beginning with Gentzen's proof of the consistency of PA from an appropriate transfinite induction principle.

Incidentally, in the sense of reverse mathematics Kruskal's theorem is quite strong; it is not provable in the system ATR$_0$, or even in the stronger system $\Pi^1_1$-CA$_0$, making it quite rare amongst standard combinatorial facts. The proof of appropriate unprovability goes by showing that Kruskal's theorem implies that certain ordinals are well-founded; these ordinals are large enough that (a la Gentzen) induction along them implies the consistency of ATR$_0$ and of $\Pi^1_1$-CA$_0$, so by Godel's incompleteness theorem neither system can prove Kruskal's theorem.

Noah Schweber
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    See also: Quickest way to understand Kruskal's Tree Theorem – Simply Beautiful Art Aug 01 '18 at 14:14
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    Is it also necessary to rule out the case that there are arbitrarily long finite games, although each finite by Kruskal's theorem? Adam P. Goucher proves in a comment section on his blog post TREE(3) and impartial games that the entire game tree of all possible sequences is finite by Kőnig's lemma. – C7X Nov 09 '23 at 04:26
  • @C7X Since I don't mention games at all in my answer, I'm not sure what your comment is asking. Can you clarify? – Noah Schweber Nov 09 '23 at 04:30
  • @NoahSchweber Sorry, by a "game" I mean a sequence $s$ of $n$-colored trees such that $s$ satisfies the node count restriction (no more than $i$ nodes in $T_i$ for any $1\leq i\leq\textrm{len}(s)$) and homeomorphic embedding property (no $1\leq i<j\leq\textrm{len}(s)$ where $T_i$ homeomorphically embeds into $T_j$). By the game tree, I mean the tree of games where a game $s$ is placed at a child node of a game $t$ iff $s$ can be obtained from the sequence $t$ by appending one $n$-colored tree to $t$. – C7X Nov 09 '23 at 05:18