An order type $\tau$ equal to its power $\tau^n, n>2$

Question

In this question we are concerned only with linear (aka total) order types. By a cardinality of an order type we understand a cardinality of an instance of this type, which obviously does not depend on selection of such particular instance. Some of the order types that occur most often and have particularly nice properties are ordinal numbers (which are order types of well-orders) and their well-known and prominent subset, finite ordinals (aka natural numbers $\mathbb{N}$).

Recall that:

• $0, 1, 2, \dots$ — unique linear order type for each finite cardinality.
• $\omega$ — the order type of $\mathbb{N}$ ordered my magnitude, the smallest infinite (denumerable) ordinal.
• $\omega_1$ — the order type of the set of all countable ordinals, ordered by "is initial segment of" relation. It is the smallest uncountable ordinal, the initial ordinal of the 2nd infinite cardinal $\aleph_1$.
• $\eta$ — the dense countable order type of rational numbers $\mathbb{Q}$ ordered my their magnitude, which is incidentally an order type of any dense denumerable linear order without first and last elements.

The sum and product of order types are natural generalizations of (and are consistent with) the sum and product of ordinals, which we consider well-known. These operations are associative, but in general, not commutative. For example: $\eta+1 \ne 1+\eta\ne\eta$, but $\eta+1+\eta=\eta+\eta=\eta=\eta\cdot\eta=\eta\cdot\omega=(\eta+1)\cdot\omega\ne\omega\cdot\eta$ (the last order being not dense). Of course $\eta\cdot\omega_1\ne\eta$ because of different cardinality, although every proper initial segment of $\eta\cdot\omega_1$ is $\eta$ or $\eta+1$. Note that $\eta\cdot\omega_1=(\eta + 1)\cdot\omega_1\ne(1+\eta)\cdot\omega_1$, and even $(1+\eta)\cdot\omega_1\ne1+\eta\cdot\omega_1$, although $(1+\eta)\cdot\omega=1+\eta$. We agree that a positive integer power of a order type is just a syntactic shortcut for repeated multiplication.

Note that there are order types satisfying $\tau^2=\tau$, for example: $0, 1, \eta, \omega\cdot\eta$ and $\omega^2\cdot\eta$.

Question: Is there a linear order type $\tau$ such that $\tau^2\ne\tau$, but $\tau^n=\tau$ for some integer $n>2$? If yes, then of what least cardinality it can be?

Answer 1

This doesn't actually answer the question asked, admittedly, but I'm posting it as an answer since it does answer a related question.

We can ask more generally, can we find an order $\tau$ such that there exist distinct $n$ and $m$ such that $\tau^n=\tau^m$, other than those satisfying $\tau=\tau^2$? Note that if we have such a $\tau$, then either there exists some $n$ such that $\tau^n=\tau^{n+1}$, or by taking an appropriate power of $\tau$ we can find an order that actually answers the original question.

Well, as I said, I don't have an example of the latter, but I do have an example of the former. Let $\tau=\omega(\eta+1)$. Since $(\eta+1)\omega=\eta$, we get $\tau^2=\omega\eta$, and similarly $\tau^3=\omega\eta=\tau^2$. But $\omega(\eta+1)\ne\omega\eta$ as the former has a final segment isomorphic to $\omega$ and the latter doesn't (as one way to see why, note that any final segment contains a copy of $\omega2$).

So at least there exists $\tau$ with $\tau^3=\tau^2$ but not $\tau^2=\tau$, though that doesn't seem to yield an answer to the original question.

Answer 2

For any ordinal $\tau$ > 1, the function $F(n) = \tau^n$ is an increasing function, so if $\tau^2 \neq \tau$ then $\tau^2 > \tau$ and so $\tau^n \neq \tau$ for any $n$.