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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 by 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 (or, equivalently, by $\in$ 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 also an order type of any dense denumerable linear order without first and last elements (e.g. the set of positive algebraic numbers).

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 an order type is just a syntactic shortcut for repeated multiplication.

There are some 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 $n>2$?

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For a negative answer, it will suffice to rule out countable order types like that, since any uncountable order type can be made countable in a forcing extension to which the calculation of $\tau^n$ is absolute. –  JDH Jul 24 '12 at 16:36
    
@JDH: Thanks, I updated the question. –  Vladimir Reshetnikov Jul 24 '12 at 17:39
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The "then it is not possible" part of your argument is exactly where I see one using a countable-elementary substructure. My argument was using provability of the claim, since it was applying the question in a forcing extension of the universe, a different universe. –  JDH Jul 26 '12 at 19:53
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@HarryAltman: $\omega \cdot \eta \ne 1 + \eta$ because the left side is not dense. $(\omega \cdot \eta)^2=(\omega \cdot \eta) \cdot (\omega \cdot \eta)=\omega \cdot ((\eta \cdot \omega) \cdot \eta)=\omega \cdot (\eta \cdot \eta)=\omega \cdot \eta$. Here we used $\eta \cdot \omega=\eta$ and $\eta \cdot \eta = \eta$. These identities are true because every countable dense linear order without ends is isomorphic to $\eta$. –  Vladimir Reshetnikov Aug 2 '12 at 16:12
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@MichaelGreinecker Thanks, I reposted it at mathoverflow.net/q/149117/9550 –  Vladimir Reshetnikov Nov 17 '13 at 2:22

3 Answers 3

up vote 3 down vote accepted

Such questions have been considered in the past. From W. Sierpiński's Cardinal and Ordinal Numbers, second edition revised, Warszawa 1965, p. 235: "We do not know so far any example of two types $\varphi$ and $\psi$, such that $\varphi^2=\psi^2$ but $\varphi^3\ne\psi^3$, or types $\gamma$ and $\delta$ such that $\gamma^2\ne\delta^2$ but $\gamma^3=\delta^3$. Neither do we know any type $\alpha$ such that $\alpha^2\ne\alpha^3=\alpha$." Also, from p. 254: "We do not know whether there exist two different denumerable order types which are left-hand divisors of each other. Neither do we know whether there exist two different order types which are both left-hand and right-hand divisors of each other." Of course, if $\tau^n=\tau$ for some integer $n\gt2$, then $\tau^2$ and $\tau=\tau^2\tau^{n-2}=\tau^{n-2}\tau^2$ are both left-hand and right-hand divisors of each other.

For what it's worth, here is a partial answer to your question, for a very special class of order types. By "order type" I mean linear order type. An order type $\xi$ is said to have a "first element" if it's the type of an ordered set with a first element, i.e., if $\xi=1+\psi$ for some $\psi$; the same goes for "last element".

Proposition. If $\alpha$ is a countable order type, and if $\alpha\xi=\alpha$ for some order type $\xi$ with no first or last element, then $\alpha\beta=\alpha$ for every countable order type $\beta\ne0$.

Corollary. If $\tau$ is a countable order type with no first or last element, and if $\tau^n=\tau$ for some integer $n\gt1$, then $\tau^2=\tau$.

The corollary is obtained by setting $\alpha=\beta=\tau$ and $\xi=\tau^{n-1}$ in the proposition.

The proposition is proved by a modified form of Cantor's back-and-forth argument. Namely, let $A$ be an ordered set of type $\alpha=\alpha\xi$, and let $B$ be an ordered set of type $\alpha\beta$. An isomorphism between $A$ and $B$ will be constructed as the union of a chain of partial isomorphisms $f_k$ of the following form. The domain of $f_k$ is $I_1\cup I_2\cup\dots\cup I_k$, where $I_1,\dots,I_k$ are intervals in $A$ of order type $\alpha$; $I_1\lt\dots\lt I_k$; the interval in $A$ between $I_j$ and $I_{j+1}$ ($1\le j\lt k$), as well as the interval to the left of $I_1$ and the interval to the right of $I_k$, have order types which are nonzero right multiples of $\alpha$. The range of $f_k$ is $J_1\cup\dots\cup J_k$ where $J_1,\dots,J_k$ are intervals in $B$ of type $\alpha$, etc. etc. etc., and $f(I_1)=J_1,\dots,f(I_k)=J_k$.

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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.

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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 $n$.

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@VladimirReshetnikov: Answer amended (but increasing just means $\geq$ rather than $>$). Btw, I've just noticed you're asking for arbitrary order types, not just ordinals. I'll have a think about that... –  Kris Jul 24 '12 at 3:47
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Because the question is not restricted to ordinals only, and indeed, ordinals is a trivial case, this is not really an answer. It might be better if you convert it into a comment. –  Vladimir Reshetnikov Jul 24 '12 at 15:20

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