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There is a theorem which says given any two ordinals $\alpha$ and $\beta$, exactly one of the following holds: $\alpha\in\beta$, or $\beta\in\alpha$, or $\alpha=\beta$.

What is the name of this theorem and where can I find it?

lyrically wicked
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    Ordinal Membership is Trichotomy - https://proofwiki.org/wiki/Ordinal_Membership_is_Trichotomy – mlerma54 Dec 20 '18 at 04:30
  • First let me clarify a bit here regarding the question/answers in the linked thread. For ordinals less than $\epsilon_0$ we often write a single ordinals in multiple ways. For example, we could write either $\omega^\omega$ or $\omega^2+\omega^\omega$ etc. This is perhaps somewhat analogous to how we can write $7$, $(5+(2 \cdot 1))$, $(2+(5 \cdot 1))$ and so on. The point in the linked question was that if we are given a "finite string" which uses the alphabet ${+,\cdot,e,(,),\omega,0,s }$ ($e$ is for exponent, $s$ for successor), then there are two further questions. – SSequence Dec 20 '18 at 05:16
  • $(1)$ Is it decidable in a (strict) algorithmic sense whether a given string from these alphabet forms a "valid" expression? $(2)$ Is it decidable whether the ordinals that two different strings represent are same or not? And if not, which one is greater? The answer to both of these questions is positive (that's all that I was trying to say in the answer to linked question). – SSequence Dec 20 '18 at 05:18
  • Given your comment in the other question, I think what you might be meaning to ask is that: "Suppose we are given two well-orders (of $\mathbb{N}$) say $W_1,W_2$ with order-types $\alpha_1, \alpha_2$ respectively. Then is there a "reasonable sense" in which we can compare $\alpha_1$ and $\alpha_2$ and determine which one is greater? And if one of them is greater, then can we find the "address" (very informally speaking ... this needs to be made precise) of the smaller ordinal?" I think the answer to that would depend on what do we mean by a "reasonable sense". – SSequence Dec 20 '18 at 05:30
  • @SSequence: but is it always possible to provide a precise description for an arbitrary ordinal using a finite string? For example, what language (and alphabet) would we use to describe this ordinal? – lyrically wicked Dec 20 '18 at 05:34
  • The answer in the linked question was specifically for ordinals less than $\epsilon_0$. – SSequence Dec 20 '18 at 05:36
  • But is this theorem valid for any pair of arbitrarily chosen ordinals? – lyrically wicked Dec 20 '18 at 05:37
  • For example, in infinite models (such as ORM) given the relations $lessthan_1: \mathbb{N}^2 \rightarrow {0,1}$ and $lessthan_2: \mathbb{N}^2 \rightarrow {0,1}$ one can determine (giving the ORM "oracle access" to $lessthan_1$ and $lessthan_2$) : $(a)$ whether they can ever represent a well-order relations on $\mathbb{N}$ (with any order-type) $(b)$ Suppose $lessthan_1,lessthan_2$ can both serve as well-order relations. So if we consider these functions as well-order relations (for well-orders $W_1,W_2$ on $\mathbb{N}$) then can we decide which of $W_1,W_2$ one has greater order-type? Yes. – SSequence Dec 20 '18 at 05:40
  • What does ORM abbreviation stand for? – lyrically wicked Dec 20 '18 at 05:43
  • @lyricallywicked "Ordinal Register Machine" – SSequence Dec 20 '18 at 06:31
  • Also, to be formally correct, I think I should have written $lessequals: \mathbb{N}^2 \rightarrow {0,1}$ (with obvious meaning) instead of $lessthan: \mathbb{N}^2 \rightarrow {0,1}$ in the comment above. I was a bit careless about this in my answer to your previous question too. – SSequence Dec 20 '18 at 07:38
  • "But is this theorem valid for any pair of arbitrarily chosen ordinals?" Yes. – Noah Schweber Dec 23 '18 at 05:14

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This is the trichotomy principle (or law of trichotomy, or etc.) for ordinals. It can also be phrased for arbitrary well-orderings, as follows:

For any two well-orderings $A,B$, exactly one of the following situations occurs: $(i)$ there is a non-surjective order-preserving embedding of $A$ into $B$; $(ii)$ there is a non-surjective order-preserving embedding of $B$ into $A$; $(iii)$ there is an order-preserving bijection between $A$ and $B$.

Intuitively, $(i)$ means $A<B$, $(ii)$ means $B<A$, and $(iii)$ means $A=B$. (Exercise: the principle above does in fact imply trichotomy for ordinals as you've phrased it.) Note also that we can replace "embedding" with "initial segment embedding" (= embedding whose image is an initial segment).

You can find proofs of the trichotomy in any decent introductory set theory textbook - for example, it's Theorem $6.3$ in Kunen's book (phrased there in terms of arbitrary well-orderings).

Note that nowhere do we need to assume that the ordinals (or well-orderings) involved are countable; this is a generally, always-applying principle. Also, to forestall a frequent (in my experience) confusion, note that it does not need the axiom of choice (rather, replacement is the key principle).

Noah Schweber
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