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Forcing notions are partial orders. In some sense each partial order is a "combination" of some well-orderings and each well-orderings is isomorphic to a unique ordinal number. Thus in some sense a partial order is a combination of ordinal numbers. Just as an intuitive correspondence it seems the following equation holds: $$\frac{\text{Composite Numbers}}{\text{Prime Numbers}}=\frac{\text{Partial Orders}}{\text{Well Orders}}$$

As there is a natural correspondence between well-orderings and ordinal numbers it seems natural to have a correspondence between partial orders and a "combination" of ordinals (which are order types of those well-orderings which one can find within a given partial order).

Definition: By "combination" of ordinals I mean some sort of operator $*:Ord\times Ord\longrightarrow V$ between ordinal numbers. By "coding partial orders using combination of ordinal numbers" I mean assigning a unique code set $c$ to a partial order $\mathbb{P}$ which is made by a "combination" of order types of well-ordering subsets of $\mathbb{P}$ using the operator $*$. For example if partial order $\mathbb{P}$ has four sub-well orderings $\mathbb{W}_1$, $\mathbb{W}_2$, $\mathbb{W}_3$ and $\mathbb{W}_4$ with order types $\alpha, \alpha, \beta, \gamma$ then the code $c$ of $\mathbb{P}$ is $\alpha*\alpha*\beta*\gamma$. (Something similar to assigning the code $2\times2\times3\times5$ to composite number $60$ which is a combination of prime numbers using $\times$ operator).

Question: Is there any known coding method for partial orders using ordinal numbers? All other known types of coding methods for partial orders via other mathematical objects different from ordinals are also welcome.

Motivation: Shelah, Foreman and Magidor announced a conjecture in their paper "$0^{\sharp}$ and Some Forcing Principles" as follows. As Asaf mentioned below it is known as Foreman's Maximality Principle (FMP).

FMP: Each set forcing notion either collapses a cardinal or adds a real.

It seems too complicated and there are few partial results about it. A source of its complexity is the fact that it is about all forcing notions of the universe and we have few tools to deal with them. A possible approach to simplify the conjecture is finding a "better equivalent version" for it.

Note that we have a fairy large amount of information about the backbone of each model of ZFC (i.e. $Ord$ axis). In the other words we know more facts about "ordinals" of a given model of ZFC than its partial orders (forcing notions). Thus if one can find a suitable coding of the partial orders to sets of ordinals then he can obtain an equivalent version of the FMP conjecture by analyzing the "formulation" of codes of partial orders and proving some ZFC-theorems as follows:

1) A set $c$ of ordinals is a code of a forcing notion iff $c$ has the property $A$.

2) A forcing notion $\mathbb{P}$ collapses a cardinal iff its associated code $c_{\mathbb{P}}$ has the property $B$.

3) A forcing notion $\mathbb{P}$ adds a real iff its associated code $c_{\mathbb{P}}$ has the property $C$.

FMP$^{*}$: Each set $c$ of ordinal numbers with property $A$ has the property $B\vee C$.

It seems easier to find a (ZFC or large cardinal consistency) proof for a statement about sets of ordinals in comparison with a statement about all partial orders of the universe. Also one can search for a possible counter example of the FMP conjecture by a diagonal argument based on finding a special set of ordinals with property $A\wedge \neg B \wedge\neg C$.

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It's not immediately clear how one would "combine" well-orders to obtain, for example, $( \mathbb{R} , < )$. –  Arthur Fischer Jun 5 at 5:22
    
This is often known as Foreman's Maximality Principle. Not as "SFM". –  Asaf Karagila Jun 6 at 11:53
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2 Answers 2

Interesting question; here's a longish comment.

There's certainly a correspondence if we replace "partial orders" with scattered orders.

$$\frac{\text{Composite Numbers}}{\text{Prime Numbers}}=\frac{\text{Scattered Orders}}{\text{Well Orders}}$$

But I think you're interested in the more general problem of coding arbitrary partial orders as (multisets of) ordinals. I think this is a very interesting question, although your particular proposal doesn't really work. To see this, take the coproduct $C$ of two copies of $\mathbb{Z}.$ Now pick two incomparable elements of $C$ and identify them, creating a new partial order $D$. Then every finite ordinal embeds into both $C$ and $D$ countably-infinitely many times, as do $\omega$ and $\omega^\dagger$. Furthermore, no other ordinals embed into $C$, nor into $D$. So, we cannot distinguish $C$ from $D$ in this way. The same observation holds if we replace "embedding" with "dense embedding," by which I mean that we're not "skipping" elements.

That being said, there might be a more "local" variant of your idea that "does the trick," so to speak.

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I'm not sure I understand the motivation of your question, but anyway the answer to your coding question is no, in general. It is consistent with $\mathsf{ZFC}$ that the set of isomorphism types of countable partial orderings does not itself admit any definable well-ordering (e.g. if $\mathbb{R}$ is not definably well-ordered; in fact this is equivalent) and if this happens then any definable "coding" function from finite sequences of ordinals to partial orderings will necessarily omit some countable partial orderings from its range, even if you only want to code them up to isomorphism.

Granted, countable partial orderings are all forcing-equivalent either to the trivial forcing, to the Cohen forcing $(2^{\mathord{<}\omega}; \subseteq)$, or to the lottery sum of these two forcings, so from the forcing point of view there is a trivial classification of posets in this case anyway. But probably if you clarified your reason for talking about forcing then the above argument could be modified to address it.


EDIT: Vopenka's theorem that every set is in a generic extension of $\text{HOD}$ makes essential use of the fact that a certain forcing poset which is not literally a set of ordinals can be coded by a set of ordinals. Your question made me think of this, although it is probably not what you're after.

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Countable forcing can be the lottery sum of a trivial forcing and a Cohen forcing, too. (Which is then equivalent to neither). (Isn't it annoying to talk to mathematicians sometimes? :-D) –  Asaf Karagila Jun 6 at 9:12
    
@Asaf Thanks. I will fix it. –  Trevor Wilson Jun 6 at 15:12
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