This question was inspired by Justin Palumbo's excellent question Cantor Bernstein for notions of forcing.

In his question, Justin considers a relation $\lhd$ on partial orders (defined below as $\lhd_1$) which begins to capture the notion of one partial order being "stronger" than another. He asks a natural question, namely whether this relation satisfies antisymmetry -- that is, if $\mathbb{Q}\lhd\mathbb{P}$ and $\mathbb{P}\lhd\mathbb{Q}$, does it follow that $\mathbb{P} \sim \mathbb{Q}$? Here, $\sim$ means forcing equivalent: two partial orders $\mathbb{P}$ and $\mathbb{Q}$ are forcing equivalent if they give rise to the same generic extensions, that is, every extension $V[G]$ by $\mathbb{P}$ can be realized as an extension $V[H]$ by $\mathbb{Q}$, and vice versa (in Joel Hamkins' terminology, we could say they are forcing equivalent if they give us access to the same neighborhood of the multiverse).

Unfortunately the given relation does not satisfy antisymmetry. In fact, the counterexamples given in the responses to the question demonstrate that $\lhd$ falls short of capturing the idea of one partial order being stronger than another. I'm trying to find the "right version" of $\lhd$ -- a relation on partial orders $\mathbb{Q} \lhd \mathbb{P}$ that captures our intuitive notion of '$\mathbb{P}$ is stronger than $\mathbb{Q}$'. A good test criteria is whether the relation satisfies antisymmetry. I will consider a number of (I hope) natural strengthenings of the $\lhd$ relation (or variations -- it's not clear that all of them are strengthenings).

Consider the following relations on partial orders:

  1. $\mathbb{Q} \lhd_1 \mathbb{P}$: Every $\mathbb{P}$ extension contains a $\mathbb{Q}$ extension. $\mathbb{Q} \lhd_1 \mathbb{P}$ if and only if whenever $V[G]$ is a forcing extension by $\mathbb{P}$, there is an $H\in V[G]$ generic for $\mathbb{Q}$ with $V[H] \subseteq V[G]$. (this is the original definition from Justin's post).

  2. $\mathbb{Q} \lhd_2 \mathbb{P}$: Every $\mathbb{P}$ extension contains a $\mathbb{Q}$ extension meeting an arbitrary condition. $\mathbb{Q} \lhd_2 \mathbb{P}$ if and only if whenever $V[G]$ is an extension by $\mathbb{P}$, and $q \in \mathbb{Q}$, there is $H$ in $V[G]$ generic for $\mathbb{Q}$ with $q \in H$.

  3. $\mathbb{Q} \lhd_3 \mathbb{P}$: Every $\mathbb{Q}$ extension is contained in a $\mathbb{P}$ extension. $\mathbb{Q} \lhd_3 \mathbb{P}$ if and only if whenever $V[H]$ is an extension by $\mathbb{Q}$, there is an extension $V[G]$ by $\mathbb{P}$ such that $V[H] \subseteq V[G]$.

  4. $\mathbb{Q} \lhd_4 \mathbb{P}$: $\mathbb{Q}$ embeds into $\mathbb{P}$. $\mathbb{Q} \lhd_4 \mathbb{P}$ if and only if there is a complete embedding of (the Boolean algebra of) $\mathbb{Q}$ into (the Boolean algebra of) $\mathbb{P}$.

  5. $\mathbb{Q} \lhd_5 \mathbb{P}$: Every $\mathbb{Q}$ extension is equal to a $\mathbb{P}$ extension. $\mathbb{Q} \lhd_5 \mathbb{P}$ if and only if whenever $V[H]$ is an extension by $\mathbb{Q}$, there is an extension $V[G]$ by $\mathbb{P}$ such that $V[H] = V[G]$.

Please note that I have ignored serious metamathematical considerations in giving these definitions, but I believe each of them has a first-order definition.

Question. Which of the relations above satisfy antisymmetry? That is, for which of the relations above do we have $\mathbb{Q}\lhd\mathbb{P}$ and $\mathbb{P}\lhd\mathbb{Q}$ implies $\mathbb{P}$ is forcing equivalent to $\mathbb{Q}$?

From the responses to Justin's question it is clear that $\lhd_1$ does not satisfy antisymmetry. On the other hand, $\lhd_5$ does satisfy antisymmetry, but is extremely restrictive in terms of the partial orders that it can compare (for example, $Add(\omega,1)$ and $Add(\omega,1)\times Add(\omega_1,1)$ are incomparable by $\lhd_5$, although the latter is apparently more powerful than the former).

At the risk of violating the suggested guidelines for keeping questions focussed and specific, I'll include two followup questions -- feel free to ignore.

Question 2. What are the relative strengths of these relations (which ones imply which others)?

For example, we (trivially) have $\mathbb{Q} \lhd_5 \mathbb{P}$ implies $\mathbb{Q} \lhd_3 \mathbb{P}$, and $\mathbb{Q} \lhd_2 \mathbb{P}$ implies $\mathbb{Q} \lhd_1 \mathbb{P}$.

Question 3. What is the "right" notion of $\lhd$ for partial orders, that captures our intuitive idea of one partial order being "stronger" or "more effective" than another?

This is "too vague", but it's the question that motivated those above.

  • 1
    $\begingroup$ The relation 2 is not reflexive. I think there are ZFC examples but only a conditional one comes to mind now: a Suslin tree with special square. After adding one branch the rest of the tree becomes special and therefore the generic branch is the only branch through that tree. $\endgroup$ – François G. Dorais Nov 25 '12 at 18:05
  • $\begingroup$ Similar to comparability of cardinals using injective functions, I think that the "right" basic notion is a reflexive relation $\trianglelefteq$. The antisymmetric relation "strictly stronger" is then naturally defined by ``$\trianglelefteq $ but not $\trianglerighteq$.'' $\endgroup$ – Goldstern Nov 25 '12 at 18:11
  • $\begingroup$ Francois, this is a good point, and to my mind this makes $\lhd_2$ much less attractive -- I agree with Goldstern that our basic notion should be reflexive, at least. $\endgroup$ – jonasreitz Nov 25 '12 at 18:52
  • $\begingroup$ Relation 2 will be reflexive if $\mathbb{P}$ is homogeneous, although I'm not sure if this is a necessary condition (motivation for the definition of relation 2 came from a comment by Joel Hamkins about homogeneous partial orders). $\endgroup$ – jonasreitz Nov 25 '12 at 18:55
  • 1
    $\begingroup$ Justin, thanks for your comments. Yes, I see that $\lhd_4$ implies $\lhd_1$ -- great! For the reverse implication, I'm worried that the map sending $q$ to $[q\in \tau]$ may not preserve incompatibility (and thus may not give a complete embedding). For example, if $\mathbb{Q}$ is the lottery sum of two posets, and $\tau$ always yields a generic for the first of them (with no mention of the second), then it will map all members of the second poset to Boolean value 0. This is the case $\endgroup$ – jonasreitz Nov 26 '12 at 3:48

I believe I have an answer to my first question.

To recap, $\lhd_1$ is not antisymmetric (as demonstrated in the answers to Justin Palumbo's question).

As Francois G. Dorais points out in a comment, $\lhd_2$ is not reflexive (and therefore not antisymmetric).

It follows directly from the definition that $\lhd_5$ is antisymmetric (it was defined to accomplish exactly this).

This leaves $\lhd_3$ and $\lhd_4$. I will show that neither of these is antisymmetric -- the same counterexample works for both.

Let $\mathbb{R}_n$ be the product that adds a Cohen generic to each of the first $n$ cardinals. Now let $\mathbb{P}$ be the lottery sum of the posets $\{ \mathbb{R}_{2n} \mid n \in \omega \}$. This partial order adds Cohen generics to the first $2n$ cardinals, where $n$ is chosen generically. Similarly, let $\mathbb{Q}$ be the lottery sum of $\{ \mathbb{R}_{2n+1} \mid n \in \omega \}$, which adds Cohen generics to the first $2n+1$ cardinals. Observe that $\mathbb{P}$ and $\mathbb{Q}$ are not forcing equivalent, since they do not share any of the same generic extensions (extensions by $\mathbb{P}$ and $\mathbb{Q}$ always differ by a Cohen generic on at least one cardinal).

However, any generic extension $V[G]$ by $\mathbb{P}$ can be extended to a generic extension $V[H]$ by $\mathbb{Q}$ by simply adding a Cohen generic to the 'next' ($2n+1^{th}$) cardinal. This shows that $\mathbb{P} \lhd_3 \mathbb{Q}$. The same argument shows that $\mathbb{Q} \lhd_3 \mathbb{P}$. Thus $\lhd_3$ is not antisymmetric.

For $\lhd_4$, observe that each $\mathbb{R}_n$ embeds completely into $\mathbb{R}_{n+1}$, and we can thus construct a complete embedding of $\mathbb{P}$ into $\mathbb{Q}$ by working term-by-term with each member of the lottery sum. The same argument allows us to construct a complete embedding of $\mathbb{Q}$ into $\mathbb{P}$. Thus $\mathbb{P} \lhd_4 \mathbb{Q}$ and $\mathbb{Q} \lhd_4 \mathbb{P}$, and so $\lhd_4$ is not antisymmetric either.

  • $\begingroup$ Antisymmetric order doesn't necessarily imply that the order is reflexive. $\endgroup$ – Eran Nov 27 '12 at 20:03

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