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Forcing with finite conditions is a common concept used by set theorists. I was thinking about its meaning, but I couldn't find any exact definition of it. At the first glance it seemed to me that maybe by finite conditions, experts mean that the conditions of a forcing notion are finite however one can naturally turn any arbitrary forcing to another one which are isomorphic as partial orders and the conditions of former are finite. On the other hand, the conditions of many forcings with finite conditions are intrinsically infinite.

By the way implicitly, forcing with finite conditions means every condition has finitely many information about generic object(from forcing point of view). My question is:

Is there any formal definition of forcing with finite condition? Or can some one give a formal definition of forcing with finite condition including well-known forcings with finite conditions?

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    $\begingroup$ My suggestion: if the generic object obtained from the forcing, when considered as a set of ordinals (or generally objects from the ground model), does not contain a countable ground model set. This means in some sense that the conditions can only give finite information about the final generic object. $\endgroup$
    – Mohammad Golshani
    Commented Dec 27, 2016 at 8:31
  • $\begingroup$ However, there might be much better interpretations that experts can answer. $\endgroup$
    – Mohammad Golshani
    Commented Dec 27, 2016 at 8:33
  • $\begingroup$ @MohammadGolshani I see. so you mean we have to consider the poset whose domain is a set of ordinals? or you mean if generic filter defines a generic set of ordinals? we may have some generic objects which obtained for example via some different operators, such as pairing or etc. $\endgroup$
    – Rahman. M
    Commented Dec 27, 2016 at 8:44
  • $\begingroup$ Are you looking for the definition of finite or compact elements of a poset? en.wikipedia.org/wiki/Compact_element $\endgroup$
    – Andrej Bauer
    Commented Dec 27, 2016 at 9:57
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    $\begingroup$ I think Mohammad's suggestion is on the right track at least. All the forcings that I know of that commonly are called "forcing with finite conditions" have the property he mentions, e.g. Baumgartner's adding a club with finite conditions; Todorcevic's $\in$-collapse; Friedman, Mitchell, Neeman, Krueger posets to add a club to $\omega_2$. Those are all strongly proper too, though there are other examples which aren't strongly proper (e.g. semiproper variations of Velickovic, Gitik-Magidor). $\endgroup$
    – Sean Cox
    Commented Dec 27, 2016 at 17:44

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My impression is that this term is used loosely, without formal definition, to refer to the common situation where we have a forcing notion consisting of a partial order whose elements are all finite functions (that is, each on a finite domain) and the forcing order is functional extension. The most canonical example is Cohen forcing $\text{Add}(\omega,1)$ to add a Cohen real or actually any number of Cohen reals. Other examples include the forcing to collapse a given cardinal to $\omega$ or the forcing to add a club set using finite conditions (not with countable conditions, which is another common way to do it). In some cases, there are examples where conditions are augmented with some other finite amount of information that still counts as forcing with finite conditions even though the order isn't literally functional extension.

Every forcing notion is equivalent, of course, to a forcing notion whose conditions are all finite, since we could replace any element $p$ in a partial order with $\{p\}$, which is a singleton set and hence finite, and then redefine the corresponding order on these singletons. So every condition in the new partial order is finite. But this literal interpretation is never what is meant by the phrase, "forcing with finite conditions," and I have never seen a formal definition of the concept going significantly beyond the loose understanding given in the previous paragraph.

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  • $\begingroup$ Is there any mathematical significance to having finite forcing conditions? And assuming there is, shouldn't then there be a purely order-theoretic way of characterizing finite conditions? $\endgroup$
    – Andrej Bauer
    Commented Dec 27, 2016 at 22:39
  • $\begingroup$ In the finite-function case, it is often important since it generally makes the forcing notion amenable to the $\Delta$-system theorem, which can be used to prove a good chain condition. Perhaps this is the main significance. $\endgroup$
    – Joel David Hamkins
    Commented Dec 27, 2016 at 22:40
  • $\begingroup$ About finding an order-theoretic characterization, I admire your proposal and attempt to do this. I am less confident, however, that it will be useful, since I think the characterization that you get will not be invariant under equivalence-of-forcing. For example, we should not expect to be able to pass the property to the Boolean completions (since those elements have infinitely much information). $\endgroup$
    – Joel David Hamkins
    Commented Dec 27, 2016 at 22:42
  • $\begingroup$ That's interesting, since almost every notion of finiteness and compactness I am aware of fall under the order-theoretic notion. It might be worthwhile drilling into finiteness-in-forcing a bit more, ot see if there's anything else there. $\endgroup$
    – Andrej Bauer
    Commented Dec 27, 2016 at 22:56
  • $\begingroup$ The finite conditions notion that I mentioned is equivalent to saying that the forcing notion embeds into the Levy collapse poset for some large enough cardinal, since every collection of finite partial functions can be seen as a suborder of that partial order. (But note that this embedding is not a complete embedding and will not respect forcing.) $\endgroup$
    – Joel David Hamkins
    Commented Dec 27, 2016 at 23:12
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It might be the case that the definition of finite elements from domain theory is what you are looking for. It makes precise the idea that an element of a poset carries finite amount of information.

Definition: Let $(P, {\leq})$ be a poset. An element $x \in P$ is finite (or compact) if it is inaccessible by directed suprema: if $D \subseteq P$ is directed and $x \leq \sup D$ then $x \leq y$ for some $y \in D$.

This is a purely order-theoretic characterization of finiteness and is therefore immune to changes in the presentation of $P$.

Looking at the list of forcing notions, it seems that the following have the property that all elements of the forcing poset are finite:

  1. Cohen forcing

  2. Levy collapsing of an uncountable cardinal $\lambda$ to $\omega$ (but not the collapsing of $\lambda$ to $\kappa$ when $\kappa$ is uncountable – although in this case we get the related notion of $\kappa$-finiteness).

  3. Shooting a club with countable conditions.

I have not checked all the other notions, but they generally seem to contain some elements that are non-finite.

In domain theory it usually does not matter so much that a poset consists of only finite elements, but rather that every element is the supremum of finite elements below it. This is known as algebraicity of the poset. Speaking somewhat off the top of my head, I would propose the following:

Tentative definition: A $B$-valued model has finite conditions if the complete Boolean algebra $B$ is algebraic (every element is the supremum of finite elements below it). A forcing notion $P$ has finite conditions if the associated complete Boolean algebra is algebraic.

There is going to be a characterization of a $P$ with finite conditions in the above sense which is intrinsic to $P$ (and certainly if $P$ only has finite elements then it will satisfy the condition). But before working out the details of that, I'd prefer to hear from the experts if the suggestion seems plausibly useful.

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  • $\begingroup$ shooting club with finite or countable? $\endgroup$
    – Rahman. M
    Commented Dec 27, 2016 at 13:25
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    $\begingroup$ No, this is not what the term means. But it is an interesting suggestion. $\endgroup$
    – Andrés E. Caicedo
    Commented Dec 27, 2016 at 14:01
  • $\begingroup$ @Rahman.M: that's not a directed supremum, si it? $\endgroup$
    – Andrej Bauer
    Commented Dec 27, 2016 at 16:54
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    $\begingroup$ I think there are no finite elements in an atomless Boolean algebra, since every element has an infinite antichain below, and then the finite joins of elements in this antichain form a directed set of conditions entirely below the given condition. So from a forcing perspective, the tentative proposal cannot be what we want, since we generally do not allow atoms in forcing notions. $\endgroup$
    – Joel David Hamkins
    Commented Dec 27, 2016 at 22:47

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