Talk:Forcing (mathematics)

message: I'm a bit new to the guidelines for formatting math in WP, so forgive me if this isn't in line with the policies. I don't have the symbol ⊩.  Seems like here would be a good place to use <math>, especially since this is a critical symbol and I can't tell what it is.--Luke Gustafson 08:08, 21 December 2005 (UTC)
replies: text: Which browser are you using? I have edited the article to use the LaTeX symbol, but generally this is not preferred for inline symbols. Have you seen the same problem on other browsers and computers? - Gauge 23:54, 7 January 2006 (UTC)
thread_title: Formatting Math

message: Some things need adding:
Some points (which are maybe obvious to everyone else, but confused the hell out of me when I studied this material):
* It needs to be clear that (mostly for useful applications) G \not \in M
* Also M \neq M[G]
* the notion of absoluteness needs to be metioned a bit more to give the flavor
* be clear that c.c.c. is not absolute!
Another thing that I found very useful was seeing forcing conditons as being analogous to Cauchy sequences forcing real numbers into existence from the rationals, using e.g. the surd symbols as things to compose names with.  I don't know how rigourous this analogy can be made, (and thus how appropriate it is) but it certainly helped me.
Some things need correcting:
Complete atomless Boolean algebra B is always uncountable and by this reason it can not be in countable transitive model M. But for application in forcing satisfying of sentence "B is complete" by model M suffices. This means that \inf(S),\sup(s)\in B for any S\in M\cap P(B).
Nedeljko Stefanovic (talk) 18:17, 11 October 2011 (UTC)
replies: text: What you're missing is that, in the Boolean-valued model approach, there ordinarily just isn't any countable transitive M in the picture.  One looks at sentences whose truth value in VB is 1 (the greatest element of B).  A countable transitive M is generally superfluous. --Trovatore (talk) 08:15, 11 October 2011 (UTC)
replies: text: The whole point of the Boolean-valued approach is that it is semantic, but does not need a countable model.  The compromise is that you have to liberalize your semantics so that they are no longer two-valued, but take truth values in B.  But having done that, there is really no point in dragging in a countable model, or at least I have never seen a reason to do so. --Trovatore (talk) 10:01, 11 October 2011 (UTC)

text: You can perhaps argue that there's no fundamental difference.  I don't think it's a good use of our time to argue that point.  Whether there's a fundamental difference or not, nevertheless the Boolean-valued approach is not treated in the literature using countable transitive models.  My basic reference is Bell, Boolean-valued models and independence proofs in set theory, which does not mention countable transitive models except perhaps once in an exercise.  This make sense to me; I don't know why you would go to all the trouble to set up Boolean-valued semantics, only to pass to a transitive countable model, which then allows you to use two-valued semantics. --Trovatore (talk) 19:21, 13 October 2011 (UTC)
thread_title:  Needs Editing 

message: I'm about to make an edit to the introduction mentioning the use of forcing in areas other than set theory, in particular in recursion theory (hopefully sometime I can even write up that article).  There has even been a bit of use of forcing in model theory even though for the most part they just define genericity directly without forcing.
However, I'm wondering if we would be better off moving all the stuff specific to set theory to it's own page and using this page as a short description of the basic concept (truth by meeting dense sets or something) and providing links to the different notions. Let me know if you object otherwise someday I might get around to doing this (though I should probably write up the recursion theory page first).Logicnazi 01:38, 30 August 2007 (UTC)
replies: text: Forcing is used in many areas, sure, but we can probably keep it in one article. The applications to recursion theory are more esoteric (in some sense) so they could probably be kept to a section in this article plus a mention in the lede. THe idea of coming up with a general definition of forcing is nice, but it isn't done much in the literature, and it needs to appear there before appearing here. — Carl (CBM · talk) 01:52, 30 August 2007 (UTC)
replies: text: As things stand, forcing is split into two categories on the disambiguation page: mathematics and recursion theory.  I think a single page on forcing, with perhaps links to more specifics on forcing in set theory, recursion theory, model theory, what have you, would be reasonable.  But if not, surely we ought not to have recursion theory portrayed as though it does not belong to mathematics, no?  I will change the disambiguation page to "Forcing (Set Theory)".
Reviving an old topic...  I changed the disambiguation page.  Anyone object to renaming this page Forcing (set theory) (i.e., having the point do the reverse of what it does now)?  Recursion theory is a part of mathematics, so this seems to me just wrong.  —Preceding unsigned comment added by Ddzhafar (talk • contribs) 15:49, 10 March 2010 (UTC)

text: I think it's reasonable to have a page called forcing (mathematics) (or possibly forcing (mathematical logic)) that treats all flavors of forcing.  The set-theoretic version will probably be the most prominent because it's the one with which the most has been done, but the others should be represented as well.  Right now forcing (recursion theory) is pretty slim and I don't know why the content couldn't be merged here.  If someone wanted to do enough with recursion-theoretic forcing to warrant a separate article, then I would treat that article as a fork, with a summary section in this article and a  pointing to the forked article. --Trovatore (talk) 20:40, 10 March 2010 (UTC)
thread_title:  Move to Forcing_(Set Theory) 

message: Am I missing something? I am not much of a mathematician, but I see a problem in the following definition:
:Name(0) = {};
:Name(α + 1) = the power set of (Name(α) × P);
:Name(λ) = ∪{Name(α) : α < λ}, for λ a limit ordinal,
The second line worries me. I always thougt that A × B = {(a,b): a ∈ A \wedge b ∈ B}. It follows from this that {} × X = {} for any X, and hence, by induction, that V(P) = ∪{Name(α) : α is an ordinal} = {}.
I can only assume that something else is meant. For example, it would work if you used {{}} or P in the first line instead of {}. The result would be different, but not significantly: every tuple would get a {} appended to the front if you used {{}} rather than P.
This may seem like a pedantic quibble, but, like a lot of people (I should think), I am struggling with the concepts anyway, and throwing in logical inconsistencies doesn't help.
Rdbenham (talk) 06:18, 2 April 2009 (UTC)
replies: text: It's true that the empty set cross anything is again the empty set.  But the powerset of the empty set is not the empty set.  So the definition works (or at least doesn't fail for the reason you think it does).  It just takes one more iteration at the beginning to get going, than maybe you were expecting. --Trovatore (talk) 09:28, 2 April 2009 (UTC)

text: Yes, thanks for putting me straight about that. I should have looked into it more carefully. I last studied mathematics 30 years ago, and I am just now trying to get back into it. For some reason I ignored the powerset operation, maybe because it was written out in words rather than symbols.
Rdbenham (talk) 15:13, 5 April 2009 (UTC)
thread_title:  Dumb Question 

message: I have a general question regarding notions of forcing that can be formulated exactly in the context of the section Cohen Forcing in this article.
I use notation as in the article. Additionally, let M denote the universe of the model of set theory under consideration.
The subset X is not an element of M as is shown in the article. I buy this.
Equivalently, the function g:\omega \rightarrow \{0,1\} constructed from G is not an element of M (my conclusion).
Now, X is constucted using axioms of set theory. From G we get g using union. If P is a notion of forcing (as defined), 
then G \subset P since G is a filter. But then we must have g = \bigcup G \in M since P \subset M. It follows that X \in M
There is the possibility (close to a certainty) that I don't understand this. There is also the possibility that there is some fuzziness in the definition of a generic filter.
If the definition of a generic filter reads G contains a filter or G behaves like a filter instead of G is a filter, then I'd be happier. I'd certainly be happier if the definition of a generic filter didn't require the subset property.
In short, I see an existing generic filter as paradoxical in this case unless the definitions are reworded. I have finally, after many years of layman interest in mathematics, acquired a real book , Set Theory by Thomas Jech. The passage in his book on this particular question (the same forcing notion is used) parallels the article more or less excatly, so that doesn't help me. YohanN7 (talk) 11:26, 21 May 2012 (UTC)
replies: text: \forall A \, \exists P \, \forall B \, [B \in P \iff \forall C \, (C \in B \Rightarrow C \in A)].
If one puts the forcing poset as A, then one might naively expect the generic set G to appear as one of the B's in P above because it is a subset of the forcing poset.
The point here is that the variable B doesn't ever evaluate to G because G isn't an element in our model. Hence G is not an element in the power set either, consistent with what we had to begin with.
(Also, I wrote above that one could the the power set of the whole model P(M). That is of course nonsense seen from within the model wether the model is a set (in a larger model) or not.)
Maybe I shouldn't have written about this at all in this talk page in the first place, but I think that the article could perhaps elaborate a little bit more than it does on the fact that the existence of generic sets isn't paradoxical. Subsets of sets aren't necessarily sets from the point of view of the model. B t w, I checked out the definition of the term subset and it says that a set is a subset if ... . What is the standard terminology? Does e.g. the generic set count as a subset of the forcing poset (in terms of terminology) even if it isn't a set in the model? YohanN7 (talk) 21:19, 30 May 2012 (UTC)
thread_title:  Cohen Forcing 

message: The current "Consistency" section says: "Less commonly seen is the approach using the "internal" definition of forcing, in which no mention of set or class models is made. This was Cohen's original method, and in one elaboration, it becomes the method of Boolean-valued analysis."
When I read the references (both Cohen's paper The Independence of the Continuum Hypothesis in the external links section, and Cohen's 1966 book), it seems to me that Cohen mentions models a lot. I can imagine how one can avoid mentioning models (by focusing on formal consistency); I don't see how that was Cohen's original method.
I am also unsure how such a method is notably different from just "forcing over V" (which the "Consistency" section mentions as a different approach). From my understanding, when forcing over V, it is impossible to define the class model V[G] in the conventional sense, since the interpretations of \in and = depend on the specific choice of G. The best one can do is to define V[G] as a Boolean-valued model. Or does "forcing over V" mean "I'm actually forcing over a model, but I'll call this model V, so everything I write is by default within this model"? Bbbbbbbbba (talk) 02:30, 23 April 2023 (UTC)
replies: text: I think "internal" is just the same as "syntactic" (you define a forcing relation inductively on formulas, then look at sentences that are forced by the empty condition).  As for "Cohen's original method", I think it's a bit confusing to bring that up at this point in the text, because Cohen originally used something called "ramified forcing", which I've never learned in detail, but I gather involved somehow recapitulating the L\,\! hierarchy while taking forcing into account.  That difference doesn't seem to be the one the text in question is talking about, which to me means the text in question could be misleading. --Trovatore (talk) 20:06, 23 April 2023 (UTC)
replies: text: My understanding of ramified forcing is: Forcing involves a lot of transfinite recursion, and whereas unramified forcing uses V_\alpha to rigorize transfinite recursion, ramified forcing uses L_\alpha. Apparently, when proving the expanded model is a model of ZFC, these two methods encounter difficulties on different axioms. I haven't looked into the detailed proofs myself, but if my references are to be trusted, then A Cheerful Introduction to Forcing and the Continuum Hypothesis says that Foundation and Choice are the really difficult ones for unramified forcing, while Cohen (1966) says that Power Set and Replacement are the non-trivial ones for ramified forcing. This does make some sense since Power Set and Replacement already causes troubles when proving the constructible universe itself is an inner model. Bbbbbbbbba (talk) 02:25, 24 April 2023 (UTC)
thread_title:  "No mention of set or class models" in Cohen's original method?