6
$\begingroup$

So I was reading about forcing in wikipedia to try to get an intuitive idea about forcing in set theory. There is this paragraph in it:

A subtle point of forcing is that, if $X$ is taken to be an arbitrary "missing subset" of some set in $M$, then the $M[X]$ constructed "within $M$" may not even be a model. This is because $X$ may encode "special" information about $M$ that is invisible within $M$ (e.g. the countability of $M$), and thus prove the existence of sets that are "too complex for $M$ to describe".

Then there is another line about how forcing avoids this problem:

Forcing avoids such problems by requiring the newly introduced set $X$ to be a generic set relative to $M$. Some statements are "forced" to hold for any generic $X$. For example, a generic $X$ is "forced" to be infinite. Furthermore, any property (describable in $M$) of a generic set is "forced" to hold under some forcing condition. The concept of "forcing" can be defined within $M$, and it gives $M$ enough reasoning power to prove that $M[X]$ is indeed a model that satisfies the desired properties.

Here is the definition of a "generic filter"

For $G$ to be "generic relative to $M$" means:
$\bullet$ If $D \in M$ is a "dense" subset of $\mathbb{P}$ $\space$(that is, for each $p \in \mathbb{P}$ , there exists a $q \in D$ such that $q \leq p$), then $G\cap D \neq \emptyset $.

Why is the definition of generic filter designed like this? (intersecting with dense subsets of $\mathbb{P}$ in $M$). What does it do to avoid the problem mentioned in the first paragraph?

Note: I don't have any background in set theory except I read some stuff from Enderton's set theory textbooks.

$\endgroup$
2
  • $\begingroup$ My understanding is that genericity does not avoid the issue in the first paragraph a priori, so the definition was not chosen to avoid the problem. The description in the first paragraph describes why we use generic filters a posteriori. $\endgroup$
    – Hanul Jeon
    Jan 29 at 15:07
  • $\begingroup$ Not an answer to this question, but there was a recent question that gives an example of what they’re talking about in the first paragraph math.stackexchange.com/questions/4841926/… $\endgroup$ Jan 29 at 17:32

3 Answers 3

4
$\begingroup$

I think the modern abstract approach to forcing obscures the underlying motivation. Cohen's original approach (as in his monograph Set Theory and the Continuum Hypothesis) is much more concrete and shows why one might think of defining forcing as we do.

I'll briefly describe a hybrid version that shows why the dense open set requirement comes up in the modern approach.

You want to have that, for every sentence $\varphi$ in the language of forcing, there's some condition $p\in G$ which decides $\varphi.$ This is how we ensure that every first-order fact about $M[G]$ depends only on some individual condition being in $G$ and so can be talked about in $M$ (rather than depending on all of $G,$ or on more of $G$ than $M$ can know about).

An equivalent way of phrasing this requirement is that for every $\varphi$ in the language of forcing, $$G\cap\{p\mid p\Vdash \varphi \text{ or }p\Vdash\neg\varphi\}\neq\emptyset.$$

(By the way, I'm oversimplifying a bit here. You need to do all this with an auxiliary notion of "strong forcing," which is defined via a somewhat elaborate transfinite induction, and then you can define the normal notion of forcing, sometimes called "weak forcing," using that. But that doesn't affect the principle here, so I won't delve into those details.)

Note that for every sentence $\varphi$ in the language of forcing, the set $\{p\mid p\Vdash \varphi \text{ or }p\Vdash\neg\varphi\}$ is a dense open set of conditions.

So if you require that $G$ meet every dense open set of conditions in $M,$ it will meet all the sets $\{p\mid p\Vdash \varphi \text{ or }p\Vdash\neg\varphi\},$ which are the dense open sets that we really care about.

But the dense open set requirement is more elegant, in that it can be described simply just based on the partial ordering, algebraically, without having to develop the forcing relation first.

So that's why we look at dense open sets in defining genericity.

$\endgroup$
6
  • $\begingroup$ Is the strong forcing a forcing defined without concerning the dense set condition, so, e.g., $p\Vdash \phi\lor\psi$ if and only if $p\Vdash \phi$ or $p\Vdash \psi$? $\endgroup$
    – Hanul Jeon
    Jan 30 at 16:12
  • $\begingroup$ @HanulJeon Yes. By the way, often strong forcing in this context is denoted by $\Vdash'$, reserving $\Vdash$ for weak forcing, which is the normal forcing relation everyone is used to. Anyway, yes, strong forcing for disjunction works just as you said. On the other hand, $p\Vdash'\neg\varphi$ iff no extension of $p$ strongly forces $\varphi.$ Universally quantified sentences are treated similarly: $p\Vdash'\forall x\varphi(x)$ iff there does not exist a name $c$ and an extension $q$ of $p$ such that $q\Vdash '\neg\varphi(c).$ (continued) $\endgroup$ Jan 30 at 18:47
  • $\begingroup$ The whole thing is kind of elaborate, with a complicated system of names, and then an induction first on limited sentences and then on all sentences. (Even the atomic formulas require care, because there's an induction on set-theoretic rank going on there.) In the end, you define the usual forcing relation $p\Vdash\varphi$ to mean $p\Vdash'\neg\neg\varphi.$ I might have some of the details wrong since I'm remembering this from 50 years ago! $\endgroup$ Jan 30 at 18:47
  • $\begingroup$ It sounds like the intuitionistic forcing whose detail is available in the last Chapter of Bell's book about the Boolean-valued models. Your approach also reminds me a sheaf-theoretic view of forcing, obtained from a usual topos-theoretic construction with the double-negation topology. $\endgroup$
    – Hanul Jeon
    Jan 30 at 19:20
  • $\begingroup$ @HanulJeon Interesting; I'm not familiar with those versions of forcing, but there is something like a pseudo-constructivist feel to the strong forcing relation. Of course, it's not really constructive, since the constructivity is all relative to the ground model — which, as a model of ZF, isn't itself constructive. (It's similar that way to L, which is sort of constructive relative to the class of ordinals that you have available to work with.) $\endgroup$ Jan 30 at 21:04
2
$\begingroup$

One way to motivate generic filters is that generic filters are hidden points (from $V$) that exist almost everywhere. Let us recall the Baire category theorem:

Baire category theorem. Let $X$ be a locally compact regular space, then every countable intersection of open dense sets of $X$ is non-empty.

Also, we can view ultrafilters as points in a topological manner (see the convergence of a filter.) Similarly, we can view forcings as topological spaces: one way of doing it is imposing the Scott topology over a forcing poset (a topology generated by sets of the form $\{q\mid q\le p\}$.) Or you may consider the Stone dual of the forcing poset (given by the set of all ultrafilters over the poset.) In either way, we may understand a forcing poset as a basis for a topological space.

Under the Scott topology, a dense set (a la partial order) $D\subseteq \mathbb{P}$ corresponds to a topologically open dense set over the space. Thus a filter (or a point) $G$ being a generic can mean that it belongs to every open dense set in the ground model $M$, so the generic point is "generic" in every open dense condition in the ground model.


But I do not think the above explanation justifies why it avoids the issues you mentioned. We can prove that $M[G]$ is a model of $\mathsf{ZFC}$ if $M$ is, but it seems not to follow from the above informal explanation. We should rather view it happening by chance. There are other model constructions (like, classical realizability) that are not forcing but also give a new model of $\mathsf{ZFC}$ from another.

$\endgroup$
6
  • $\begingroup$ Instead of by chance, I think of the fact "$M[G]\vDash ZFC$ if $M\vDash ZFC$" as a consequence of the fact that $M$ builds the $\mathbb{P}-$names that become the elements in $M[G]$. Because $M$ models all the $ZFC$ axioms, it can use the strength of these axioms to build names that also satisfy $ZFC$ and I think of this construction process as $M[G]$ inheriting $ZFC$ from $M$. The properties of the generic filter allow us to prove that this intuition works out $\endgroup$
    – blark
    Feb 17 at 1:41
  • $\begingroup$ @blark Weak set theories like $\mathsf{KP}$ can also form the class of $\mathbb{P}$-names. Of course, the generic extension of $\mathsf{KP}$ will also satisfy $\mathsf{KP}$, and it is correct that axioms of $\mathsf{ZFC}$ allow to prove a generic extension of a model of $\mathsf{ZFC}$ also satisfies $\mathsf{ZFC}$. $\endgroup$
    – Hanul Jeon
    Feb 17 at 1:50
  • $\begingroup$ @blark But your explanation does not address why the universe of $\mathbb{P}$-names must satisfy the axioms of $\mathsf{ZFC}$ (if we work over $\mathsf{ZFC}$, of course.) It may not be true for weaker variants of $\mathsf{ZFC}$ like $\mathsf{ZC}$ (I am unsure if $\mathsf{ZC}$ even could define $V^\mathbb{P}$, though.) Also, you need more justification on why the generic filter "allows" us to see why what we expect works. $\endgroup$
    – Hanul Jeon
    Feb 17 at 1:53
  • $\begingroup$ I just mean to give the main idea for the proof, since it's easy to find a full treatment in Jech or Kunnen's books, for instance. In my answer to the original post, I give some motivation for the defining features of generic filters. Although this is more directed towards how we can define the forcing relation, Kunnen uses genericity via the forcing relation in his proof that $M[G]\vDash ZFC$ if $M\vDash ZFC$. Again, my focus is to give a perspective that could help OP read a formal proof rather than give a full account of the argument $\endgroup$
    – blark
    Feb 17 at 8:35
  • $\begingroup$ @blark I think we have different opinions on the question: I thought the OP wanted to hear was why it works. You focus on how it works. Spector's answer placed in somewhere between (I think it is closer to how, but it addresses why we have seen so many dense set arguments when we handle forcing.) $\endgroup$
    – Hanul Jeon
    Feb 17 at 23:34
1
$\begingroup$

To answer your question, I think it's worth asking why we want the ground model $M$ to be able to describe the forcing relation in the first place. One justification for this is that we as mathematicians cannot perform infinite tasks without assistance from the assumed existence of a model in the first place. It's similar to the slogan that "programmers borrow randomness from the universe" to describe the inability to write an algorithm that yields a truly random output and the solution to sample things like time or temperature or wind patterns to use as inputs, as these can function as some natural form of randomness that we can steal from the world around us. As a mathematician, for instance, I cannot build a wellordering of the real line; I rely on the fact that such a wellordering already exists in a model of $ZFC$, as a consequence of the axioms that hold in that model. Similarly, I cannot build a model of $ZF(C)$ from scratch. This is why we deal in terms of relative consistency: given that a model $M$ of $ZF(C)$ exists, how can I use this to build a model $N$ of $ZF(C)$ with the properties that I would like to show to be consistent. In the context of forcing, this gives some intuitive justification for wanting $M$ to be able to describe the forcing relation, since we are using it to construct the new model.

With this motivation in place, we can begin to address your question. The ordinals of a model $M$ serve to describe all possible index-lengths of processes that $M$ can carry out; by a process, I mean one with a well-defined "next" step in the process at each point, so the indexing must be wellordered. This is one way to justify the constructible universe $L^M\subseteq M$: $L^M_\alpha$ is the class of sets that $M$ can build using defining formulae over some transfinite process of length $\alpha$, and $L^M=\bigcup_{\alpha\in OR^M}L^M_\alpha$ is the collection of such constructions over all such process lengths that $M$ is aware of (with these process lengths given by $OR^M$). When dealing with forcing over a poset $\mathbb{P}$, the $\mathbb{P}-$names give a way to describe all possible constructions that $M$ can describe (hence, indexed over the ordinals from $M$ as in the case for $L^M$) with reference to an arbitrary new set $G\subseteq\mathbb{P}$. The $\mathbb{P}-$names are built by $M$ without knowing yet what $G$ will be, but with a clear idea for how to determine what each one becomes once $G$ is given.

Let's pause and take a different approach: we can think of $\mathbb{P}$ as a collection of pieces of information $p\in\mathbb{P}$. Our goal is to sew the information together into a coherent new set of information from $\mathbb{P}$ that $M$ does not yet know about, namely $G\subseteq\mathbb{P}$. Critically, we need two things: first, that the new set of information from $\mathbb{P}$ is coherent. This is why we require $G$ to be a filter of information; if $p$ is a piece of information in $G$, then $G$ contains all weaker information than $p$, and any two pieces of information $p$ and $q$ from $G$ must be compatible. The second thing we need relates to the generic condition about meeting dense subsets. If a new subset $X\subseteq\mathbb{P}$ is considered, $M$ in general has no way of knowing what arbitrary route through $\mathbb{P}$ was taken to build $X$. As finite mathematicians who borrow infinite processes from $M$, we cannot ourselves build some new $X$ to put in, and now neither can $M$, so how are we supposed to gather knowledge about the new structure $M[X]$? This dilemma is captured by your first quote: if $X$ encodes some information that makes it so $M$ cannot describe it without extra outside help, we're at a dead end. The solution is genericity. $M$ knows about each partial fragment $p$ of $G$ in its construction and this simple condition on meeting dense subsets gives us a mechanism to predict just enough about how $G$ is built past a point $p$ to make claims about what statements must be true of $M[G]$ given that $p\in G$. Although we still cannot know ultimately which points $p\in\mathbb{P}$ are added to build $G$, we can make predictions as to what statements must be true given that a point $p$ has been included in $G$, i.e. what sentences $\phi$ that are forced to be true if $p$ is a piece of information included in your construction of $G$.

How does the generic condition on dense subsets let us predict enough of the construction of $G$ beyond a point $p\in\mathbb{P}$ to meet this goal? A dense subset $D$ can be thought of as representing some property (for instance, $D$ could be the set of points that claim that the range of a function we are building includes the value $7$). If $D$ is dense in $\mathbb{P}$, at any partial point $p$ in the construction of $G$, it is still possible to extend and include a point from $D$ since there is some $d\in D$ with $d\leq p$. The dense sets represent properties that it is never too late to throw into your construction, and genericity states that if it is never too late to include such a point from $D$, then it is guaranteed that this happens at some point in the construction. This is an extremely strong predictive tool for what happens beyond a point $p$ in the construction of $G$! In everyday terms, at any given point in the day, it is possible for me to wash my dishes, yet I may still not complete the task. If this were a generic process, the possibility of washing my dishes at any point in the day would guarantee that the task be completed at some point before midnight. With this predictive strength for what must be included in $G$ beyond a point $p$ in its construction, we can make conclusions about certain types of pieces of information from $\mathbb{P}$ that must and must not be added to $G$, in a way that is sufficient to conclude which sentences $\phi$ must be true in $M[G]$ based solely on the fact that $p\in G$. Thus, $M$ knows about each possible point $p$ in the construction of $G$, and can use this generic condition to make predictions of this form, which allows $M$ to fully define the forcing relation and make claims about what would necessarily be the case if it came to know about some new generic filter $G\subseteq\mathbb{P}$ with any particular $p\in G$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .