Does one need V not equal to L in order to do Cohen's forcing?

[nomadreid]

TL;DR Summary

Do I understand correctly that Cohen's technique forces sets from V which are not in L into the gap between the naturals and its power set? If so, does one use the implicit assumption that V is not equal to L?

Cohen's forcing essentially takes Gödel constructible universe L and adds sets from the larger universe of sets V until it provides a set having a cardinality between that of the naturals N and its power set. But in order to do that, you must assume that V has sets to add that are not in L, because if V=L, the continuum hypothesis CH holds. If this is right so far, then does one assume something like the existence of Ramsey cardinals to do the proof? I suspect I am missing something here, as I have never seen such an assumption in expositions of forcing.

 

[SSequence]

Unfortunately I have no idea about forcing whatsoever, so I don't think I can add much to help you. Very very vaguely I think its about building various models (that satisfy the axioms of set theory). But how, I don't have any idea.

Given some degree of (informal) familiarity with L, a couple of quite basic comments. One thing to note is that by default to get V=L (as a model) one must add an extra axiom (constructibility) to ZF. Quite roughly the axiom is about adding sets in a particular manner at each stage $L_\alpha$ to $L_{\alpha+1}$ while ensuring that the whole of V satisfies the set theory axioms.

But in order to do that, you must assume that V has sets to add that are not in L, because if V=L, the continuum hypothesis CH holds.

This sounds about right to me.

More specifically, it seems to me (pure speculation on my part and it may be wrong) that to move from CH being satisfied in a model to ~CH (in another model), one would definitely have to add many more real numbers. Also seemingly, the new reals would have to added in such a way so that none of the new reals that are added are able to code a well-order for the $\omega_1$ of the original model.

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Generally speaking, it is complicated though. One complication is that the notion of ordinals (such as $\omega_1$) isn't absolute between various models. In others words one can have two different models (both satisfying the the set-theory axioms) such that $\omega_1$ of first model is completely different from $\omega_1$ of second model.

Why I think it might matter (even for L) is that it makes the notion of L in itself "relative" in a certain sense I think. Because if one has two different models for "ZF+constructibility" with different $\omega_1$'s, one would have more reals in one model compared to the other one (both models satisfying CH ofc).

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I don't have much more to add, so hopefully is was at least a little bit helpful.

 

[nomadreid]

Thanks for your input, SSequence. Indeed, one of the trickier technicalities of Cohen's forcing is about not collapsing cardinals. Perhaps someone with more familiarity with forcing can then see if this point is relevant to my question.

 

[Citan Uzuki]

Generally, forcing starts with a countable transitive model $M$ of ZFC. Since it is countable, $M$ will not even be close to containing all subsets of the naturals, so it is possible to add a new generic set $G$ to $M$. $G$ will not be in the $L$ of $M[G]$, since $L^{M[G]}=L^M \subseteq M \subset M[G]$, but may be in the actual constructible universe $L$, since there are more ordinals in reality than there are in $M[G]$

 

[nomadreid]

Citan Uzuki, thanks very much for this. This raises a couple of questions, but I will start with a central one that may be leading me to my confusion. As I understand it, L is a minimal model of ZFC. As such, any such model M would be isomorphic to an elementary extension of L, no?

 

[Citan Uzuki]

No, L is not the minimal model of ZFC, it is merely the smallest transitive inner model that contains all the ordinals.

 

[nomadreid]

Super! Thanks very much, Citan Uzuki. You have found my central misunderstanding which had led me to my question, and corrected it. So, question answered.

 

[SSequence]

@Citan Uzuki
I don't understand your post#4 but I have few questions. In post#4, when given a model $M$, one writes $L^M$, I have two basic questions:
(i) Will $L^M$ also always be guaranteed to be a model (of set-theory) given that we know $M$ to be a model of set-theory.
(ii) Second question, is related to what I want to understand (to see if I have a misunderstanding here). Is $L^M$ "completely determined" by the ordinals contained in $M$ or not?

@nomadreid
To be honest, I don't quite understand the discussion in preceding posts. However, it seems that there should be many models (of set-theory) that will contain less sets than $L$. For example, there are countable models $M$ of set-theory (given its congruity). It seems to me, that any such model $M$ will not contain many of the ordinals contained in $L$ (in particular, many countable ordinals of $V$). That's what it seems like to me (I could be wrong).

 

[nomadreid]

SSequence, to answer the questions directed at me. You are correct; I mistakenly believed that L was the minimal model of ZFC, but as Citan Uzuki kindly pointed out, this belief was incorrect, probably having come from truncated the part about it being minimal if one requires all the ordinals to be in the model. So, yes, there are many countable transitive models M of ZF such that L has sets which are not in L. I was simply wrong.

I will be interested in Citan Uzuki's answer to your other questions.

 

[Citan Uzuki]

Hi SSequence:

(i) Will $L^M$ also always be guaranteed to be a model (of set-theory) given that we know $M$ to be a model of set-theory.

Yes.

(ii) Second question, is related to what I want to understand (to see if I have a misunderstanding here). Is $L^M$ "completely determined" by the ordinals contained in $M$ or not?

If $M$ is a transitive model, then yes. For general models, the answer is no, see this stackexchange thread: https://math.stackexchange.com/ques...there-is-a-isomorphism-between-their-ordinals

 

[SSequence]

@nomadreid
It seems to me that there is something that needs to be pointed out w.r.t. what I wrote in post#8. I am not sure it is worth bumping the thread for, but I wasn't aware of it at least. It seems that when I wrote in post#8:

For example, there are countable models $M$ of set-theory (given its congruity).

there are some further conditions that are required.

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I was looking at the MSE question linked in previous post. Looking at one of the suggested questions:
https://math.stackexchange.com/questions/85196/how-many-countable-models-of-zfc-are-there

One thing that the answer states is that for a model in which the statement ~con(ZFC) is true, inside that model there would be no countable model of ZFC at all (admittedly, I don't know why). After that the answer states a stronger condition (for a model) inside which a countable model is guaranteed.

I am guessing you would be aware of this already anyway. But it seems like useful enough point to add.

 

[nomadreid]

Thanks, SSequence. That information is worthwhile, as I am still struggling with the key elementary ideas (no pun intended) about forcing.

One thing that the answer states is that for a model in which the statement ~con(ZFC) is true, inside that model there would be no countable model of ZFC at all (admittedly, I don't know why).

That seems to be answered here
https://math.stackexchange.com/ques...of-zfc-inside-which-zfc-does-not-have-a-model

So, while I am on the subject and have the attention of people who understand these things, I have another question. (In for a penny, in for a pound; as well to be hung for a sheep as for a lamb, etc.) In Section 5 of "A beginner’s guide to forcing" by Timothy Y. Chow (https://arxiv.org/pdf/0712.1320.pdf), the author gives a brief outline of the key arguments. One point that I find puzzling is the following: “…construct a function F from the Cartesian product $\aleph$₂^M X $\aleph$₀ into the set 2={0,1}. We may interpret F as a sequence of functions from $\aleph$₀ into 2.”

I am a little puzzled by that last “interpret”: in which model? If we are interpreting it in M, then the Cartesian product would have cardinality $\aleph$₂^M, which in M would not be the same as $\aleph$₀. Of course, from a more powerful model, $\aleph$₂^M has the same cardinality as $\aleph$₀, but I don’t see why he would be saying that we would be interpreting it in a more powerful model.

I am guessing you would be aware of this already

No, best to assume I know nothing, and you will never be far wrong.

 

[SSequence]

I think the thread can be suitably be labeled as "advanced".

Yes, I have also learned a number of things from the thread since my knowledge of the topic is limited to reading first six chapters derek goldrei's book on sets (without doing any of the exercises) about seven years ago. Since its been a long time and I didn't make any notes while studying I can't re-call a good number of things that I read. But some of the more basic ideas, I can re-call.

That and skimming a large number of questions/answers on MSE/MO (for the most part somewhat randomly last year in '19) to try to understand whatever small bits and pieces that I could get (to make some sense of).

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I think you would better wait from a more informed reply to be sure. Here is my (informal) thought upon looking at this (I will use "$\omega$" instead of "$\aleph$").

First of all, I think it helps to forget about models momentarily and just look at an analogous function $G:\omega_2 \times \omega \rightarrow\{0,1\}$. Now I think it is reasonably clear that this function $G$ encodes $\omega_2$ many functions $G_\alpha:\omega \rightarrow\{0,1\}$ (with $\alpha<\omega_2$). For any arbitrary $\alpha<\omega_2$ we can define $G_\alpha$ as:
$G_\alpha(x)=G(\alpha,x)$ for all $x \in \omega$

Now to ensure that we have actually encoded $\omega_2$ many functions we also have to verify that (here we have $\alpha,\beta<\omega_2$):
$G_\alpha \neq G_\beta$ whenever $\alpha \neq \beta$

=================

Now let's come back to the function $F:\omega^M_2 \times \omega \rightarrow \{0,1\}$. In an analogous way, we can define an $F_\alpha$ for all $\alpha<\omega^M_2$. This function $F$ encodes functions $F_\alpha:\omega \rightarrow \{0,1\}$ (with $\alpha<\omega^M_2$). Now for any arbitrary $\alpha<\omega^M_2$ we can define:
$F_\alpha(x)=F(\alpha,x)$ for all $x\in\omega$

Now we want a similar condition as before to be true (here we have $\alpha,\beta<\omega^M_2$):
$F_\alpha \neq F_\beta$ whenever $\alpha \neq \beta$

=================

Now coming back to what you wrote:

I am a little puzzled by that last “interpret”: in which model? If we are interpreting it in M, then the Cartesian product would have cardinality $\aleph$₂^M, which in M would not be the same as $\aleph$₀. Of course, from a more powerful model, $\aleph$₂^M has the same cardinality as $\aleph$₀, but I don’t see why he would be saying that we would be interpreting it in a more powerful model.

Assuming that we would have $\omega^M=\omega$ (it is quite unclear to me whether this has to be kept as a necessary condition or not ... the text in linked document doesn't seem to mention this), as long as $F$ is contained in $M$ then [within $M$] $F$ would encode $\omega^2_M$ many distinct functions from $\omega^M$ to $\{0,1\}$. So, in that case, CH will be false in $M$.

 

[nomadreid]

Thanks very much for your helpful answer, SSequence. It is helping me to advance. I think I see my central problem (but I will need to work on it, so this may not be my last post); that I was looking at the 2^$\omega$₂ number of elements rather than the $\omega$₂ number of functions.

I shall go back to Square One and try again. Once again, thanks for your help and patience.