# Is the Forcing Technique an additional independent axiom to ZFC?

Can the Forcing Technique introduced by Cohen be considered to be an axiom or is it a 'technique' with no additional assumptions to ZFC. So does Forcing introduce new objects that are not in V ? The two references below seem to show that Forcing does and it does not create 'new' objects, like real numbers, to ZFC.

I read Wikipedia Forcing https://en.wikipedia.org/wiki/Forcing_(mathematics) which says :

"Intuitively, forcing consists of expanding the set theoretical universe $$V$$ to a larger universe $$V^*$$. In this bigger universe, for example, one might have many new real numbers, identified with subsets of the set $$\mathbb {N}$$ of natural numbers, that were not there in the old universe, and thereby violate the continuum hypothesis."

However https://plato.stanford.edu/entries/set-theory/#For says :

"The first problem we face is that M may contain already all subsets of ω. Fortunately, by the Löwenheim-Skolem theorem for first-order logic, M has an elementary submodel which is isomorphic to a countable transitive model N. So, since we are only interested in the statements that hold in M, and not in M itself, we may as well work with N instead of M, and so we may assume that M itself is countable. Then, since P(ω) is uncountable, there are plenty of subsets of ω that do not belong to M. But, unfortunately, we cannot just pick any infinite subset r of ω that does not belong to M and add it to M."

Any clarification will be greatly welcomed.

• Can you link to definitions of either? I suspect from reading this that the axiom you mean is this, but I'm not an expert.
– J.G.
Feb 15 at 13:49
• @J.G. no, those 2 axioms are related but not quite what the question is asking about. I'll answer the question in the when I get home (unless someone else will do it before) and it will hopefully be more clear then
– ℋolo
Feb 15 at 15:33
• The two quotes, concatenated and read, seem to serve as a good explanation to me. Intuitively, forcing is “extending the universe”, but this doesn’t make sense literally/rigorously. One way of making sense of it is the approach of using countable models which there are no obvious obstacles to extending. Feb 15 at 16:20
• @spaceisdarkgreen - any chance you could expand on your comment, particularly "Intuitively, forcing is “extending the universe”, but this doesn’t make sense literally/rigorously. ". I cant see how this is related to your next sentence " One way of making sense of it is the approach of using countable models which there are no obvious obstacles to extending.". Does it have something to do with there being no definite unique model of ZFC, that we could say "yes that is the model of ZFC"?
– user239186
Feb 15 at 21:01
• @LittleCheese I don’t understand the confusion. What is the universe but a giant model of ZFC? What is a countable model of ZFC but a tiny universe? Feb 15 at 23:29

## 1 Answer

Usually when doing forcing you start by assuming that there is a transitive (countable) set $$M$$ such that $$(M,∈)$$ satisfy all of the axioms of $$ZF(C)$$.

Then you extend $$M$$, intuitively we are extending the universe, but in reality we start with a set $$M$$ and we define a new set $$M[G]$$.

On face value this is indeed something stronger than $$ZF(C)$$, we need the set $$M$$ to exists, this is an assumption that is stronger than just $$ZF(C)$$.

That being said, it is possible to go around this problem and do everything inside of $$ZF(C)$$. $$ZF(C)$$ has a theorem called The Reflection Principle, which states that while there isn't a model of $$ZF(C)$$, there is a $$V_α$$ that satisfy arbitrary large finite fragment of $$ZF(C)$$.

So we start with large enough fragment $$T$$ of $$ZF(C)$$ to have all of the axioms we need (only finitely many), then use the LS theorem to get a countable elementary substructure of the $$V_α$$, then use Mostowski Collapse to get a transitive countable model of $$T$$, then we can do forcing on this fragment, all of this is done in $$ZF(C)$$, no extra assumptions needed.

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. Feb 20 at 21:21