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.
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.