Forcing (mathematics) - Cohen Forcing

Cohen Forcing

The simplest nontrivial forcing poset is ( Fin(ω,2), ⊇, 0 ), the finite partial functions from ω to 2={0,1} under reverse inclusion. That is, a condition p is essentially two disjoint finite subsets p−1 and p−1 of ω, to be thought of as the "yes" and "no" parts of p, with no information provided on values outside the domain of p. q is stronger than p means that q ⊇ p, in other words, the "yes" and "no" parts of q are supersets of the "yes" and "no" parts of p, and in that sense, provide more information.

Let G be a generic filter for this poset. If p and q are both in G, then p∪q is a condition, because G is a filter. This means that g=⋃G is a well-defined partial function from ω to 2, because any two conditions in G agree on their common domain.

g is in fact a total function. Given n ∈ ω, let D_n={ p : p(n) is defined }, then D_n is dense. (Given any p, if n is not in p’s domain, adjoin a value for n, the result is in D_n.) A condition p ∈ G∩D_n has n in its domain, and since p ⊆ g, g(n) is defined.

Let X=g−1, the set of all "yes" members of the generic conditions. It is possible to give a name for X directly. Let X = { ( nˇ, p ) : p(n)=1 }, then val( X, G ) = X. Now suppose A⊆ω in V. We claim that X≠A. Let D_A = { p : ∃n, n∈dom(p) and p(n)=1 if and only if n∉A }. D_A is dense. (Given any p, if n is not in p’s domain, adjoin a value for n contrary to the status of "n∈A".) Then any p∈G∩D_A witnesses X≠A. To summarize, X is a new subset of ω, necessarily infinite.

Replacing ω with ω×ω₂, that is, consider instead finite partial functions whose inputs are of the form (n,α), with n<ω and α<ω₂, and whose outputs are 0 or 1, one gets ω₂ new subsets of ω. They are all distinct, by a density argument: given α<β<ω₂, let D_α,β={p:∃n, p(n,α)≠p(n,β)}, then each D_α,β is dense, and a generic condition in it proves that the αth new set disagrees somewhere with the βth new set.

This is not yet the falsification of the continuum hypothesis. One must prove that no new maps have been introduced which map ω onto ω₁ or ω₁ onto ω₂. For example, if one considers instead Fin(ω,ω₁), finite partial functions from ω to ω₁, the first uncountable ordinal, one gets in V a bijection from ω to ω₁. In other words, ω₁ has collapsed, and in the forcing extension, is a countable ordinal.

The last step in showing the independence of the continuum hypothesis, then, is to show that Cohen forcing does not collapse cardinals. For this, a sufficient combinatorial property is that all of the antichains of this poset are countable.