Editing Forcing (mathematics) (section)

== Cohen forcing ==

The simplest nontrivial forcing poset is <math> (\operatorname{Fin}(\omega,2),\supseteq,0) </math>, the finite partial functions from <math> \omega </math> to <math> 2 ~ \stackrel{\text{df}}{=} ~ \{ 0,1 \} </math> under ''reverse'' inclusion. That is, a condition <math> p </math> is essentially two disjoint finite subsets <math> {p^{-1}}[1] </math> and <math> {p^{-1}}[0] </math> of <math> \omega </math>, to be thought of as the "yes" and "no" parts of {{nowrap|<math> p </math>,}} with no information provided on values outside the domain of <math> p </math>. "<math> q </math> is stronger than <math> p </math>" means that <math> q \supseteq p </math>, in other words, the "yes" and "no" parts of <math> q </math> are supersets of the "yes" and "no" parts of <math> p </math>, and in that sense, provide more information.

Let <math> G </math> be a generic filter for this poset. If <math> p </math> and <math> q </math> are both in <math> G </math>, then <math> p \cup q </math> is a condition because <math> G </math> is a filter. This means that <math> g = \bigcup G </math> is a well-defined partial function from <math> \omega </math> to <math> 2 </math> because any two conditions in <math> G </math> agree on their common domain.

In fact, <math> g </math> is a total function. Given <math> n \in \omega </math>, let <math> D_{n} = \{ p \mid p(n) ~ \text{is defined} \} </math>. Then <math> D_{n} </math> is dense. (Given any <math> p </math>, if <math> n </math> is not in <math> p </math>'s domain, adjoin a value for <math> n </math>—the result is in <math> D_{n} </math>.) A condition <math> p \in G \cap D_{n} </math> has <math> n </math> in its domain, and since <math> p \subseteq g </math>, we find that <math> g(n) </math> is defined.

Let <math> X = {g^{-1}}[1] </math>, the set of all "yes" members of the generic conditions. It is possible to give a name for <math> X </math> directly. Let

:<math> \underline{X} = \left \{ \left (\check{n},p \right ) \mid p(n) = 1 \right \}.</math>

Then <math>\operatorname{val}(\underline{X},G) = X.</math> Now suppose that <math> A \subseteq \omega </math> in <math> V </math>. We claim that <math> X \neq A </math>. Let

:<math> D_{A} = \{ p \mid (\exists n)(n \in \operatorname{Dom}(p) \land (p(n) = 1 \iff n \notin A)) \}.</math>

Then <math>D_A</math> is dense. (Given any <math> p </math>, find <math> n </math> that is not in its domain, and adjoin a value for <math> n </math> contrary to the status of "<math> n \in A </math>".) Then any <math> p \in G \cap D_A</math> witnesses <math> X \neq A </math>. To summarize, <math> X </math> is a "new" subset of <math> \omega </math>, necessarily infinite.

Replacing <math> \omega </math> with <math> \omega \times \omega_{2} </math>, that is, consider instead finite partial functions whose inputs are of the form <math> (n,\alpha) </math>, with <math> n < \omega </math> and <math> \alpha < \omega_{2} </math>, and whose outputs are <math> 0 </math> or <math> 1 </math>, one gets <math> \omega_{2} </math> new subsets of <math> \omega </math>. They are all distinct, by a density argument: Given <math> \alpha < \beta < \omega_{2} </math>, let

:<math> D_{\alpha,\beta} = \{ p \mid (\exists n)(p(n,\alpha) \neq p(n,\beta)) \},</math>

then each <math> D_{\alpha,\beta} </math> is dense, and a generic condition in it proves that the αth new set disagrees somewhere with the <math> \beta </math>th new set.

This is not yet the falsification of the continuum hypothesis. One must prove that no new maps have been introduced which map <math> \omega </math> onto <math> \omega_{1} </math>, or <math> \omega_{1} </math> onto <math> \omega_{2} </math>. For example, if one considers instead <math> \operatorname{Fin}(\omega,\omega_{1}) </math>, finite partial functions from <math> \omega </math> to <math> \omega_{1} </math>, the [[first uncountable ordinal]], one gets in <math> V[G] </math> a bijection from <math> \omega </math> to <math> \omega_{1} </math>. In other words, <math> \omega_{1} </math> 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 [[antichain]]s of the forcing poset are countable.