Editing Forcing (mathematics) (section)

=== Examples ===

Let <math>S</math> be any infinite set (such as <math>\mathbb{N}</math>), and let the generic object in question be a new subset <math>X \subseteq S</math>. In Cohen's original formulation of forcing, each forcing condition is a ''finite'' set of sentences, either of the form <math>a \in X</math> or <math>a \notin X</math>, that are self-consistent (i.e. <math>a \in X</math> ''and'' <math>a \notin X</math> for the same value of <math>a</math> do not appear in the same condition). This forcing notion is usually called '''Cohen forcing'''.

The forcing poset for Cohen forcing can be formally written as <math> (\operatorname{Fin}(S,2),\supseteq,0) </math>, the finite partial functions from <math> S </math> to <math> 2 ~ \stackrel{\text{df}}{=} ~ \{ 0,1 \} </math> under ''reverse'' inclusion. Cohen forcing satisfies the splitting condition because given any condition <math>p</math>, one can always find an element <math>a \in S</math> not mentioned in <math>p</math>, and add either the sentence <math>a \in X</math> or <math>a \notin X</math> to <math>p</math> to get two new forcing conditions, incompatible with each other.

Another instructive example of a forcing poset is <math> (\operatorname{Bor}(I),\subseteq,I) </math>, where <math> I = [0,1] </math> and <math> \operatorname{Bor}(I) </math> is the collection of [[Borel subset]]s of <math> I </math> having non-zero [[Lebesgue measure]]. The generic object associated with this forcing poset is a '''random real number''' <math>r \in [0, 1]</math>. It can be shown that <math>r</math> falls in every Borel subset of <math>[0, 1]</math> with measure 1, provided that the Borel subset is "described" in the original unexpanded universe (this can be formalized with the concept of ''Borel codes''). Each forcing condition can be regarded as a random event with probability equal to its measure. Due to the ready intuition this example can provide, probabilistic language is sometimes used with other divergent forcing posets.