Forcing Orders, Names, and Generic Extensions: Examples and Counterexamples

Two conditions in G have a common extension and hence agree on the intersection of their domains. Thus their union c is a binary partial function. For each n, $D_n=\{s:\mathrm{length}(s)>n\}$ is a dense set in M: extend any short sequence with zeros to length n+1. Genericity supplies a condition of G of length above n, so the domain of c is all omega.

For each binary sequence $r\in M$, the set $E_r=\{s:\mathrm{\exists }n<\mathrm{length}(s)s(n)\mathrm{\ne }r(n)\}$ belongs to M and is dense. Given s, if it already disagrees it is in E_r; otherwise append the bit $1-r(\mathrm{length}(s))$. A condition of $G\cap E_r$ witnesses $c\mathrm{\ne }r$.

Internal Replacement and Union over the set of finite sequences and their finitely many coordinates form tau in M. Every entry has a name as its first coordinate, so tau is a name. F1 makes its value exactly $\{⟨n,s(n)⟩:s\in G,n<\mathrm{length}(s)\}$, the graph of c by step 1.1. Each graph coordinate is included by a condition covering n, and any selected entry agrees with c.

In the supplied M use $P=2^{<\omega }$ with extension order. Its finite sequences and order are the actual ones by transitivity and actual omega. Below each s the sequences formed by appending 0 and 1 are incompatible, so P is atomless. Apply F2 through the empty condition to obtain an M-generic filter G.

F1 now gives $G\mathrm{\notin }M$, refuting the universal claim. Equivalently, F3 makes $\bigcup G$ a binary sequence not in M, whereas G in M would put its actual union in M by internal Union and transitivity. The counterexample remains conditional on the supplied model; ZF does not here prove that such a model exists. Singleton forcing still has a ground-model generic filter, so the false claim is not replaced by an unconditional assertion about all forcing orders.

In this algebra join is union, meet is intersection and complement is relative to $\{0,1\}$. Every family has these bounds, so the algebra is complete. The empty atomic clauses give $E(e,e)=1_B$ and $I(e,e)=0_B$. Thus $I(e,t)=b\wedge 1_B=b$, while $E(t,e)=(\mathrm{\neg }b\vee 0_B)\wedge 1_B=\mathrm{\neg }b$.

The equality clause for t with itself has in both factors the single term $\mathrm{\neg }b\vee I(e,t)=\mathrm{\neg }b\vee b=1_B$, so $E(t,t)=1_B$. The only subname of t is e, whose valuation is empty. Since $b\in U_0$ and $b\mathrm{\notin }{U}_1$, the valuation rule gives respectively the singleton of empty and the empty set. These U_i are filters deciding each subset by whether it contains i; no ultrafilter-extension principle or generic truth theorem is used.