Preservation, Cohen Forcing, and the Continuum: Examples and Counterexamples

For $n\in \omega$, let $A_n$ be the conditions $p$ with $p(\xi ,n)=1$ and no other coordinates in their domains. In this displayed canonical version $A_n$ is the singleton containing $\{((\xi ,n),1)\}$, hence an antichain; equivalently one may use any maximal antichain deciding that bit and retain only its value-$1$ part. Thus ${\dot{c}}_{\xi }=\{⟨\check{n},p⟩:n\in \omega ,p\in A_n\}$ is nice.

By name evaluation, $n\in ({\dot{c}}_{\xi }{)}_G$ exactly when some $p\in G$ sets $(\xi ,n)$ to $1$. Directedness makes this equivalent to $(\bigcup G)(\xi ,n)=1$, which is precisely $n\in c_{\xi }$.

Fix $0<\alpha <\theta$. For $n\in \omega$ let $D_n=\{p:(\alpha ,n)\in \mathrm{dom}p\}$, and for $\beta <\alpha$ let $E_{\beta }=\{p:\mathrm{\exists }np(\alpha ,n)=\beta \}$. Extend a finite condition at the requested coordinate, choosing a fresh $n$ for $E_{\beta }$, so both families are dense. A generic meets them all, and $g_{\alpha }(n)=(\bigcup G)(\alpha ,n)$ is a surjection $\omega \twoheadrightarrow \alpha$.

F1 gives $\theta$-cc, hence preserves the cardinal $\theta$. Since every infinite ordinal below it is countable by step 1.1, $\theta$ is the least uncountable cardinal in the extension, namely ${\mathrm{\aleph }}_1$.

Map $p$ to $(r_0,r_1)$, where $r_i(0,n)=p(i,n)$ whenever $(i,n)\in \mathrm{dom}p$. Each $r_i$ is a condition in $\mathrm{Add}(\omega ,1)$; conversely recover $p$ from $(r_0,r_1)$ by $p(i,n)=r_i(0,n)$. These maps are inverse and preserve reverse inclusion and compatibility. F1 applied to the partition yields the three model equalities and mutual genericity.

For distinctness, below any pair of finite conditions choose a fresh $n$ and extend the first with bit $0$ and the second with bit $1$. The resulting dense set is met, so the coordinate-union definition in F1 gives $c_0(n)\mathrm{\ne }{c}_1(n)$.

Let $P=\mathrm{Add}(\omega ,1)$. It is countable, hence ccc, as also recorded in F1. Define $p_n=\{((0,k),0):k<n\}$. Then $p_{n+1}\le p_n$, but a common lower bound would contain ${\bigcup }_n{p}_n$, an infinite function, and so would not be a condition. Therefore $P$ is not countably closed.

The explicit descending chain already refutes the claim without appealing to a B-page example. Under F1's strict $<\kappa$ convention, its assertion that this forcing is $\omega$-closed concerns only finite descending sequences and is not the advertised countable closure.