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Suppose $S \subset \omega_{1}$ is stationary co-stationary. Then there is a forcing notion $P_{S}$ which shoots a closed unbounded $C \subset S$ without collapsing cardinals (or changing cofinalities).

Let $P$ = $Levy$($\aleph_{0}$, $< \aleph_{1}$). So P is essentially adding $\aleph_{1}$ Cohen reals. If $G_{P} \subset P$ is (directed and) generic over V, then $G_{P}$ is also generic over $L$ (the constructible universe) as $P \in L$ and also over $L[S]$. By Theorem I 6.7 the forcing P satisfies the countable chain condition and $\aleph_{1}^{L[G_{P}]} = \aleph_{1}^{V} = \aleph_{1}^{V[G_{P}]}$ and $V$, $V[G_{P}]$ have the same cardinals and cofinalities. Let

$Q$ = {$C$ : $C$ is a closed bounded subset of $S$ which belongs to $L[S, G_{P}]$} and $C_{1} < C_{2}$ iff $C_{1} = C_{2} \cap (Max(C_{1}) + 1)$.

Clearly $Q$ is a forcing notion of power $\aleph_{1}$, so it cannot collapse cardinals or regularity of cardinals except possibly $\aleph_{1}$ (all finite subsets of $S$ belong to $Q$). So we shall prove that $P * \dot{Q}$ does not collapse $\aleph_{1}$. So let (in $V$) $N \prec (H(\lambda), \epsilon)$, $N$ countable, $(p, \dot{q}) \in P * \dot{Q} \in N$, $(p, \dot{q}) \in N$ , $\delta := N \cap \omega_{1} \in S$ and suppose $G_{P} \subset P$ is generic over $V$ and $p \in G_{P}$. Note that as $S$ is a stationary subset of $\omega_{1}$, there is such $N$ (in $V$). So it is enough to find $(p',\dot{q'}) \geq (p, \dot{q})$ which is $(N, P * \dot{Q})$-generic. As $P$ satisfies the countable chain condition, p is $(N, P)$-generic (by 2.9), hence $N[G_{P}] \cap V = N$. Clearly $\dot{Q}[G_{P}] \in L[S, G_{P}] \subset V[G_{P}]$, now $N[G_{P}]$ does not necessarily belong to $L[S, G_{P}]$, but $N[G_{P}] \cap L[S, G_{P}]$ is $N[G_{\gamma}] \cap L_{\delta}[S, G_{P}] = L_{\delta}[S, G_{P}] \in L[S, G_{P}]$ and is a countable set in $L[S, G_{P}]$.

In $L[S, G_{P}]$ we have an enumeration of $\dot{Q}[G_{P}] \cap N[G_{P}]$ (of length $\omega$), say $\langle q_{n} : n < \omega \rangle$ (but not of the set of dense subsets); in fact we have it even in $L[S, G_{P} \upharpoonright(\delta + 1)]$ (and as we use $Levy(\aleph_{0}, < \aleph_{1})$ not $Levy(\aleph_{0}, < \kappa)$ even in $L[S, G_{P} \upharpoonright \delta]$). Now in $L[S, G_{P}]$ there is a Cohen generic real over $V[G_{P} \upharpoonright (\delta + 1)]$ say $r^{*} \in$ $^{\omega}{\omega}$ and we use it to construct a sequence $\bar{C} = \langle C_{n} : n < \omega \rangle$ such that $C_{n} \in \dot{Q}[G_{P}]$ in $L[S, G_{P}]$ i.e. $\bar{C} \in L[S, G_{P}]$, e.g. we choose $C_{n}$ by induction on $n$; we let $C_{0} = \dot{q}[G_{P}]$ and we let $C_{n+1}$ be $q_{m(n)}$ where $m(n)$ is the first natural number $m$ such that: $m \geq r^{*}(n)$ and $\dot{Q}[G_{P}] \models$ "$C_{n} \leq q_{m}$". Let

$G$ := {$q$ : $q \in \dot{Q}[G_{P}]$, and $q \in N[G_{P}]$ and for some $n$, $\dot{Q}[G_{P}] \models$ "$q \leq C_{n}$"}.

Clearly $G \subset N[G_{P}] \cap \dot{Q}[G_{P}]$ is generic over $V[G_{P} \upharpoonright (\delta + l)]$. So $q^{\dagger} = (\bigcup_{n} C_{n}) \cup \{\delta\} \in \dot{Q}[G_{P}]$ is $(N[G_{P}], \dot{Q}[G_{P}])$-generic. Going to names we finish.

A few questions regarding the proof above (Theorem 4.4 of Shelah's PIF). Here '$p \geq q$' means '$p$ extends $q$'.

First let $A \in N[G_{P}]$ be a maximal antichain of $\dot{Q}[G_{P}]$.

  1. How do we ensure that the $G$ constructed at the end of the proof intersect $N[G_{P}] \cap A$?
  2. If we cannot ensure the above, how do we show that $q^{\dagger}$ is ($N[G_{P}]$, $\dot{Q}[G_{P}]$)-generic?

I think there is some crucial properties of Cohen reals that I am missing here, such as how Cohen-genericity translate to $\dot{Q}[G_{P}]$-genericity, if at all.

Much thanks in advance.

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up vote 3 down vote accepted

Sorry this is sketchy, but I hope it helps!

A Cohen condition $p$ of length $n$ is going to determine $\langle C_m:m<n\rangle$. We know there's going to be an $i<\omega$ such that $q_i$ extends both $C_{n-1}$ and a member of $N[G_P]\cap A$. Extend $p$ to a Cohen condition $p'$ with $p'(n)=i$. If $r^*$ is any Cohen generic extending $p'$, then running the construction with $r^*$ guarantees that $C_n$ extends a member of $N[G_P]\cap A$. So a density argument tells us the construction is forced to work.

The point is that in Shelah's construction, $m(n)$ is going to be equal to $r^*(n)$ very often because $r^*$ is Cohen.

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Answered this quickly as I'm leaving work. If on the way home I realize I've said something silly, I'll edit! –  Todd Eisworth May 9 '13 at 20:59
    
Ah I get it! Thank you so much. Just one more question. To conclude that $P * \dot{Q}$ is proper from the argument in the proof, I reckon we would need {$N$ : $N$ is a countable subset of $H(\lambda, \epsilon)$ and $N \cap \omega_{1} \in S$} to contain a club. But I cannot seem to show this. Is this true or am I going in a wrong direction altogether? –  Zoorado May 10 '13 at 14:12
    
It won't be proper if $S$ is co-stationary (as the stationarity of the complement of $S$ is destroyed), but it will preserve $\omega_1$ because you get generics for "enough" elementary submodels. –  Todd Eisworth May 10 '13 at 15:19
    
Ok I see. Thanks again! –  Zoorado May 11 '13 at 11:51
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