Main Proof (part 3)

March 8, 2011 at 13:43 | Posted in Uncategorized | Leave a comment

We will now define certain increasing sequences {\langle \alpha_n:n<\omega\rangle} and {\langle i_n:n<\omega\rangle} as in the last post, along with another sequence {\langle \epsilon_n:n<\omega\rangle}.

Start by setting {\alpha_0=\min({\rm nacc}(C_\delta))} and {i_0=0}. Given {\alpha_n} and {i_n}, we let {\alpha_{n+1}} be the least element of {{\rm nacc}(C_\delta)} above {\alpha_n} with the property that

\displaystyle  \gamma(\alpha_n)<\sup(C_{\alpha_{n+1}}), \ \ \ \ \ (1)

and define

\displaystyle  \epsilon_n = \min(C_{\alpha_{n+1}}\setminus\gamma(\alpha_n). \ \ \ \ \ (2)

What is the point? Note that this arrangement ensures that

\displaystyle  f_{\epsilon_n}\in \mathcal{N}(\alpha_{n+1}, i) \ \ \ \ \ (3)

for any {i<\kappa}. Since {\gamma(\alpha_n)\leq\epsilon}, it follows that we can choose {i_{n+1}\geq i_n} such that “{f_{\epsilon_n}} majorizes {g(\alpha_n, i_n)} beyond {\mu_{i_{n+1}}}”, i.e.,

\displaystyle  \sup(\mathcal{N}^+(\alpha_n, i_n)\cap\theta)<f_{\epsilon_n}(\theta) \ \ \ \ \ (4)

whenever {\theta\in \mathfrak{a}_{\alpha_n, i_n}\setminus \mu_{i_{n+1}}}.

Now define

\displaystyle  N=\bigcup_{n<\omega}\mathcal{N}(\alpha_n, i_n), \ \ \ \ \ (5)

and

\displaystyle  N^+=\bigcup_{n<\omega}\mathcal{N}^+(\alpha_n, i_n). \ \ \ \ \ (6)

We know the following facts from previous posts:

  • {N\in M^*\cap [\mu]^{<\mu}=\mathcal{P}},
  • {N\subseteq N^+}, and
  • {a\subseteq N^+}.

Thus, the theorem will be done provided we establish

\displaystyle  N\cap\mu = N^+\cap \mu. \ \ \ \ \ (7)

Here we use the “Elementary submodel argument” from February 12, which tells us that we have what we want provided

\displaystyle  \theta\in N\cap\mu\cap{\sf Reg}\Longrightarrow \sup(N^+\cap\theta)=\sup(N\cap\theta). \ \ \ \ \ (8)

The above is satisfied trivially for {\theta\leq\mu_{i^*}} as {\mu_{i^*}\subseteq N}; thus, we need only verify things for {\theta\in N\cap\mu\cap{\sf Reg}} which are greater than {\mu_{i^*}}. Furthermore, we know {N\subseteq N^+}, so

\displaystyle  \sup(N\cap\theta)\leq\sup(N^+\cap\theta) \ \ \ \ \ (9)

in any case.

Assume now by way of contradiction that we have {\theta} such that

  • {\theta\in N\cap\mu\cap{\sf Reg}},
  • {\mu_{i^*}<\theta}, and
  • {\sup(N\cap\theta)<\sup(N^+\cap\theta)}.

Choose {n<\omega} so large that

  • {\sup(N\cap\theta)<\sup(\mathcal{N}^+(\alpha_n, i_n)\cap\theta)}, and
  • {\theta\in \mathcal{N}(\alpha_n, i_n)}.

Since {\mu_{i_{n+1}}<\mu_{i^*}} and {\theta\in\mathfrak{a}_{\alpha_n, i_n}}, it follows from the definition of {\epsilon_n} that

\displaystyle  \sup(\mathcal{N}^+(\alpha_n, i_n)\cap\theta)<f_{\epsilon_n}(\theta). \ \ \ \ \ (10)

But

\displaystyle  f_{\epsilon_n}(\theta)\leq \sup(\mathcal{N}(\alpha_{n+1}, i_{n+1})\cap\theta) \ \ \ \ \ (11)

as both {\mathfrak{a}_{\alpha_n, i_n}} and {f_{\epsilon_n}} are in the latter model. Thus, on the one hand we have

\displaystyle  \sup(N\cap\theta)\leq \sup(\mathcal{N}^+(\alpha_n, i_n)\cap\theta)<f_{\epsilon_n}(\theta), \ \ \ \ \ (12)

and on the other,

\displaystyle  f_{\epsilon_n}(\theta)\leq \sup(\mathcal{N}(\alpha_{n+1}, i_{n+1})\cap\theta)\leq \sup(N\cap\theta). \ \ \ \ \ (13)

Clearly, this is absurd.

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