Main Proof (part 2)

February 21, 2011 at 11:34 | Posted in Uncategorized | Leave a comment

Proposition 1 Suppose {\langle\alpha_n:n<\omega\rangle} is increasing in {{\rm nacc}(C_\delta)} and {\langle i_n:n<\omega\rangle} is increasing in {\kappa}. Then

\displaystyle  \bigcup_{n<\omega}\mathcal{N}(\alpha_n,i_n)\in M^*. \ \ \ \ \ (1)

Proof: Define

  • {\alpha^*:=\sup\{\alpha_n:n<\omega\}},
  • {C^*=\bigcup_{n<\omega}C_{\alpha_n}}, and
  • {i^*:=\sup\{i_n:n<\omega\}}.

Here is where the choice of {\langle C_\alpha:\alpha<\mu^+\rangle} comes to our rescue. Since {\alpha_n\in{\rm nacc}(C_\delta)} for each {n}, it follows that

\displaystyle  C_{\alpha_n}=C_\delta\cap\alpha_n \ \ \ \ \ (2)

for each {n}, and therefore

\displaystyle  C_{\alpha_n} = C_{\alpha_{n+1}}\cap\alpha_n\subseteq C_{\alpha_{n+1}}, \ \ \ \ \ (3)

and

\displaystyle  C^*=C_\delta\cap\alpha^*. \ \ \ \ \ (4)

Now it follows easily that

\displaystyle  \bigcup_{n<\omega}\mathcal{N}(\alpha_n, i_n)={\rm Sk}_{\mathfrak{A}}(\mu_{i^*}\cup\{f_\xi:\xi\in C^*\}). \ \ \ \ \ (5)

Next, we define

\displaystyle  \epsilon:=\min({\rm nacc}(C_\delta)\setminus\alpha^*). \ \ \ \ \ (6)

Given our definition of {\langle M_\alpha:\alpha<\mu^+\rangle}, it is not hard to see

  • {\epsilon\cup\mu\subseteq M_\epsilon}, and
  • every proper initial segment of {\langle f_\xi:\xi<\epsilon\rangle} is in {M_\epsilon}.

In particular,

\displaystyle  \mu_{i^*}\cup\{f_\xi:\xi\in C^*\}\subseteq M_\epsilon. \ \ \ \ \ (7)

Note that {C^*\in M_{\epsilon+1}}, as it is definable from {\bar{C}}, {\epsilon}, and {\alpha^*}:

\displaystyle  C^*= C_\epsilon\setminus \{\alpha^*\}, \ \ \ \ \ (8)

and so

\displaystyle  \bigcup_{n<\omega}\mathcal{N}(\alpha_n, i_n)\in M_{\epsilon+1}\subseteq M^*, \ \ \ \ \ (9)

as it can be computed in {M_{\epsilon+1}} by taking the Skolem hull of {\mu_{i^*}\cup\{f_\xi:\xi\in C^*\}} in the model {M_\epsilon}.

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