## Summary of first project

So what have we learned? The main thing is that we’ve now got a correct(ed) proof of the following result of Shelah, which we can state without any ambiguity:

Theorem 1 If ${\mu}$ is singular of uncountable cofinality ${\kappa}$, then

$\displaystyle {\rm pp}(\mu)=\mu^+ \Longleftrightarrow{\rm cov}(\mu,\mu,\kappa^+, 2)=\mu^+. \ \ \ \ \ (1)$

We’ve also got a pretty good idea of how this fits into the rest of Shelah’s work on “cov vs pp”. Most importantly though, forcing myself to work through his proof in this format has suddenly made most of his work on this problem seem much more accessible: the myriad arguments for partial results present in “Cardinal Arithmetic” are all just variations on the pattern of proof we’ve seen in the preceding posts. The next part of this project will consist of demonstrating this by giving streamlined (and cleaned up!) proofs of several more such results. (The point is that everything is much clearer in hindsight, so lots of very difficult proofs can be simplified in light of later discoveries.)

Note: I still want to go back and cross-reference the blog posts, and organize the titles and tags in a coherent way.

## Main Proof (part 3)

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)

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)

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)

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.