Thursday, June 13, 2013

Around Observation 5.5

\def\REG{\sf {REG}}

I apologize for the lack of context for this post, but I'm just trying to get some ideas down.

Proposition: Assume \(\cf(\lambda)<\theta=\cf(\theta)<\lambda\).    Then

$(1)\quad\quad \pp(\lambda)=\pp_\theta(\lambda)\Longrightarrow\cov(\pp(\lambda),\lambda,\theta^+,\theta)=\pp(\lambda).$

Proof:   We know $\lambda<\pp(\lambda)$ so $\lambda<\cov(\pp(\lambda),\lambda,\theta^+,\theta)$. Thus, by the cov vs. pp theorem (Theorem 5.4 on page 87 of The Book) it suffices to prove the following:

If $\lambda\leq\mu\leq\pp(\lambda)$ and $\cf(\mu)=\theta$, then $\pp_{\Gamma(\theta)}(\mu)\leq\pp(\lambda)$.

Clearly for such a $\mu$ we have $\pp_{\Gamma(\theta)}(\mu)\leq\pp(\mu)$, so we must show

$(2)\quad\quad \lambda\leq\mu\leq\pp(\lambda)\wedge\cf(\mu)=\theta\Longrightarrow\pp(\mu)\leq\pp(\lambda)$

But  we know the following:

  • $\pp(\mu)=\pp_\theta(\mu)$ by definition.
  • $\pp(\lambda)=\pp_\theta(\lambda)$ by assumption, and
  • $\pp_\theta(\mu)\leq\pp_\theta(\lambda)$ by "inverse monotonicity" of $\pp_\theta$  (Conclusion 2.3(2) on page 57 of The Book)
Putting all these together gives us what we need.   Q.E.D.

As far as I know, it is still unknown if the hypothesis of the preceding proposition can fail.  We do know the following:

  • If $\lambda$ is a strong limit and $\cf(\lambda)\leq\theta=\cf(\theta)<\lambda$, then $\pp(\lambda)=\pp_\theta(\lambda)$.
  • Actually, instead of assuming $\lambda$ is a strong limit, it suffices to assume that $\pp_\theta(\mu)<\lambda$ for all sufficiently large $\mu<\lambda$ with $\cf(\mu)\leq\theta$.  This is part of Corollary 1.6 on page 321.
  • We could also assume $\pp(\lambda)<\lambda^{+(\cf(\lambda))^+}$ and obtain the same conclusion.
What is the point?  In the first place, the argument given above allows us to deduce Observation 5.5 on page 404.  In the second place...well, I'll write the next piece out tomorrow.   

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