A quick thought

April 26, 2011 at 11:13 | Posted in Uncategorized | Leave a comment

This particular post is tangential to the task at hand. I just want play around with some of the ideas in yesterday’s posts to sharpen them a bit, and maybe prove a fact or two. I’m still not sure if I’m making things too difficult here by bringing in all the pcf stuff; it may be the case that these sorts of facts are well-known and provable by other methods. And once again: I’m writing these entries quickly, just trying to get ideas down, and there’s always the possibility I’m making silly mistakes somewhere. But that’s the point of a blog — I can work out things as I go along, and make corrections where I need to.

In the statement of the following theorem, {<^*} refers to the natural “modulo a bounded set” order, i.e. for {f} and {g} in {\prod\mathfrak{a}}, we have {f<^* g} if and only if the set of {\theta} for which {g(\theta)\leq f(\theta)} is bounded in {\mathfrak{a}}.

Theorem 1 Let {\mathfrak{a}} be a progressive set of regular cardinals. Then there is a {\xi<\sup\mathfrak{a}} such that

\displaystyle  \max{\rm pcf}(\mathfrak{a}\setminus\xi)={\rm cf}\left(\prod\mathfrak{a}, <^*\right). \ \ \ \ \ (1)

Recall that

\displaystyle  \max{\rm pcf}(\mathfrak{a}\setminus\xi)={\rm cf}\left(\prod (\mathfrak{a}\setminus\xi), <\right), \ \ \ \ \ (2)

where “{f<g}” means “{f(\theta)<g(\theta)} for all {\theta\in\mathfrak{a}}”, and so really the theorem gives us a {\xi<\sup\mathfrak{a}} such that

\displaystyle  {\rm cf}\left(\prod (\mathfrak{a}\setminus\xi), <\right)={\rm cf}\left(\prod\mathfrak{a}, <^*\right). \ \ \ \ \ (3)

Proof: It is clear that

\displaystyle  {\rm cf}\left(\prod\mathfrak{a}, <^*\right)\leq {\rm cf}\left(\prod (\mathfrak{a}\setminus\xi), <\right) \ \ \ \ \ (4)

for any {\xi<\sup\mathfrak{a}}, as a cofinal set in the right hand partial order yields a cofinal set in the left hand one.

Now let us define

\displaystyle  \tau:={\rm cf}\left(\prod\mathfrak{a}, <^*\right)^+ \ \ \ \ \ (5)

and consider the ideal {J} on {\mathfrak{a}} generated by {J_{<\tau}[\mathfrak{a}]} together with the bounded subsets of {\mathfrak{a}}.

We claim that {\mathfrak{a}\in J}. Why? Well, assume {J} is a proper ideal on {\mathfrak{a}}, and let {D} be an ultrafilter on {\mathfrak{a}} disjoint to {J}. Since {J_{<\tau}[\mathfrak{a}]\subseteq J}, we know

\displaystyle  {\rm cf}\left(\prod\mathfrak{a}, <^*\right)<\tau\leq{\rm cf}\left(\prod\mathfrak{a}, <_D\right). \ \ \ \ \ (6)

But {D} is also disjoint to the bounded ideal, so Conclusion 3.2 on page 62 of Cardinal Arithmetic tells us, we have

\displaystyle  {\rm cf}\left(\prod\mathfrak{a}, <_D\right)\leq{\rm cf}\left(\prod\mathfrak{a}, <^*\right) \ \ \ \ \ (7)

and we have a contradiction.

So, it follows that there is a {\xi<\sup(\mathfrak{a})} such that {\mathfrak{a}\setminus\xi\in J_{<\tau}[\mathfrak{a}]}. But

\displaystyle  \mathfrak{a}\setminus\xi\in J_{<\tau}[\mathfrak{a}]\Longleftrightarrow\max{\rm pcf}(\mathfrak{a}\setminus\xi)<\tau, \ \ \ \ \ (8)

and so

\displaystyle  \max{\rm pcf}(\mathfrak{a}\setminus\xi)\leq{\rm cf}\left(\prod\mathfrak{a}, <^*\right) \ \ \ \ \ (9)

as desired. \Box

One obvious corollary of the above is the following result on scales.

Corollary 2 Suppose {\mu} is singular, and {\langle \mu_i:i<{\rm cf}(\mu)\rangle} is an increasing sequence of regular cardinals carrying a scale of length {\mu^+}. Then there is an {i_0<{\rm cf}(\mu)} such that {\langle \mu_i:i_0\leq i<{\rm cf}(\mu)\rangle} carries a scale {\langle f_\alpha:\alpha<\mu^+\rangle} with the property that for any {g\in\prod_{i_0\leq i<{\rm cf}(\mu)}\mu_i}, there is an {\alpha<\mu^+} such that {g(i)<f(i)} whenever {i_0\leq i<{\rm cf}(\mu)}.

The above is easy to arrange by brute force if {\mu} is of countable cofinality (as the increasing sequence of regular cardinals has {\max{\rm pcf}} equal to {\mu^+}) but I don’t see immediately that the brute force argument generalizes to {\mu} of uncountable cofinality, where {\max{\rm pcf}} could be large by virtue of an initial segment of {\langle \mu_i:i<{\rm cf}(\mu)\rangle}.

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