## A non fixed point by any other name…

This is a silly post, but looking ahead to Conclusion 3.2A reminded me of one of the frustrations I had when first trying to read The Book. One of the major themes of pcf theory is that it tells us quite a lot about cardinal arithmetic at singular cardinals ${\mu}$ that are not fixed points of the ${\aleph}$ function. The problem is that this is equivalent to lots of other conditions, and very often Shelah will use a different formulation of this property in the statement of a theorem, so it isn’t clear right away that what he’s really talking about is a cardinal that isn’t a fixed point. This little proposition is just an elementary exercise in set theory, but it’s useful to keep in mind when reading The Book. (And let’s hope I haven’t flubbed it since I’m typing from home while babysitting…)

Proposition 1 Let ${\mu}$ be a singular cardinal. Then the following are equivalent:

1. ${\mu}$ is not a fixed point, i.e., ${\mu<\aleph_\mu}$.
2. ${\mu=\aleph_{\xi+\zeta}}$ where ${\zeta}$ is a limit ordinal ${<\aleph_\xi}$.
3. There is a ${\tau<\mu}$ such that ${(\tau,\mu)\cap{\sf Reg}}$ is progressive.

Proof: Suppose ${\mu<\aleph_\mu}$, and fix ${\delta<\mu}$ such that ${\mu=\aleph_\delta}$. Since ${\mu}$ is a limit cardinal, we know ${|\delta|^+<\mu}$, so choose ${\xi<\delta}$ such that

$\displaystyle |\delta|^+=\aleph_\xi. \ \ \ \ \ (4)$

Now fix ${\zeta}$ such that

$\displaystyle \delta=\xi+\zeta. \ \ \ \ \ (5)$

It is clear that ${\zeta}$ must be a limit (as ${\delta}$ is), and the fact that

$\displaystyle \mu=\aleph_\delta=\aleph_{\xi+\zeta} \ \ \ \ \ (6)$

is completely trivial. For the last bit, we know

$\displaystyle |\zeta|\leq |\delta|<|\delta|^+=\aleph_\xi, \ \ \ \ \ (7)$

and we are done with proving that (1) implies (2)

Now assume (2) holds, set ${\tau=\aleph_\xi}$, and define

$\displaystyle \mathfrak{a}=(\tau,\mu)\cap{\sf Reg}. \ \ \ \ \ (8)$

Now note

$\displaystyle |\mathfrak{a}|\leq |(\tau,\mu)|=|\zeta|<\tau<\tau^+=\min(\mathfrak{a}), \ \ \ \ \ (9)$

and we have (3).

Moving on to the last implication, let us suppose ${\tau<\mu}$ and ${\mathfrak{a}=(\tau,\mu)\cap{\sf Reg}}$ is progressive, i.e., ${|\mathfrak{a}|\leq\tau}$. But then ${|(\tau,\mu)\cap{\sf Card}|\leq\tau}$ as well, and hence

$\displaystyle |\mu\cap{\sf Card}|\leq\tau. \ \ \ \ \ (10)$

Thus, we have

$\displaystyle \mu\leq\aleph_{\tau^+}<\aleph_\mu, \ \ \ \ \ (11)$

as required.

$\Box$