Are equivalent (in ZFC) the following two statements, for any infinite cardinal $\mu$?


(i) For every infinite cardinal $\kappa$, $|\{ \lambda \in \kappa : \lambda \textrm{ is a singular cardinal and} \operatorname{pp} (\lambda) \geq \kappa \}| \leq \mu$.


(ii) For every infinite cardinal $\kappa$, $|\{ \lambda \in \kappa : \lambda \textrm{ is a singular cardinal and} \operatorname{pp} (\lambda) > \kappa \}| \leq \mu$.


The statement (i), with $\mu = \aleph\_0$, is the Shelah's Weak Hypothesis (SWH). See for instance page 360 on



> 
> S. Shelah. *Cardinal Arithmetic*,
>  volume 29 of Oxford Logic Guides. Oxford University Press, New York, 1994.
> 
> 
> 


The statements (ii) appear in the final section of



> 
> Moti Gitik. [*Short Extenders Forcings II*](http://www.tau.ac.il/~gitik/short%20extenders%20forcing%202-mar2013.pdf), preprint, July 24, 2013.
> 
> 
> 



Some variants of the Shelah's Weak Hypothesis
They are equivalent, though it took me a while to see it. This is perhaps my fifth attempt at getting a proof, so caveat lector.


Clearly (i) implies (ii), so assume by way of contradiction that (i) fails while (ii) holds.
Choose $\kappa$ least such that
$$|\{\lambda<\kappa:\rm{pp}(\lambda)\geq\kappa\}|>\mu.$$


We assume that (ii) holds, so we can show


* $\kappa$ is singular of cofinality $\mu^+$, and
* $\{\lambda<\kappa:\rm{pp}(\lambda)=\kappa\}$ is unbounded in $\kappa$ of order-type $\mu^+$


Our next move is to recall some facts about $\rm{pp}\_\theta(\lambda)$ where $\lambda$ is a singular cardinal. (Remember that $\rm{pp}\_\theta(\lambda)$ is just like $\rm{pp}(\lambda)$ except that we are allowed to use sets of size $\theta$ instead of sets of size $\rm{cf}(\lambda)$; thus $\rm{pp}(\lambda)=\rm{pp}\_{\rm{cf}(\lambda)}(\lambda)$.)


What we need is the following:


If $\rm{cf}(\lambda)\leq\theta<\lambda$, then


* $\rm{pp}(\lambda)\leq\rm{pp}\_\theta(\lambda)$,
* $\rm{cf}(\rm{pp}\_\theta(\lambda))>\theta$,
* if $\sigma<\lambda$ is singular of cofinality at most $\theta$ and $\rm{pp}\_\theta(\sigma)\geq\lambda$, then $\rm{pp}\_\theta(\lambda)\leq\rm{pp}\_\theta(\sigma)$, and
* $\rm{pp}\_\theta(\lambda)=\rm{pp}(\lambda)$ if $\rm{pp}\_\theta(\sigma)<\lambda$ for all sufficiently large singular $\sigma<\lambda$ of cofinality at most $\theta$


The first two of these are easy (famous last words...), the third is "Inverse Monotonicity" (see page 57 of Cardinal Arithmetic) while the last is one of the main conclusions of [Sh:371]: see $\otimes\_2$ on page 313 of Cardinal Arithmetic.


Since $\kappa$ has cofinality $\mu^+$, we find


$$\rm{pp}(\lambda)=\kappa\Longrightarrow \rm{pp}(\lambda)<\rm{pp}\_{\mu^+}(\lambda).$$


and therefore $\{\lambda<\kappa:\kappa<\rm{pp}\_{\mu^+}(\lambda)\}$ is unbounded in $\kappa$.


Suppose now that $\alpha<\kappa$, and let $\lambda\_\alpha$ be the least $\lambda>\alpha$ such that $\rm{pp}\_{\mu^+}(\lambda)>\kappa$.


If $\alpha<\sigma<\lambda\_\alpha$ and $\sigma$ is singular, then
$$\rm{pp}\_{\mu^+}(\sigma)<\kappa$$
by our choice of $\lambda\_\alpha$.


But then we know 
$$\rm{pp}\_{\mu^+}(\sigma)<\lambda\_\alpha$$
as otherwise inverse monotonicity would imply
$$\rm{pp}\_{\mu^+}(\lambda\_\alpha)\leq\rm{pp}\_{\mu^+}(\sigma)<\kappa$$
contradicting our choice of $\lambda\_\alpha$.


But now [Sh:371] applies to $\lambda\_\alpha$, and we conclude
$$\kappa<\rm{pp}(\lambda\_\alpha)=\rm{pp}\_{\mu^+}(\lambda\_\alpha)$$.


Thus, $\{\lambda<\kappa:\kappa<\rm{pp}(\lambda)\}$ is unbounded in $\kappa$ hence of cardinality at least $\mu^+$. This violation of (ii) gives us our contradiction.


