What else does ZFC prove about the "spectrum" of a cardinal number?

147 Views Asked by At

An auxiliary definition:

Definition 0. Given an infinite set $X$ and a filter $\mathcal{F}$ on $X$, let $\sim_\mathcal{F}$ denote the equivalence relation on $\mathcal{P}(X)$ defined as follows: $A \sim_\mathcal{F} B$ iff we can find $K \in \mathcal{F}$ such that $A \cap K = B \cap K$.

Now for the main one:

Definition 1. By the spectrum of a cardinal number $\kappa$, I simply mean the set of all infinite cardinal numbers $\nu$ such that there exists a filter $\mathcal{F}$ on $\mathcal{P}(\kappa)$ satisfying $|\mathcal{P}(\kappa)/\!\sim_{\mathcal{F}}\!| = \nu$.

Write $\mathrm{Spec}(\kappa)$ for the spectrum of $\kappa$.

Clearly, if $\kappa$ is an infinite cardinal number, then $2^\kappa \in \mathrm{Spec}(\kappa).$ But honestly, that's about all I can show!

Questions. What else does ZFC prove about $\mathrm{Spec}(\kappa)$? In particular:

  1. Given $\nu \in \mathrm{Spec}(\kappa)$ and $\nu'$ in the interval $[\nu,2^\kappa]$, can we conclude that $\nu' \in \mathrm{Spec}(\kappa)$?

  2. Is it a general fact that $\kappa^+ \in \mathrm{Spec}(\kappa)$? (Note that this follows from GCH.)

  3. Is it a general fact that $\kappa \in \mathrm{Spec}(\kappa)$? If not, does this imply an outright contradiction?

I'm also curious to know whether the existence of cardinal numbers that do not belong to $\mathrm{Spec}(\kappa)$ for any $\kappa$ is consistent with ZFC.

1

There are 1 best solutions below

6
On BEST ANSWER

It's not too hard to show that $\operatorname{Spec}(\kappa)$ contains no cardinals smaller than $2^{\aleph_0}$. Indeed, let $F$ be any filter on $\mathcal{P}(\kappa)$ such that the quotient $B=\mathcal{P}(\kappa)/F$ is infinite. Since $B$ is an infinite Boolean algebra, we can find an infinite sequence of disjoint nonzero elements of $B$, and by induction we can find representatives $a_n\in\mathcal{P}(\kappa)$ of these elements which are actually disjoint in $\mathcal{P}(\kappa)$. For any $S\subseteq \omega$, let $a_S=\bigcup_{n\in S} a_n$. Then for $S\neq T$, $a_S\neq a_T$ mod $F$, since their symmetric difference contains some $a_n$. Thus $S\mapsto a_S$ is an injective homomorphism $\mathcal{P}(\omega)\to B$.