\DeclareMathOperator{\pp}{pp}
\DeclareMathOperator{\tcf}{tcf}
\DeclareMathOperator{\pcf}{pcf}
\DeclareMathOperator{\cov}{cov}
\def\cf{\rm{cf}}
\DeclareMathOperator{\otp}{otp}
\def\REG{\sf {REG}}
\def\restr{\upharpoonright}
\def\bd{\rm{bd}}
\def\subs{\subseteq}
\def\cof{\rm{cof}}
\def\ran{\rm{ran}}
\DeclareMathOperator{\ch}{Ch}
\DeclareMathOperator{\PP}{pp}
\DeclareMathOperator{\Sk}{Sk}
\)
Granted, the title of this post is not very illuminating, as there are several theorems due to Gitik and Shelah. What I have in mind is the following result:
Theorem 1.5 of [GiSh:412]
If $\mu>\kappa$, then the set $\{\cov(\mu,\kappa,\theta,\omega_1):\theta<\kappa\}$ is finite.
The result is in [GiSh:412], and is described as a ``non-GCH version'' of the Hajnal-Shelah theorem stating the set $\{\mu^\theta:2^\theta<\mu\}$ is finite. It is a lovely illustration of the power of Shelah's cov vs. pp Theorem, and it is the use of this theorem that requires the fourth parameter in cov to be uncountable.
The proof in [GiSh:412] is very short, but it doesn't quite work as written. We'll take a look at their proof and show it works except in the case where $\kappa$ is singular of uncountable cofinality, and then give an argument reducing this case to the one where $\kappa$ is regular.
Lemma
If $\mu>\kappa$ and $\kappa$ is not singular of uncountable cofinality, then the set $\{\cov(\mu,\kappa,\theta,\omega_1):\theta<\kappa\}$ is finite.
Proof:
Assume this fails. By monotonicity properties of cov, we can find an increasing sequence $\langle\theta_n:n<\omega\rangle$ of cardinals such that
$$\cov(\mu,\kappa,\theta_n,\omega_1)<\cov(\mu,\kappa,\theta_{n+1},\omega_1)$$
for each $n<\omega$. Without loss of generality, we may assume each of these covering numbers is greater than $\mu$.
Since we have $\omega_1$ appearing as the fourth component, we can apply the Cov vs. pp Theorem (Theorem 5.4 of Chapter II of Cardinal Arithmetic), which tells us that if $\cov(\mu,\kappa,\theta,\omega_1)>\mu$, then
$$\cov(\mu,\kappa,\theta,\omega_1)=\sup\{\pp_{\Gamma(\theta,\omega_1)}(\chi): \kappa\leq\chi\leq\mu \text{ and }\cf(\chi)<\theta\}$$
In particular, for each $n<\omega$ we have
$$\mu<\cov(\mu,\kappa,\theta_n,\omega_1)=\sup\{\pp_{\Gamma(\theta_n,\omega_1)}(\chi): \kappa\leq\chi\leq\mu \text{ and }\cf(\chi)<\theta_n\}.$$
Given $n$, let $\chi_n$ be the least cardinal satisfying
- $\kappa\leq\chi_n\leq\mu$,
- $\omega_1\leq\cf(\chi_n)\leq\theta_n$, and
- $\pp_{\Gamma(\theta_n,\omega_1)}(\chi_n)>\mu$.
Notice that $\chi_n$ must exist as $\cov(\mu,\kappa,\theta_n,\omega_1)>\mu$, and furthermore $\kappa<\chi_n$, because we have assumed $\kappa$ is not singular of uncountable cofinality -- this is the point where we use that assumption.
Since the sequence $\langle \chi_n:n<\omega\rangle$ is non-increasing, we can drop a few terms and assume that it is constant, say with value $\chi^*$. Thus, there is a single cardinal $\chi^*$ such that for each $n<\omega$, $\chi^*$ is the least cardinal satisfying
- $\kappa<\chi^*\leq\mu$,
- $\omega_1\leq \cf\chi^*<\theta_n$, and
- $\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*)>\mu$.
Now suppose $\kappa<\chi<\chi^*$ and $\omega_1\leq\cf(\chi)<\theta_n$. Since $\chi^*$ is the least cardinal with the enumerated properties, we know
$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi)\leq\mu.$$
But we can then conclude
(*) $\pp_{\Gamma(\theta_n,\omega_1)}(\chi)<\chi^*$,
as otherwise Inverse Monotonicity (see part (4) of the No Hole Conclusion 2.3 of Chapter II of Cardinal Arithmetic) forces us to conclude
$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*)\leq\pp_{\Gamma(\theta_n,\omega_1)}(\chi),$$
which would contradict our choice of $\chi^*$.
Since we have (*) holding for all relevant $\chi<\chi^*$, we can now apply a generalization of one of the main results of [Sh:371]. The actual theorem we need is Lemma 1.2 of the published version of [Sh:410], which tells us that since (*) holds for all sufficiently large $\chi<\chi^*$, we have
(**) $\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*)=\pp_{\Gamma((\cf \chi^*)^+,\omega_1)}(\chi^*).$
(There are some issues with the proof of Lemma 1.2 in [Sh:410] that I will deal with in a future post, but the result still holds.)
Thus, the sequence $\langle\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*):n<\omega\rangle$ is actually constant with value some $\lambda^*$. The following claim immediately yields a contradiction:
Claim: For all $n<\omega$, $\cov(\mu,\kappa,\theta_n,\omega_1)=\lambda^*$.
Proof of claim:
Let $n<\omega$ be given. Recall that by the cov vs. pp Theorem, we know
$$\cov(\mu,\kappa,\theta_n,\omega_1)=\sup\{\pp_{\Gamma(\theta_n,\omega_1)}(\chi): \kappa\leq\chi\leq\mu \text{ and }\cf(\chi)<\theta_n\}$$
so in particular $$\lambda^*\leq\cov(\mu,\kappa,\theta_n,\omega_1).$$
Let $\chi$ be one of the cardinals used in computing the supremum. By our assumptions on $\kappa$, we know $\kappa\neq\chi$ so $\kappa<\chi$. If $\chi<\chi^*$, then by (*) we know
$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi)<\lambda^*$$
On the other hand, if $\chi^*<\chi$ then because $\chi<\mu<\pp_{\Gamma(\theta_n, \omega_1)}(\chi^*)$, we know
$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi)\leq \pp_{\Gamma(\theta_n, \omega_1)}(\chi^*)=\lambda^*$$
by inverse monotonicity.
Thus, $\cov(\mu,\kappa,\theta_n,\omega_1)=\lambda^*$. $_\square$
$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi)\leq \pp_{\Gamma(\theta_n, \omega_1)}(\chi^*)=\lambda^*$$
by inverse monotonicity.
Thus, $\cov(\mu,\kappa,\theta_n,\omega_1)=\lambda^*$. $_\square$
Now where did we use the assumption that $\kappa$ is not a singular cardinal of uncountable cofinality? It was used when we needed to know $\kappa<\chi^*$, so that we can conclude (*) holds for all sufficiently large $\chi<\chi^*$ of the right cofinality.
Now how do we deal with this last case? Let us assume that $\kappa$ is singular of uncountable cofinality, and find the sequence $\langle \theta_n:n<\omega\rangle\rangle$. For each $n<\omega$ there is a cardinal $\kappa_n<\kappa$ such that
$$\cov(\mu,\kappa,\theta_n, \omega_1)=\cov(\mu,\kappa^*,\theta_n,\omega_1)$$
for all $\kappa^*\geq\kappa_n$ (as the covering numbers are non-increasing as we increase $\kappa^*$ to $\kappa$).
Since $\kappa$ has uncountable cofinality, there is a single $\kappa^*<\kappa$ (which can be taken to be regular) such that
$$\cov(\mu,\kappa,\theta_n,\omega_1)=\cov(\mu,\kappa^*,\theta_n,\omega_1)\text{ for all }n<\omega.$$
Now how do we deal with this last case? Let us assume that $\kappa$ is singular of uncountable cofinality, and find the sequence $\langle \theta_n:n<\omega\rangle\rangle$. For each $n<\omega$ there is a cardinal $\kappa_n<\kappa$ such that
$$\cov(\mu,\kappa,\theta_n, \omega_1)=\cov(\mu,\kappa^*,\theta_n,\omega_1)$$
for all $\kappa^*\geq\kappa_n$ (as the covering numbers are non-increasing as we increase $\kappa^*$ to $\kappa$).
Since $\kappa$ has uncountable cofinality, there is a single $\kappa^*<\kappa$ (which can be taken to be regular) such that
$$\cov(\mu,\kappa,\theta_n,\omega_1)=\cov(\mu,\kappa^*,\theta_n,\omega_1)\text{ for all }n<\omega.$$
Now we generate a contradiction as in the preceding case. $_\square$
