Shelah proved that it is consistent that GCH holds below $\aleph_\omega$, while $2^{\aleph_\omega}=\aleph_{\omega+\alpha+1}$ for any countable ordinal $\alpha$ you care to choose. (See Theorem 36.5 of Jech's book, for example).

In such a model, ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$ as well. Now if you add $\aleph_{\omega_1+1}$ Cohen reals (which has no effect on the pcf structure) you end up with a model where

• $|\prod A| = \aleph_{\omega_1+1}$, and
• ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$.

So ${\rm max pcf} (A)$ could potentially be any successor cardinal below $\aleph_{\omega_1}$.

(Of course, it's still unknown if $\aleph_{\omega_1}\leq{\rm max pcf }(A)$ is possible, so this is the best answer we can hope for given our current knowledge.)

I don't know the answer to your "evens and odds" question, but certainly you can split $A$ up into two disjoint pieces whose "gap" is as large as possible:

Let $\tau$ denote ${\rm max pcf}(A)$, and suppose $\aleph_{\omega+1}<\tau$.

We know there exists an unbounded $B\subseteq A$ such that $\prod B$ contains a scale (mod finite) of length $\aleph_{\omega+1}$. This implies ${\rm max pcf}(B)=\aleph_{\omega+1}$.

The set $A\setminus B$ cannot be in the ideal $J_{<\tau}[A]$ (otherwise, we contradict ${\rm max pcf}(A)=\tau$, and so we must conclude ${\rm max pcf}(A\setminus B)=\tau$ as well.

Shelah proved that it is consistent that GCH holds below $\aleph_\omega$, while $2^{\aleph_\omega}=\aleph_{\omega+\alpha+1}$ for any countable ordinal $\alpha$ you care to choose. (See Theorem 36.5 of Jech's book, for example).

In such a model, ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$ as well. Now if you add $\aleph_{\omega_1+1}$ Cohen reals (which has no effect on the pcf structure) you end up with a model where

• $|\prod A| = \aleph_{\omega_1+1}$, and
• ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$.

So ${\rm max pcf} (A)$ could potentially be any successor cardinal below $\aleph_{\omega_1}$.

(Of course, it's still unknown if $\aleph_{\omega_1}\leq{\rm max pcf }(A)$ is possible, so this is the best answer we can hope for given our current knowledge.)

I don't know the answer to your "evens and odds" question, but certainly you can split $A$ up into two disjoint pieces whose "gap" is as large as possible:

Let $\tau$ denote ${\rm max pcf}(A)$, and suppose $\aleph_{\omega+1}<\tau$.

We know there exists an unbounded $B\subseteq A$ such that $\prod B$ contains a scale (mod finite) of length $\aleph_{\omega+1}$. This implies ${\rm max pcf}(B)=\aleph_{\omega+1}$.

The set $A\setminus B$ cannot be in the ideal $J_{<\tau}[A]$ (otherwise, we contradict ${\rm max pcf}(A)=\tau$), and so we must conclude ${\rm max pcf}(A\setminus B)=\tau$ as well.

Shelah proved that it is consistent that GCH holds below $\aleph_\omega$, while $2^{\aleph_\omega}=\aleph_{\omega+\alpha+1}$ for any countable ordinal $\alpha$ you care to choose. (See Theorem 36.5 of Jech's book, for example).

In such a model, ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$ as well. Now if you add $\aleph_{\omega_1+1}$ Cohen reals (which has no effect on the pcf structure) you end up with a model where

• $|\prod A| = \aleph_{\omega_1+1}$, and
• ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$.

So ${\rm max pcf} (A)$ could potentially be any successor cardinal in the range of interest to you.

I don't know the answer to your "evens and odds" question, but certainly you can split $A$ up into two disjoint pieces whose "gap" is as large as possible:

Let $\tau$ denote ${\rm max pcf}(A)$, and suppose $\aleph_{\omega+1}<\tau$.

We know there exists an unbounded $B\subseteq A$ such that $\prod B$ contains a scale (mod finite) of length $\aleph_{\omega+1}$. This implies ${\rm max pcf}(B)=\aleph_{\omega+1}$.

The set $A\setminus B$ cannot be in the ideal $J_{<\tau}[A]$ (otherwise, we contradict ${\rm max pcf}(A)=\tau$, and so we must conclude ${\rm max pcf}(A\setminus B)=\tau$ as well.

Shelah proved that it is consistent that GCH holds below $\aleph_\omega$, while $2^{\aleph_\omega}=\aleph_{\omega+\alpha+1}$ for any countable ordinal $\alpha$ you care to choose. (See Theorem 36.5 of Jech's book, for example).

In such a model, ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$ as well. Now if you add $\aleph_{\omega_1+1}$ Cohen reals (which has no effect on the pcf structure) you end up with a model where

• $|\prod A| = \aleph_{\omega_1+1}$, and
• ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$.

So ${\rm max pcf} (A)$ could potentially be any successor cardinal below $\aleph_{\omega_1}$.

(Of course, it's still unknown if $\aleph_{\omega_1}\leq{\rm max pcf }(A)$ is possible, so this is the best answer we can hope for given our current knowledge.)

I don't know the answer to your "evens and odds" question, but certainly you can split $A$ up into two disjoint pieces whose "gap" is as large as possible:

Let $\tau$ denote ${\rm max pcf}(A)$, and suppose $\aleph_{\omega+1}<\tau$.

We know there exists an unbounded $B\subseteq A$ such that $\prod B$ contains a scale (mod finite) of length $\aleph_{\omega+1}$. This implies ${\rm max pcf}(B)=\aleph_{\omega+1}$.

The set $A\setminus B$ cannot be in the ideal $J_{<\tau}[A]$ (otherwise, we contradict ${\rm max pcf}(A)=\tau$, and so we must conclude ${\rm max pcf}(A\setminus B)=\tau$ as well.