On singular products of cardinal numbers

Question

I want to know whether it is possible to show in $ZFC$ that there exist a limit ordinal $\lambda$, a strictly increasing sequence of cardinal numbers $\langle \mu_\alpha : \alpha \in \lambda\rangle$ and a cofinal subset $A \subseteq\lambda$, such that $$\prod_{\alpha\in A} \mu_\alpha < \prod_{\alpha\in \lambda} \mu_\alpha.$$

It is easy to show that $\lambda$ cannot be regular. It is also not so hard to find examples using some extra assumptions on the cardinal arithmetic.

Answer 1

This is not provable in ZFC, and in fact it is disproved by GCH.

Assuming GCH, we have, for $\kappa$ singular, that $\kappa^{cf(\kappa)}=2^\kappa=\kappa^+=\kappa^\kappa$ (this last equality because $\kappa^+= 2^\kappa\le\kappa^\kappa\le (2^\kappa)^\kappa=2^\kappa=\kappa^+$). But this forces both products in the question to equal $\kappa^+$ (where $\kappa=\sup\mu_\alpha$).

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EDIT: To clarify why the (a priori smaller) left hand side must be $\kappa^+$: clearly it's enough to show that it's $>\kappa$, since it's at most $\kappa^{cf(\kappa)}=\kappa^+$. Fix a cofinal sequence $\mu_\alpha$. Given any set $A\subseteq \kappa$, let $A_\alpha=A\cap \mu_\alpha$. Then any set $A$ is determined by the sequence $(A_\alpha: \alpha<cf(\kappa))$. This is an element of $$\prod_{\alpha<cf(\kappa)}2^{(\mu_\alpha)}=\prod_{\alpha<cf(\kappa)}\mu_\alpha^+,$$ so the set of all such sequences has size at least $2^\kappa$.

Answer 2

${}$ Hi Ramiro! I looked at very similar questions in my undergraduate thesis (see here).

Your question is related to a conjecture of Tarski, in

A. Tarski, Quelques théorèmes sur les alephs, Fund. Math. 7 (1925), 1-14.

In that paper, he proves that $$ \prod_{\alpha<\lambda}\aleph_\alpha=\aleph_\lambda^{|\lambda|}$$ for any limit ordinal $\lambda$, and conjectures that if $\lambda$ is a limit ordinal and $(\mu_\alpha\mid\alpha<\lambda)$ is a strictly increasing sequence of infinite cardinals with limit $\mu$, then $$\prod_{\alpha<\lambda}\mu_\alpha=\mu^{|\lambda|}.$$ It is a nice exercise to prove that this holds assuming the singular cardinals hypothesis. Recall that this is the claim that for all $\kappa$, $$2^\kappa=\kappa^++\kappa^{{\mathrm cf}(\kappa)}.$$ It follows easily from this that the existence of a sequence and an $A$ satisfying the inequality as you suggest violates the singular cardinals hypothesis.

Violations of Tarski's conjecture were studied in some detail by Jech and Shelah (using pcf theory), see

MR1070525 (91j:03066). Jech, Thomas; Shelah, Saharon. On a conjecture of Tarski on products of cardinals. Proc. Amer. Math. Soc. 112 (1991), no. 4, 1117–1124.

Among others, they provide a proof of the "nice exercise" above, and show that if the conjecture fails, then there is a counterexample with $\lambda=\omega_1+\omega$. Indeed, they show that we can find a limit ordinal $\gamma>\omega_1$ such that $$\aleph_\gamma^{\aleph_1}>\aleph_{\gamma+\omega}^{\aleph_0},$$ and this can in turn be used to produce an example as you require.