# How Rowbottom can $\aleph_{\omega}$ be?

A current unknown hypothesis is whether or not $\aleph_\omega$ can be Rowbottom. It is well known that the consistency strength of such a conjecture is stronger than that of the existence of a measurable cardinal, however there is no known upper bound of consistency strength (as far as I am aware of).

A possible way to find the consistency strength of this is by approximating it in little steps, generalizing Rowbottom-ness to a weaker form. This led me to this question.

Let $RW=\min\{n\in\omega:\aleph_\omega\not\rightarrow[\omega_\omega]_{\aleph_n,<\aleph_1}^{<\omega}\}$.

By definition, $\aleph_\omega$ is Rowbottom if and only if $RW$ does not exist, so if it is inconsistent for there to be a measurable cardinal, then $RW$ exists, and if $0^{\#}$ does not exist, then $RW$ exists.

Clearly, $RW>0$. If $RW=0$, then it would mean that there is a partition $f:[\omega_\omega]^{<\omega}\rightarrow\omega$ such that $f"H$ is uncountable for some $H$, meaning $\omega$ is uncountable.

Futhermore, Kanamori in his "The Higher Infinite" provides a theorem sufficient to prove that $RW=\min\{n\in\omega\setminus\{0\}:(\aleph_\omega,\aleph_n)\not\twoheadrightarrow(\aleph_\omega,\aleph_0)\}$. By the definition of the square bracket partition property, $RW=\min\{n\in\omega:\exists f:[\omega_\omega]^{<\omega}\rightarrow\omega_n\forall H\in[\omega_\omega]^{\aleph_\omega}(|f"H|>\aleph_0)\}$.

Kanamori also provides in the same text that if $\aleph_\omega\rightarrow[\omega_\omega]^{<\omega}_{\aleph_n,<\aleph_1}$, then $\aleph_1$ is inaccessible in $L$. Therefore $V=L\rightarrow RW=1$ and $RW^L=1$, and therefore as well, $Con(\text{ZFC}+RW>1)\rightarrow Con(\text{ZFC}+\text{inaccessible})$.

Quick recap:

• $\aleph_\omega$ is Rowbottom if and only if $RW$ does not exist
• $\neg\text{Con}(\text{ZFC}+\text{measurable})\rightarrow RW$ exists
• $\neg(0^{\#}$ exists$)\rightarrow RW$ exists
• $RW>0$
• $RW=\min\{n\in\omega\setminus\{0\}:(\aleph_\omega,\aleph_n)\not\twoheadrightarrow(\aleph_\omega,\aleph_0)\}$
• $RW=\min\{n\in\omega:\exists f:[\omega_\omega]^{<\omega}\rightarrow\omega_n\forall H\in[\omega_\omega]^{\aleph_\omega}(|f"H|>\aleph_0)\}$
• $RW>1\rightarrow\aleph_1$ is inaccessible in $L$
• $V=L\rightarrow RW=1$, $L\models (RW=1)$

Questions: If $RW>1$, then does $0^{\#}$ exist? Is there some $n$ such that if $RW>n$, then $\aleph_\omega$ is Rowbottom? What are some upper bounds on consistency strength for some of these?

• I don't see your last comment. Compactness way produce non-standard natural numbers. – Mohammad Golshani Dec 16 '17 at 13:25
• @Keith That is irrelevant to the issue raised by Mohammad. – Andrés E. Caicedo Dec 16 '17 at 17:17
• Your compactness comment is indeed incorrect. All compactness would tell you is that there is a model of ZFC which satisfies $RW>k$ for every true natural number $k$. However, this model may itself have nonstandard natural numbers, so it won't satisfy "$RW>n$ for every natural number $n$." – Noah Schweber Dec 16 '17 at 17:59
• So?${{{}}}{}{}$ – Andrés E. Caicedo Dec 16 '17 at 17:59
• If you drop choice and assume the axiom of determinacy, then $\aleph_\omega$ is Rowbottom. (See Kleinberg's book.) Not sure if this is relevant. – William Dec 16 '17 at 20:04