# A question regarding a proof pertaining to: for all $n\in\omega$, $(\aleph_{n+1})^{\aleph_0}\leq\aleph_{n+1}\cdot(\aleph_n)^{\aleph_0}$

I would like to show:

for all $n\in\omega$, $(\aleph_{n+1})^{\aleph_0}\leq\aleph_{n+1}\cdot(\aleph_n)^{\aleph_0}$

Actually I see this problem can be solved using the approach Brian Scott uses here:

For every $n < \omega$, $\aleph_n^{\aleph_0} = \max(\aleph_n,\aleph_0^{\aleph_0})$

Most unfortunately, Brian hasn't been here for over a year (hope he's OK), so perhaps someone would explain one aspect of his proof:

Now consider a function $\varphi:\aleph_0\to\aleph_m$; $\aleph_m$ is an uncountable regular cardinal, so $\sup\{\varphi(k):k\in\aleph_0\}<\aleph_m$, and $\varphi$ actually maps $\aleph_0$ into $\eta$ for some ordinal $\eta<\aleph_m$.

My question is why, since $\aleph_m$ is an uncountable regular cardinal, its cofinality is $\aleph_m$, so why is the $\text{sup}$ less than $\aleph_m$?

• By definition of cofinality – Maxime Ramzi Apr 29 '18 at 17:41
• So the inequality is actually an equality. – DanielWainfleet Apr 30 '18 at 2:49
• @DanielWainfleet I was concerned with an intermediary step. These notes caicedoteaching.files.wordpress.com/2009/04/… on p. 9 attribute it to Tarski. "Intro to Cardinal Arithmetic" p. 64 attributes it to Hausdorff. – user12802 Apr 30 '18 at 14:06
• I suspected that was the case. I said it anyway – DanielWainfleet Apr 30 '18 at 14:55
• @DanielWainfleet Appreciated; happy to receive all the help I can get. – user12802 Apr 30 '18 at 14:58

Suppose for a contradiction that $\sup \{\varphi(k):k\in\omega\}$ is $\aleph_m$. Then, the sequence of sets $\bigcup_{k\in\omega}\{\varphi(j):j\leq k\}$ is cofinal in $\aleph_m$ and has length $\aleph_0$. This contradicts the regularity of $\aleph_m$ ($m\neq 0$ since $\aleph_m$ is an uncountable cardinal).
• Thanks for this. I have been studying it for a while, and maybe I could ask some follow-up questions. First, the length $\aleph_0$ that you mention comes from $\bigcup_{k\in\omega}$? So is that really the key, that with a domain of $\omega$ you will never reach $\aleph_m$? – user12802 Apr 29 '18 at 19:52
• Yes, that's the idea of cofinality: since $\aleph_m$ is regular, it is its own cofinality, so you need a sequence of $\aleph_m$, and no less, sets smaller than it to cover it. – Leo163 Apr 29 '18 at 21:44