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I have proved that if $\aleph_n \leq \aleph_0^{\aleph_0}$, then $\aleph_n^{\aleph_0} \leq \max(\aleph_n,\aleph_0^{\aleph_0})$. Clearly $\aleph_n \leq \aleph_n^{\aleph_0}$ and $\aleph_0^{\aleph_0} \leq \aleph_n^{\aleph_0}$. Therefore, in order to establish the desired equality, I only need to show that if $\aleph_0^{\aleph_0} \leq \aleph_n$, then $\aleph_n^{\aleph_0} \leq \max(\aleph_n,\aleph_0^{\aleph_0})$.

I guess that might be easy, but I just don't see it. Any comments are welcome.

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I think it can be derived by repeated use of this result: $\aleph_{\alpha+1}^{\aleph_\beta}=\aleph_{\alpha+1}.\aleph_\alpha^{\aleph_\beta}‌​$ See e.g. Drake: Set theory: an introduction to large cardinals, p.59. (I have to admit that this result is new to me - I just tried to look up some results on cardinal exponentiation, which could help here and this one seemed reasonable.) – Martin Sleziak Nov 16 '11 at 22:58
Note that $\aleph_0^{\aleph_0}=2^{\aleph_0}$. We are, if so, asked to show: $$\aleph_n^{\aleph_0}=\aleph_n\cdot 2^{\aleph_0}=\max\{\aleph_n,2^{\aleph_0}\}$$ This is known as Bernstein's formula, Brian's answer gives the needed. – Asaf Karagila Nov 17 '11 at 7:39
@AsafKaragila: Thank you very much for this extra information. – ragrigg Nov 17 '11 at 16:05
up vote 6 down vote accepted

Suppose that there is an $n$ such that $\aleph_0^{\aleph_0}\le\aleph_n$ and $\aleph_n^{\aleph_0}>\aleph_n$, and let $m$ be the least such $n$; clearly $m>0$. 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 $\eta<\aleph_m$. Writing $^AB$ for the set of functions from $A$ into $B$, we have $$^{\aleph_0}\aleph_m=\bigcup_{\eta<\aleph_m}{^{\aleph_0}\eta}$$ and hence $$\left|^{\aleph_0}\aleph_m\right|=\left|\bigcup_{\eta<\aleph_m}{^{\aleph_0}\eta}\right|\;.$$

For each $\eta<\aleph_m$, $|\eta|\le\aleph_{m-1}$, so $$\aleph_m^{\aleph_0}=\left|^{\aleph_0}\aleph_m\right|=\left|\bigcup_{\eta<\aleph_m}{^{\aleph_0}\eta}\right|\le\aleph_m\cdot\aleph_{m-1}^{\aleph_0},$$ and $\aleph_{m-1}^{\aleph_0}\le\aleph_{m-1}$ by the minimality of $m$, so $$\aleph_m^{\aleph_0}\le\aleph_m\cdot\aleph_{m-1}^{\aleph_0}\le\aleph_m\cdot\aleph_{m-1}=\aleph_m\;,$$

contradicting the choice of $m$. Thus, $\aleph_m^{\aleph_0}\le\aleph_m=\max\{\aleph_m,\aleph_0^{\aleph_0}\}$ whenever $\aleph_0^{\aleph_0}\le\aleph_m$.

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@brian-m-scott: Interesting proof! The key idea was the regularity of $\aleph_m$ for $m < \omega$. One more question: how do you know that $\left|\bigcup_{\eta<\aleph_m}{^{\aleph_0}\eta}\right|\le\aleph_m\cdot\aleph_{m-‌​1}$? All I see is that $\left|\bigcup_{\eta<\aleph_m}{^{\aleph_0}\eta}\right|\le\aleph_m\cdot\aleph_{m-‌​1}^{\aleph_0}$. – ragrigg Nov 17 '11 at 4:06
I think I see it now. One can use your argument to prove by induction on $\omega \backslash \{0\}$ that, for every $k \in \omega \backslash \{0\}$, if $\aleph_0^{\aleph_0} \leq \aleph_k$, then $\aleph_k^{\aleph_0} \leq \aleph_k$. – ragrigg Nov 17 '11 at 4:33
@ggirgar: The induction can include $0$ trivially as well. – Asaf Karagila Nov 17 '11 at 6:54
@ggirgar: My apologies: I’d intended to fill in more detail and forgot. I’ve now done so. As you say, it’s an induction argument, though it’s simpler to phrase it in terms of least counterexample. – Brian M. Scott Nov 17 '11 at 8:47

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