I want to show that, assuming AC, that if $\kappa$ is a regular cardinal then $\kappa^{<\kappa} = \max\{\kappa, 2^{<\kappa}\}$ where $\kappa^{<\lambda}$ is defined by $$\kappa^{<\lambda} = \sup\{\kappa^{\nu}\mid\nu \in Card , \nu < \lambda\}.$$
If $\kappa= \aleph_0$ then $\aleph_0^{<\aleph_0} = \sup\{\aleph_0^n\mid n \in \omega\} =\aleph_0^{\aleph_0} = 2^{\aleph_0} = \max\{\kappa, 2^{\kappa}\} = \max\{\kappa,2^{<\kappa}\}$.

If $\kappa$ is a successor cardinal, then $\kappa = \aleph_{S(\alpha)} = \aleph_{\alpha + 1}$ and, using Hausdorff's formula, we have $$\aleph_{\alpha+1}^{<\aleph_{\alpha +1}} = \sup\{\aleph_{\alpha+1}^{\aleph_{\beta}} \mid \beta < \alpha + 1\} = \sup\{\max\{\aleph_{\alpha+1},\aleph_{\alpha}^{\aleph_{\beta}}\} \mid\beta < \alpha + 1\} = \max\{ \aleph_{\alpha+1}, \sup\{{\aleph_{\alpha}^{\aleph_{\beta}}\mid\beta < \alpha + 1\}}\} = \max\{\aleph_{\alpha+1},\aleph_{\alpha}^{\aleph_{\alpha}}\}$$ and, since $\aleph_{\alpha}^{\aleph_{\alpha}} = 2^{\aleph_{\alpha}} = 2^{<\kappa}$ this case is proved.

I cannot prove the result for $\kappa$ limit. If $\kappa$ is limit and regular cardinal then it is weakly inaccessible, i.e. $\kappa = \aleph_\kappa$. I'm asking an hint for this part and if the rest of the proof is correct.

  • 2
    $\begingroup$ Note that $2^{<\kappa}\ge\kappa$, so all you are claiming is that if $\kappa$ is regular, then $\kappa^{<\kappa}=2^{<\kappa}$. $\endgroup$ Jan 16, 2018 at 16:35

1 Answer 1


Your proof about $\aleph_0$ is definitely not correct. Note that $\aleph_0^n=\aleph_0$ for all $n>0$. Therefore $\sup\{\aleph_0^n\mid n\in\omega\}$ is literally $\sup\{1,\aleph_0\}=\max\{1,\aleph_0\}$. And I will let you figure that part out on your own.

Now, for the successor case, note that since this function is increasing, $\aleph_{\alpha+1}^{<\aleph_{\alpha+1}}=\aleph_{\alpha+1}^{\aleph_\alpha}$. This is equal to $\aleph_{\alpha+1}\cdot 2^{\aleph_\alpha}$, and incidentally $2^{\aleph_\alpha}=2^{<\aleph_{\alpha+1}}$, so the result follows.

But actually this proof also translates to the inaccessible case. If $\kappa$ is regular, limit or otherwise, and $\lambda<\kappa$, then every function from $\lambda$ to $\kappa$ is bounded. So $\kappa^\lambda$ is exactly $\sup\{\alpha^\lambda\mid\alpha<\kappa\}$.

Now, if $\kappa$ is a successor say $\lambda^+$, then this just $\kappa\cdot\lambda^\lambda=\kappa\cdot2^\lambda=\kappa\cdot2^{<\kappa}$; but if $\kappa$ is a limit cardinal, then we get $\kappa^\lambda$ to be $\sup\{\mu^\lambda\mid\mu<\kappa\text{ regular}\}$ which by the inductive hypothesis is eventually (read: for $\mu>\lambda$) equals to either $\mu$ or $2^\lambda$; since there are $\kappa$ many options for $\lambda$, we get exactly $\sup\{2^\lambda\mid\lambda<\kappa\}$ which is at least $\kappa$, but also exactly $2^{<\kappa}$ by definition.

  • $\begingroup$ I understand my mistake for $\aleph_0$-case but when i consider $\aleph_0^n$ for $n=0$, that is the cardinality of the set of all function from the empty set to $\aleph_0$, but in that set there's the empty function, so its cardinality is 1, isn't it? so $\aleph_0^{< \aleph_0} = max{1, \aleph_0}$ and $2^{<\aleph_0} = \aleph_0 = k $, am i right? $\endgroup$
    – M.B.
    Jan 16, 2018 at 16:01
  • $\begingroup$ Err, yeah, that's a typo. Sorry. And yes for the second part. $\endgroup$
    – Asaf Karagila
    Jan 16, 2018 at 16:02
  • $\begingroup$ oh ok, thank you so much! It was really helpful! $\endgroup$
    – M.B.
    Jan 16, 2018 at 16:06

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