I'll ask my question and then explain the background. Thanks in advance.

Question: Is it the case that $$ \beth_{\omega}\rightarrow (\beth_{\omega})_{{\aleph_{0}}}^{n} $$ for finite $n$? If not, what more generally can be said about cardinals $\kappa$ such that $$ \kappa\rightarrow (\kappa)_{{\aleph_{0}}}^{n} $$ holds, for finite $n$?


Wikipedia succinctly states the Erdős–Rado theorem as follows: $$ \beth_{n}^{+}\rightarrow (\aleph_{1})_{{\aleph_{0}}}^{{n+1}} $$ In words and paraphrasing slightly this says: if we countably colour the $n{+}1$-size subsets of a set $X$ of cardinality greater than $\beth_{n}$, then there exists a homogeneous subset of uncountable cardinality.

Then it re-states a more general form: $$ \exp_{n}(\kappa )^{+}\longrightarrow (\kappa^{+})_{\kappa }^{{n+1}} $$ An $\aleph_0$-colouring is also a $\kappa$-colouring for $\kappa\geq\aleph_0$, and $\beth_\omega > \exp_n(\beth_i)$ for finite $i$, so from this it seems to me that we can derive for every finite $i$ that $$ \beth_{\omega}\longrightarrow (\beth_i^{+})_{\aleph_0}^{{n+1}} . $$ This suggests that we might try to take a limit as $i$ rises towards $\omega$ --- and we arrive at my question.

Advice welcome, thank you.

  • $\begingroup$ The answer to your first question is no, even for $n=1$, as $\beth_\omega\not\to(\beth_\omega)^1_{\aleph_0}$. $\endgroup$
    – bof
    Jun 16, 2022 at 9:41
  • $\begingroup$ To start with, if $\kappa$ is an infinite cardinal such that $\kappa\to(\kappa)^2_2$, then $\kappa$ must be strongly inaccessible; and increasing the number of colours from $2$ to $\aleph_0$ makes no difference except in the case $\kappa=\aleph_0$. $\endgroup$
    – bof
    Jun 16, 2022 at 9:48
  • 2
    $\begingroup$ en.wikipedia.org/wiki/Weakly_compact_cardinal $\endgroup$
    – bof
    Jun 16, 2022 at 9:50
  • $\begingroup$ Thank you @bof, That nails the answer. $\endgroup$
    – Jim
    Jun 17, 2022 at 12:35


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