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In the language of partition calculus, I want to know if the following is true:

Let $n$ be an integer, $n \geq 2$. Then $\aleph_n \rightarrow (\aleph_0)^2_{\aleph_0}$.

I'm new to infinitary combinatorics, and it's possible that I used the wrong notation, so I will explain with words. Let $G$ be a graph such that the number of vertices is $\aleph_n$, $n \geq 2$. We color each edge of $G$ with countably many colors. Then I can find a countably infinite set $S$ of vertices, and each edge between two vertices in $S$ will be the same color.

The reason I require $n \geq 2$ is that the interval of real numbers $[0,1]$ has cardinality $|[0,1]| \geq \aleph_1$, but if I color each edge between real numbers in the interval using the rule

$\{a,b\}$ has the color $k \in \bf{Z^{\geq 0}}$ if and only if $|a-b| \in (\big (\dfrac{1}{2}\big)^{k+1}, \big(\dfrac{1}{2} \big )^k ]$

then I can color each edge with countably many colors while avoiding monochromatic triangles. So I can demonstrate that my initial statement is false for $n = 1$. However, I am not sure about graphs with higher cardinalities of vertices, and I am unable to find a result in the most commonly referenced works about partition calculus. I'd like to know if this is a known result and where I can find a proof.

I'm also curious about the following kinds of statements, and any help would be appreciated.

$\aleph_3 \rightarrow (\aleph_0)^2_{\aleph_1}$

$\aleph_{n+2} \rightarrow (\aleph_0)^2_{\aleph_n}$

$\aleph_{n+2} \rightarrow (\aleph_n)^2_{\aleph_n}$

$\kappa \rightarrow (\aleph_0)^2_{\lambda}$ where $\kappa$ is an inaccessible cardinal and $\lambda$ is an accessible cardinal.

Thank you.

  • You have actually shown that $$2^{\aleph_0}\not\to(3)^2_{\aleph_0}$$ Are you assuming GCH? – bof Sep 24 '17 at 02:28
  • This is purely for my own curiosity, so I'd be excited to see anything that uses GCH. –  Sep 24 '17 at 02:30
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  • Sorry if this is a silly question, but in most cases the Erdos Rado theorem makes a statement about coloring subsets of size larger than two. If I want to answer the original question for $n = 5$ of $n = 27$, will I still be able to approach the problem with this theorem? –  Sep 24 '17 at 03:02
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    The case $r=1$ (as the theorem is stated in the Wikipedia page) says that $$(2^\kappa)^+\to(\kappa^+)^2_\kappa.$$ Thus $$\aleph_n\to(\aleph_0)^2_{\aleph_0}\text{ iff }\aleph_n\gt2^{\aleph_0}$$ and, in particular, $$\aleph_2\to(\aleph_0)^2_{\aleph_0}\text{ iff }2^{\aleph_0}=\aleph_1$$ – bof Sep 24 '17 at 03:30
  • It's all clear to me now; I don't know why I didn't see that before. Thanks a lot! –  Sep 24 '17 at 03:50

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