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Edit 2: solved!

In this post, "we" proved that exist infinite $R_1$, such that $f$ is constant on elements of the form $\{n_1,r\}$ where $r\in R_1$. By the same considerations we can show that if we fix $n_2\in R_1$, there is exist infinite $R_2\subseteq R_1-\{n_2\}$ such that $f$ is constant on sets of the form $\{n_2,r\}$ where $r\in R_2$. Now saying that $f(\{\{n_1,r\}:r\in R_1\})=t_1$ and $f(\{\{n_2,r\}:r\in R_2\})=t_2$. I need to show via induction that we can find a sequence of naturals $n_1,n_2,n_3,...$, and a sequence of $t_1,t_2,t_3,...$ where $t_n\in\{1,2\}$ for all $n$, and if $i<j$, then $f(\{n_i,n_j\})=t_i$

Thank you!

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It is better to make your question self-contained. –  Thomas Andrews Nov 15 '12 at 14:05
How are those sequences defined? Clearly you don't just mean to choose $n_i=i$, for example. –  Thomas Andrews Nov 15 '12 at 14:09
@Thomas: I think the OP is asking how the sequences can be defined. –  Cameron Buie Nov 15 '12 at 15:14
If it would help... From the above I must to deduce link –  17SI.34SA Nov 15 '12 at 15:31
I'm glad the problem is solved. May I encourage you to post the solution, for the benefit of those who come after us. –  Gerry Myerson Nov 15 '12 at 22:39

1 Answer 1

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Applying the argument mention here - Constant $f:[\mathbb{N}]^2\to \{1,2\}$. - we can produce sets $ R_1,R_2,...,R_n,...$, which they all infinite and subsets of $\mathbb{N}$. Now, in each step, by creating(proving that tjey are infinite) these sets we also create a sequence $n_1,n_2,n_3,...$, which is infinite as well; we know that $f(\{\{n_1,r\}:r\in R_1\})=t_1$, $f(\{\{n_2,r\}:r\in R_2\})=t_2$, $f(\{\{n_3,r\}:r\in R_3\})=t_3$,... We want also to show that, $f(\{\{n_i,n_j\})=t_i$ for $i<j$, but that is obtained from $$ f(\{\{n_i,r\}:r\in R_i\})=t_i $$

Noticing the properties of $R_i$: $R_i$ is not containing the set $\{a_1,a_2,...,a_i \}$, hence $f(\{n_i,n_j \})=t_i$ for $i<j$.

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