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In the following paper http://www.math.ucsd.edu/~ronspubs/74_01_van_der_waerden.pdf, just in the first paragraph the author defines what $l$-equivalence for two m-tuples $\in [0,l]^m$ means. Can somebody please give me a more precise definition of what he means? I am not even sure what $[0,l]^m$ stands for, although I expect its just $\{0,1,\dots,l\}^m$.

Thanks

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I think it just means that (a,b,x) and (a,b,y) are 2-equivalent. I'vve not been able to understand the proof yet. –  user50336 Nov 27 '12 at 18:12

1 Answer 1

According to the first sentence of the paper, $[0,\ell]$ is indeed $\{0,1,\dots,\ell\}$, and it’s clear that $[0,\ell]^m$ is, as usual, the set of $m$-tuples of elements of $[0,\ell]$. Suppose that $\langle x_1,\dots,x_m\rangle$ and $\langle x_1',\dots,x_m'\rangle$ are $m$-tuples in $[0,\ell]^m$. Suppose first that $\ell$ occurs at least once in each of these $m$-tuples. Let $k$ be the largest index such that $x_k=\ell$; $x_k$ is the last occurrence of $\ell$ in $\langle x_1,\dots,x_m\rangle$. Similarly, let $x_{k'}'$ be the last occurrence of $\ell$ in $\langle x_1',\dots,x_m'\rangle$. Then $\langle x_1,\dots,x_m\rangle$ and $\langle x_1',\dots,x_m'\rangle$ are $\ell$-equivalent iff $k=k'$, and $x_i=x_i'$ for $i=1,\dots,k$. If neither $\langle x_1,\dots,x_m\rangle$ nor $\langle x_1',\dots,x_m'\rangle$ contains an $\ell$, the agreement requirement is vacuous, so they are automatically $\ell$-equivalent.

For example, the sequences $\langle 2,1,2,0,1\rangle$ and $\langle 2,1,2,1,0\rangle$ in $[0,2]^5$ are $2$-equivalent, and the $2$-equivalence class of these two sequences also contains $\langle 2,1,2,0,0\rangle$ and $\langle 2,1,2,1,1\rangle$; it is in fact

$$\Big\{\langle 2,1,2,a,b\rangle:\langle a,b\rangle\in[0,1]^2\Big\}\;.$$

The $2$-equivalence class of $\langle 1,1,0,1,0\rangle\in[0,2]^5$, which doesn’t contain $2$ at all, is $[0,1]^5$: the $5$-tuples that don’t contain a $2$ are the ones that agree with $\langle 1,1,0,1,0\rangle$ up through the last occurrence of $2$.

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