Editing Van der Waerden's theorem (section)

== Ergodic theory ==

Furstenberg and Weiss proved an equivalent form of the theorem in 1978, using ergodic theory.<ref name=":0">{{Cite book |last=Petersen |first=Karl E. |url=https://www.cambridge.org/core/books/ergodic-theory/4F50E2830B2812125F24D4A2CE7318D0 |title=Ergodic Theory |date=1983 |publisher=Cambridge University Press |isbn=978-0-521-38997-6 |series=Cambridge Studies in Advanced Mathematics |location=Cambridge |chapter=Chapter 2.}}</ref>
{{Math theorem|name=multiple Birkhoff recurrence theorem|note=Furstenberg and Weiss, 1978|math_statement=

If <math display="inline">X</math> is a compact metric space, and <math display="inline">T_1, \dots, T_N: X \to X</math> are homeomorphisms that commute, then <math display="inline">\exists x\in X</math>, and an increasing sequence <math display="inline">n_1 < n_2 < \cdots</math>, such that <math display="block">\lim_{j} d(T_i^{n_j} x, x) = 0, \quad \forall i \in 1:N</math>
}}The proof of the above theorem is delicate, and the reader is referred to.<ref name=":0" /> With this recurrence theorem, the van der Waerden theorem can be proved in the ergodic-theoretic style.{{Math theorem|name=Theorem|note=van der Waerden, 1927|math_statement= 

If <math display="inline">\Z</math> is partitioned into finitely many subsets <math display="inline">S_1, \dots, S_n</math>, then one of them <math display="inline">S_k</math> contains infinitely many arithmetic progressions of arbitrarily long length 

<math display="block">
\forall N, N', \; \exists |a|  \geq N', \exists r \geq 1: \{a + ir\}_{i \in 1:N} \subset S_k</math>
}}

{{Math proof|title=Proof|note=Furstenberg and Weiss, 1978|proof=

It suffices to show that for each length <math display="inline">N</math>, there exist at least one partition that contains at least one arithmetic progression of length <math display="inline">N</math>.Once this is proved, we can cut out that arithmetic progression into <math display="inline">N</math> singleton sets, and repeat the process to create another arithmetic progression, and so one of the partitions contain infinitely many arithmetic progressions of length <math display="inline">N</math>. Then we can repeat this process to find that there exists at least one partition that contains infinitely many progressions of length <math display="inline">N</math>, for infinitely many <math display="inline">N</math>, and that is the partition we want.

Consider the state space <math display="inline">\Omega = (1:N)^\Z</math>, which is compact under the metric (in fact, ultrametric) <math display="block">
      d( (x_i), (y_i)) = \max\{2^{-\vert{i}\vert} : x_i \neq y_i\}.
      </math> Since the sets <math display="inline">S_1, \dots, S_n</math> partition <math display="inline">\Z</math>, we have a well-defined sequence <math display="inline">z = (\dots, z_{-1}, z_0, z_1, \dots) = (z_i)_i</math> with <math display="inline">i\in S_{z_i}</math> for all <math display="inline">i</math>. 

Let <math display="inline">T: \Omega \to \Omega</math> be the shift map <math display="block">T((x_i)_i) = (x_{i+1})_i,</math> 
and let <math display="inline">X = cl(\{T^r z : r \in \Z\})</math> be the closure of all shifts of the sequence <math display="inline">z</math>. By the multiple Birkhoff recurrence theorem (for the maps <math display="inline">T, T^2, \dots, T^N</math>), there exist a sequence <math display="inline">x \in X</math> and an integer <math display="inline">s \geq 1</math> such that <math display="block">d(T^s x, x), d(T^{2s} x, x), \dots,  d(T^{Ns} x, x) < \frac{1}{4}.</math> 

Since <math display="inline">X</math> is the closure of shifts of <math display="inline">z</math>, and <math display="inline">T</math> is continuous, there exists a shift <math display="inline">T^m z</math> such that simultaneously, <math display="inline">x</math> is very close to <math display="inline">T^m z</math>, and <math display="inline">T^s x</math> is very close to <math display="inline">T^{s+m} z</math>, and so on: <math display="block">d( x, T^{m} z), d(T^{s} x, T^{m+s} z), \dots, d(T^{Ns} x, T^{m + Ns} z) < \frac{1}{4}.</math> 

By the triangle inequality, we then immediately have
<math display="inline">d(T^{m + is} z, T^{m + js} z) < \frac{3}{4}</math> for <math display="inline">i, j = 0, \dots, N</math>. But by construction, any sequences <math display="inline">y, y'\in \Omega</math> with <math display="inline">d(y, y') < 1</math> must have <math display="inline">y_0 = {y'}\!\!{}_0</math>. Thus <math display="inline">z_m = z_{m+s} = \dots = z_{m+Ns}</math>, and so all <math display="inline">m, m+s, \dots, m+Ns</math> lie in the partition <math display="inline">S_{z_m}</math>.

}}