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- teorema (ca)
- kombinatorikai állítás (hu)
- theorem on arithmetic progressions (en)
- математичне твердження у комбінаториці (uk)
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- Weisstein, Eric W. (en)
- O'Bryant, Kevin (en)
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- If is partitioned into finitely many subsets , then one of them contains infinitely many arithmetic progressions of arbitrarily long length (en)
- Assume is known for one value of and all possible dimensions . Then you can bound MinN for length .
: (en)
- If is a compact metric space, and are homeomorphisms that commute, then , and an increasing sequence , such that (en)
- Assume is known for a given lengths for all dimensions of arithmetic progressions with benefits up to . This formula gives a bound on when you increase the dimension to :
let , then
: (en)
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- Theorem (en)
- Lemma 1 (en)
- Lemma 2 (en)
- multiple Birkhoff recurrence theorem (en)
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| dbp:note
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- Furstenberg and Weiss, 1978 (en)
- van der Waerden, 1927 (en)
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| dbp:proof
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- It suffices to show that for each length , there exist at least one partition that contains at least one arithmetic progression of length .Once this is proved, we can cut out that arithmetic progression into singleton sets, and repeat the process to create another arithmetic progression, and so one of the partitions contain infinitely many arithmetic progressions of length . Then we can repeat this process to find that there exists at least one partition that contains infinitely many progressions of length , for infinitely many , and that is the partition we want.
Consider the state space , which is compact under the metric Since the sets partition , we have a well-defined sequence with for all .
Let be the shift map
and let be the closure of all shifts of the sequence . By the multiple Birkhoff recurrence theorem , there exist a sequence and an integer such that
Since is the closure of shifts of , and is continuous, there exists a shift such that simultaneously, is very close to , and is very close to , and so on:
By the triangle inequality, we then immediately have
for . But by construction, any sequences with must have . Thus , and so all lie in the partition . (en)
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| dbp:title
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- Proof (en)
- Van der Waerden Number (en)
- van der Waerden's Theorem (en)
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- vanderWaerdenNumber (en)
- vanderWaerdensTheorem (en)
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| rdfs:label
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- Van der Waerden's theorem (en)
- Satz von van der Waerden (de)
- Théorème de van der Waerden (fr)
- ファン・デル・ヴェルデンの定理 (ja)
- Stelling van Van der Waerden (nl)
- Теорема ван дер Вардена (uk)
- Теорема ван дер Вардена (ru)
- 范德瓦尔登定理 (zh)
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