Special cases of Szemeredi's Theorem?
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Are there examples of special cases of Szemeredi's theorem which one can give, which are slightly non-trivial?
To clarify, I'm looking for sets of integers where we can show that they contain infinitely many arithmetic progressions of any finite length. When I say "slightly non-trivial", this is a non-well-defined condition which attempts to remove commenters saying things like "the even numbers" or "all infinite arithmetic progressions" - I'm looking for something like a proof for the special case of the square-free integers.
(Final comment: This is less of a question that I particularly need answered but more of an attempt to collate a collection of such "slightly non-trivial" proofs for my and others' appreciation.)
combinatorics number-theory soft-question alternative-proof arithmetic-progressions
asked Jan 3 at 15:15
Isky MathewsIsky Mathews
863314
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add a comment |
$begingroup$
Are there examples of special cases of Szemeredi's theorem which one can give, which are slightly non-trivial?
To clarify, I'm looking for sets of integers where we can show that they contain infinitely many arithmetic progressions of any finite length. When I say "slightly non-trivial", this is a non-well-defined condition which attempts to remove commenters saying things like "the even numbers" or "all infinite arithmetic progressions" - I'm looking for something like a proof for the special case of the square-free integers.
(Final comment: This is less of a question that I particularly need answered but more of an attempt to collate a collection of such "slightly non-trivial" proofs for my and others' appreciation.)
combinatorics number-theory soft-question alternative-proof arithmetic-progressions
asked Jan 3 at 15:15
Isky MathewsIsky Mathews
863314
$endgroup$
$begingroup$
Similar question: mathoverflow.net/questions/201494/…
$endgroup$
– Dap
Jan 20 at 17:59
add a comment |
$begingroup$
Are there examples of special cases of Szemeredi's theorem which one can give, which are slightly non-trivial?
To clarify, I'm looking for sets of integers where we can show that they contain infinitely many arithmetic progressions of any finite length. When I say "slightly non-trivial", this is a non-well-defined condition which attempts to remove commenters saying things like "the even numbers" or "all infinite arithmetic progressions" - I'm looking for something like a proof for the special case of the square-free integers.
(Final comment: This is less of a question that I particularly need answered but more of an attempt to collate a collection of such "slightly non-trivial" proofs for my and others' appreciation.)
combinatorics number-theory soft-question alternative-proof arithmetic-progressions
asked Jan 3 at 15:15
Isky MathewsIsky Mathews
863314
$endgroup$
Are there examples of special cases of Szemeredi's theorem which one can give, which are slightly non-trivial?
To clarify, I'm looking for sets of integers where we can show that they contain infinitely many arithmetic progressions of any finite length. When I say "slightly non-trivial", this is a non-well-defined condition which attempts to remove commenters saying things like "the even numbers" or "all infinite arithmetic progressions" - I'm looking for something like a proof for the special case of the square-free integers.
(Final comment: This is less of a question that I particularly need answered but more of an attempt to collate a collection of such "slightly non-trivial" proofs for my and others' appreciation.)
combinatorics number-theory soft-question alternative-proof arithmetic-progressions
combinatorics number-theory soft-question alternative-proof arithmetic-progressions
asked Jan 3 at 15:15
Isky MathewsIsky Mathews
863314
asked Jan 3 at 15:15
Isky MathewsIsky Mathews
863314
asked Jan 3 at 15:15
Isky MathewsIsky Mathews
863314
asked Jan 3 at 15:15
Isky MathewsIsky Mathews
863314
asked Jan 3 at 15:15
Isky MathewsIsky Mathews
863314
863314
$begingroup$
Similar question: mathoverflow.net/questions/201494/…
$endgroup$
– Dap
Jan 20 at 17:59
add a comment |
$begingroup$
Similar question: mathoverflow.net/questions/201494/…
$endgroup$
– Dap
Jan 20 at 17:59
$begingroup$
Similar question: mathoverflow.net/questions/201494/…
$endgroup$
– Dap
Jan 20 at 17:59
$begingroup$
Similar question: mathoverflow.net/questions/201494/…
$endgroup$
– Dap
Jan 20 at 17:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The Green-Tao theorem comes to mind. It states that the primes contain arithmetic progressions of arbitrary length. If you want to search for other 'non-trivial' special cases, starting with this Erdos conjecture might be a good idea.
There are also really interesting special cases of Szemeredi's Theorem that were proved before and are more accessible such as Roth's Theorem. This theorem states that any subset of integers of positive density contains a $3$ term arithmetic progression. This theorem is particularly interesting because there has been work on improving the density value. A good survey article is here.
answered Jan 3 at 15:42
Sandeep SilwalSandeep Silwal
5,89811337
$endgroup$
$begingroup$
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
$endgroup$
– Isky Mathews
Jan 3 at 20:09
add a comment |
$begingroup$
I am not sure if this qualifies under your idea 'slightly non-trivial', but van der Waerden's Theorem is a special case of Szemerédi's Theorem as any finite partition of $mathbb{Z}$ will necessarily have at least one partition class that has positive upper density, hence guaranteeing the existence of arbitrarily long arithmetic progressions in that partition class.
Although the proof (laid out here rather tersely by McCutcheon in $S~2.1$) of van der Waerden using combinatorial arguments is certainly easier and shorter than a proof of Roth's Theorem (at least using Furstenberg's Ergodic approach, I am personally not familiar with Roth's original treatment), it would certainly constitute either a rather long MSE post or necessitate a very terse exposition.
answered Feb 6 at 12:15
Walt van AmstelWalt van Amstel
771518
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
The Green-Tao theorem comes to mind. It states that the primes contain arithmetic progressions of arbitrary length. If you want to search for other 'non-trivial' special cases, starting with this Erdos conjecture might be a good idea.
There are also really interesting special cases of Szemeredi's Theorem that were proved before and are more accessible such as Roth's Theorem. This theorem states that any subset of integers of positive density contains a $3$ term arithmetic progression. This theorem is particularly interesting because there has been work on improving the density value. A good survey article is here.
answered Jan 3 at 15:42
Sandeep SilwalSandeep Silwal
5,89811337
$endgroup$
$begingroup$
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
$endgroup$
– Isky Mathews
Jan 3 at 20:09
add a comment |
$begingroup$
The Green-Tao theorem comes to mind. It states that the primes contain arithmetic progressions of arbitrary length. If you want to search for other 'non-trivial' special cases, starting with this Erdos conjecture might be a good idea.
There are also really interesting special cases of Szemeredi's Theorem that were proved before and are more accessible such as Roth's Theorem. This theorem states that any subset of integers of positive density contains a $3$ term arithmetic progression. This theorem is particularly interesting because there has been work on improving the density value. A good survey article is here.
answered Jan 3 at 15:42
Sandeep SilwalSandeep Silwal
5,89811337
$endgroup$
$begingroup$
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
$endgroup$
– Isky Mathews
Jan 3 at 20:09
add a comment |
$begingroup$
The Green-Tao theorem comes to mind. It states that the primes contain arithmetic progressions of arbitrary length. If you want to search for other 'non-trivial' special cases, starting with this Erdos conjecture might be a good idea.
There are also really interesting special cases of Szemeredi's Theorem that were proved before and are more accessible such as Roth's Theorem. This theorem states that any subset of integers of positive density contains a $3$ term arithmetic progression. This theorem is particularly interesting because there has been work on improving the density value. A good survey article is here.
answered Jan 3 at 15:42
Sandeep SilwalSandeep Silwal
5,89811337
$endgroup$
The Green-Tao theorem comes to mind. It states that the primes contain arithmetic progressions of arbitrary length. If you want to search for other 'non-trivial' special cases, starting with this Erdos conjecture might be a good idea.
There are also really interesting special cases of Szemeredi's Theorem that were proved before and are more accessible such as Roth's Theorem. This theorem states that any subset of integers of positive density contains a $3$ term arithmetic progression. This theorem is particularly interesting because there has been work on improving the density value. A good survey article is here.
answered Jan 3 at 15:42
Sandeep SilwalSandeep Silwal
5,89811337
answered Jan 3 at 15:42
Sandeep SilwalSandeep Silwal
5,89811337
answered Jan 3 at 15:42
Sandeep SilwalSandeep Silwal
5,89811337
answered Jan 3 at 15:42
Sandeep SilwalSandeep Silwal
5,89811337
5,89811337
$begingroup$
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
$endgroup$
– Isky Mathews
Jan 3 at 20:09
add a comment |
$begingroup$
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
$endgroup$
– Isky Mathews
Jan 3 at 20:09
$begingroup$
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
$endgroup$
– Isky Mathews
Jan 3 at 20:09
$begingroup$
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
$endgroup$
– Isky Mathews
Jan 3 at 20:09
add a comment |
$begingroup$
I am not sure if this qualifies under your idea 'slightly non-trivial', but van der Waerden's Theorem is a special case of Szemerédi's Theorem as any finite partition of $mathbb{Z}$ will necessarily have at least one partition class that has positive upper density, hence guaranteeing the existence of arbitrarily long arithmetic progressions in that partition class.
Although the proof (laid out here rather tersely by McCutcheon in $S~2.1$) of van der Waerden using combinatorial arguments is certainly easier and shorter than a proof of Roth's Theorem (at least using Furstenberg's Ergodic approach, I am personally not familiar with Roth's original treatment), it would certainly constitute either a rather long MSE post or necessitate a very terse exposition.
answered Feb 6 at 12:15
Walt van AmstelWalt van Amstel
771518
$endgroup$
add a comment |
$begingroup$
I am not sure if this qualifies under your idea 'slightly non-trivial', but van der Waerden's Theorem is a special case of Szemerédi's Theorem as any finite partition of $mathbb{Z}$ will necessarily have at least one partition class that has positive upper density, hence guaranteeing the existence of arbitrarily long arithmetic progressions in that partition class.
Although the proof (laid out here rather tersely by McCutcheon in $S~2.1$) of van der Waerden using combinatorial arguments is certainly easier and shorter than a proof of Roth's Theorem (at least using Furstenberg's Ergodic approach, I am personally not familiar with Roth's original treatment), it would certainly constitute either a rather long MSE post or necessitate a very terse exposition.
answered Feb 6 at 12:15
Walt van AmstelWalt van Amstel
771518
$endgroup$
add a comment |
$begingroup$
I am not sure if this qualifies under your idea 'slightly non-trivial', but van der Waerden's Theorem is a special case of Szemerédi's Theorem as any finite partition of $mathbb{Z}$ will necessarily have at least one partition class that has positive upper density, hence guaranteeing the existence of arbitrarily long arithmetic progressions in that partition class.
Although the proof (laid out here rather tersely by McCutcheon in $S~2.1$) of van der Waerden using combinatorial arguments is certainly easier and shorter than a proof of Roth's Theorem (at least using Furstenberg's Ergodic approach, I am personally not familiar with Roth's original treatment), it would certainly constitute either a rather long MSE post or necessitate a very terse exposition.
answered Feb 6 at 12:15
Walt van AmstelWalt van Amstel
771518
$endgroup$
I am not sure if this qualifies under your idea 'slightly non-trivial', but van der Waerden's Theorem is a special case of Szemerédi's Theorem as any finite partition of $mathbb{Z}$ will necessarily have at least one partition class that has positive upper density, hence guaranteeing the existence of arbitrarily long arithmetic progressions in that partition class.
Although the proof (laid out here rather tersely by McCutcheon in $S~2.1$) of van der Waerden using combinatorial arguments is certainly easier and shorter than a proof of Roth's Theorem (at least using Furstenberg's Ergodic approach, I am personally not familiar with Roth's original treatment), it would certainly constitute either a rather long MSE post or necessitate a very terse exposition.
answered Feb 6 at 12:15
Walt van AmstelWalt van Amstel
771518
answered Feb 6 at 12:15
Walt van AmstelWalt van Amstel
771518
answered Feb 6 at 12:15
Walt van AmstelWalt van Amstel
771518
answered Feb 6 at 12:15
Walt van AmstelWalt van Amstel
771518
771518
add a comment |
add a comment |
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$begingroup$
Similar question: mathoverflow.net/questions/201494/…
$endgroup$
– Dap
Jan 20 at 17:59