Is there a set of positive integers of density 1 which contains no infinite arithmetic progression?What is the shortest route to Roth's theorem?Inverse Length 3 Arithmetic Progression Problem for sets with positive upper densityDo there exist sets of integers with arbitrarily large upper density which contains infinitely many elements that are not in an arithmetic progression of length 3?Arithmetic progressions in power sequencesAre there infinite primes among powerful order terms of Dirichlet arithmetic progressions?every arithmetic progression contains a sequence of $k$ “consecutive” primes for possibly all natural numbers $k$?On the upper Banach density of the set of positive integers whose base-$b$ representation misses at least one prescribed digitAdditivity of upper densities with respect to arithmetic progressions of integersExistence of Arithmetic Progression from density inequalityDiscrepancy in non-homogeneous arithmetic progressions
Is there a set of positive integers of density 1 which contains no infinite arithmetic progression?
What is the shortest route to Roth's theorem?Inverse Length 3 Arithmetic Progression Problem for sets with positive upper densityDo there exist sets of integers with arbitrarily large upper density which contains infinitely many elements that are not in an arithmetic progression of length 3?Arithmetic progressions in power sequencesAre there infinite primes among powerful order terms of Dirichlet arithmetic progressions?every arithmetic progression contains a sequence of $k$ “consecutive” primes for possibly all natural numbers $k$?On the upper Banach density of the set of positive integers whose base-$b$ representation misses at least one prescribed digitAdditivity of upper densities with respect to arithmetic progressions of integersExistence of Arithmetic Progression from density inequalityDiscrepancy in non-homogeneous arithmetic progressions
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Let $V$ be a set of positive integers whose natural density is 1. Is it necessarily true that $V$ contains an infinite arithmetic progression?—i.e., that there are non-negative integers $a,b,nu$ with $0leq bleq a-1$ so that: $$left an+b:ngeqnuright subseteq V$$
In addition to an answer, any references on the matter would be most appreciated.
nt.number-theory additive-combinatorics
asked 8 hours ago
MCSMCS
247110
$endgroup$
add a comment |
$begingroup$
Let $V$ be a set of positive integers whose natural density is 1. Is it necessarily true that $V$ contains an infinite arithmetic progression?—i.e., that there are non-negative integers $a,b,nu$ with $0leq bleq a-1$ so that: $$left an+b:ngeqnuright subseteq V$$
In addition to an answer, any references on the matter would be most appreciated.
nt.number-theory additive-combinatorics
asked 8 hours ago
MCSMCS
247110
$endgroup$
14
$begingroup$
Enumerate all the arithmetic progressions as $A_n$. Now remove from $mathbb N$ the $2^n$-th element of $A_n$ for each $n$.
$endgroup$
– Wojowu
8 hours ago
1
$begingroup$
So sets that contradict this exist? Excellent! Thank you. :)
$endgroup$
– MCS
8 hours ago
3
$begingroup$
Generalizing Wojowu’s construction shows that “density 1” can be achieved in the strongest possible way: for any function $f$ that tends to infinity, no matter how slowly, there is a set of non-negative integers containing at least $n-f(n)$ integers between $1$ and $n$ for each $n$ but no infinite arithmetic progression.
$endgroup$
– Henry Cohn
7 hours ago
add a comment |
$begingroup$
Let $V$ be a set of positive integers whose natural density is 1. Is it necessarily true that $V$ contains an infinite arithmetic progression?—i.e., that there are non-negative integers $a,b,nu$ with $0leq bleq a-1$ so that: $$left an+b:ngeqnuright subseteq V$$
In addition to an answer, any references on the matter would be most appreciated.
nt.number-theory additive-combinatorics
asked 8 hours ago
MCSMCS
247110
$endgroup$
Let $V$ be a set of positive integers whose natural density is 1. Is it necessarily true that $V$ contains an infinite arithmetic progression?—i.e., that there are non-negative integers $a,b,nu$ with $0leq bleq a-1$ so that: $$left an+b:ngeqnuright subseteq V$$
In addition to an answer, any references on the matter would be most appreciated.
nt.number-theory additive-combinatorics
nt.number-theory additive-combinatorics
asked 8 hours ago
MCSMCS
247110
asked 8 hours ago
MCSMCS
247110
asked 8 hours ago
MCSMCS
247110
asked 8 hours ago
MCSMCS
247110
asked 8 hours ago
MCSMCS
247110
247110
14
$begingroup$
Enumerate all the arithmetic progressions as $A_n$. Now remove from $mathbb N$ the $2^n$-th element of $A_n$ for each $n$.
$endgroup$
– Wojowu
8 hours ago
1
$begingroup$
So sets that contradict this exist? Excellent! Thank you. :)
$endgroup$
– MCS
8 hours ago
3
$begingroup$
Generalizing Wojowu’s construction shows that “density 1” can be achieved in the strongest possible way: for any function $f$ that tends to infinity, no matter how slowly, there is a set of non-negative integers containing at least $n-f(n)$ integers between $1$ and $n$ for each $n$ but no infinite arithmetic progression.
$endgroup$
– Henry Cohn
7 hours ago
add a comment |
14
$begingroup$
Enumerate all the arithmetic progressions as $A_n$. Now remove from $mathbb N$ the $2^n$-th element of $A_n$ for each $n$.
$endgroup$
– Wojowu
8 hours ago
1
$begingroup$
So sets that contradict this exist? Excellent! Thank you. :)
$endgroup$
– MCS
8 hours ago
3
$begingroup$
Generalizing Wojowu’s construction shows that “density 1” can be achieved in the strongest possible way: for any function $f$ that tends to infinity, no matter how slowly, there is a set of non-negative integers containing at least $n-f(n)$ integers between $1$ and $n$ for each $n$ but no infinite arithmetic progression.
$endgroup$
– Henry Cohn
7 hours ago
14
14
$begingroup$
Enumerate all the arithmetic progressions as $A_n$. Now remove from $mathbb N$ the $2^n$-th element of $A_n$ for each $n$.
$endgroup$
– Wojowu
8 hours ago
$begingroup$
Enumerate all the arithmetic progressions as $A_n$. Now remove from $mathbb N$ the $2^n$-th element of $A_n$ for each $n$.
$endgroup$
– Wojowu
8 hours ago
1
1
$begingroup$
So sets that contradict this exist? Excellent! Thank you. :)
$endgroup$
– MCS
8 hours ago
$begingroup$
So sets that contradict this exist? Excellent! Thank you. :)
$endgroup$
– MCS
8 hours ago
3
3
$begingroup$
Generalizing Wojowu’s construction shows that “density 1” can be achieved in the strongest possible way: for any function $f$ that tends to infinity, no matter how slowly, there is a set of non-negative integers containing at least $n-f(n)$ integers between $1$ and $n$ for each $n$ but no infinite arithmetic progression.
$endgroup$
– Henry Cohn
7 hours ago
$begingroup$
Generalizing Wojowu’s construction shows that “density 1” can be achieved in the strongest possible way: for any function $f$ that tends to infinity, no matter how slowly, there is a set of non-negative integers containing at least $n-f(n)$ integers between $1$ and $n$ for each $n$ but no infinite arithmetic progression.
$endgroup$
– Henry Cohn
7 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a concrete counterexample (where $mathbbN$ is the set of positive integers):
$$V:=mathbbNsetminusa^2b^2+b:a,binmathbbN.$$
It is straightforward to see that $V$ has density $1$, but it does not contain an infinite arithmetic progression.
answered 7 hours ago
GH from MOGH from MO
61k5154233
$endgroup$
add a comment |
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$begingroup$
Here is a concrete counterexample (where $mathbbN$ is the set of positive integers):
$$V:=mathbbNsetminusa^2b^2+b:a,binmathbbN.$$
It is straightforward to see that $V$ has density $1$, but it does not contain an infinite arithmetic progression.
answered 7 hours ago
GH from MOGH from MO
61k5154233
$endgroup$
add a comment |
$begingroup$
Here is a concrete counterexample (where $mathbbN$ is the set of positive integers):
$$V:=mathbbNsetminusa^2b^2+b:a,binmathbbN.$$
It is straightforward to see that $V$ has density $1$, but it does not contain an infinite arithmetic progression.
answered 7 hours ago
GH from MOGH from MO
61k5154233
$endgroup$
add a comment |
$begingroup$
Here is a concrete counterexample (where $mathbbN$ is the set of positive integers):
$$V:=mathbbNsetminusa^2b^2+b:a,binmathbbN.$$
It is straightforward to see that $V$ has density $1$, but it does not contain an infinite arithmetic progression.
answered 7 hours ago
GH from MOGH from MO
61k5154233
$endgroup$
Here is a concrete counterexample (where $mathbbN$ is the set of positive integers):
$$V:=mathbbNsetminusa^2b^2+b:a,binmathbbN.$$
It is straightforward to see that $V$ has density $1$, but it does not contain an infinite arithmetic progression.
answered 7 hours ago
GH from MOGH from MO
61k5154233
answered 7 hours ago
GH from MOGH from MO
61k5154233
answered 7 hours ago
GH from MOGH from MO
61k5154233
answered 7 hours ago
GH from MOGH from MO
61k5154233
61k5154233
add a comment |
add a comment |
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14
$begingroup$
Enumerate all the arithmetic progressions as $A_n$. Now remove from $mathbb N$ the $2^n$-th element of $A_n$ for each $n$.
$endgroup$
– Wojowu
8 hours ago
1
$begingroup$
So sets that contradict this exist? Excellent! Thank you. :)
$endgroup$
– MCS
8 hours ago
3
$begingroup$
Generalizing Wojowu’s construction shows that “density 1” can be achieved in the strongest possible way: for any function $f$ that tends to infinity, no matter how slowly, there is a set of non-negative integers containing at least $n-f(n)$ integers between $1$ and $n$ for each $n$ but no infinite arithmetic progression.
$endgroup$
– Henry Cohn
7 hours ago