Lebesgue measure of a set of real numbers well-approximated by rationals
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Consider the set $A$ of real numbers $r$ such that there exists a constant $C$ and sequence $frac{p_n}{q_n}$ of rational numbers (where $p_n$ and $q_n$ are integers) with $q_n rightarrow infty$ and
$$leftvert r - frac{p_n}{q_n} rightvert < frac{C}{q_n^3}.$$
How do I show the Lebesgue measure of $A$ is zero?
measure-theory diophantine-approximation
asked Aug 5 '15 at 1:12
Philip HoskinsPhilip Hoskins
1,450715
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add a comment |
$begingroup$
Consider the set $A$ of real numbers $r$ such that there exists a constant $C$ and sequence $frac{p_n}{q_n}$ of rational numbers (where $p_n$ and $q_n$ are integers) with $q_n rightarrow infty$ and
$$leftvert r - frac{p_n}{q_n} rightvert < frac{C}{q_n^3}.$$
How do I show the Lebesgue measure of $A$ is zero?
measure-theory diophantine-approximation
asked Aug 5 '15 at 1:12
Philip HoskinsPhilip Hoskins
1,450715
$endgroup$
add a comment |
$begingroup$
Consider the set $A$ of real numbers $r$ such that there exists a constant $C$ and sequence $frac{p_n}{q_n}$ of rational numbers (where $p_n$ and $q_n$ are integers) with $q_n rightarrow infty$ and
$$leftvert r - frac{p_n}{q_n} rightvert < frac{C}{q_n^3}.$$
How do I show the Lebesgue measure of $A$ is zero?
measure-theory diophantine-approximation
asked Aug 5 '15 at 1:12
Philip HoskinsPhilip Hoskins
1,450715
$endgroup$
Consider the set $A$ of real numbers $r$ such that there exists a constant $C$ and sequence $frac{p_n}{q_n}$ of rational numbers (where $p_n$ and $q_n$ are integers) with $q_n rightarrow infty$ and
$$leftvert r - frac{p_n}{q_n} rightvert < frac{C}{q_n^3}.$$
How do I show the Lebesgue measure of $A$ is zero?
measure-theory diophantine-approximation
measure-theory diophantine-approximation
asked Aug 5 '15 at 1:12
Philip HoskinsPhilip Hoskins
1,450715
asked Aug 5 '15 at 1:12
Philip HoskinsPhilip Hoskins
1,450715
asked Aug 5 '15 at 1:12
Philip HoskinsPhilip Hoskins
1,450715
asked Aug 5 '15 at 1:12
Philip HoskinsPhilip Hoskins
1,450715
asked Aug 5 '15 at 1:12
Philip HoskinsPhilip Hoskins
1,450715
1,450715
add a comment |
add a comment |
1 Answer
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Lets begin by noticing that we only have to concern ourselves with $rin [0, 1]$: any number that can be so approximated will differ by an an integer from a number in that interval. Now, consider the sets
$$A_q = left[0, frac{C}{q^3} right) cup left( 1 - frac{C}{q^3}, 1 right]cup bigcup_{i=1}^{q-1} left( frac{i}{q} - frac{C}{q^3}, frac{i}{q} + frac{C}{q^3} right).$$
A point satisying the condition must be in an infinite number of sets of the form $A_q$. Put in terms of sets, this is saying that if$$ B_n = bigcup_{q=n}^infty A_q $$then the set of points satisfying the hypothesis in $[0, 1]$ is$$ X = bigcap_{n=1}^infty B_n.$$ To get the result, note that $lambda(A_q) leq 2C/q^2$, $lambda(B_n) leq 2C sum_{i=n}^infty 1/i^2$ and so $$lambda(X) < liminf_n B_n = 0$$ as required.
answered Aug 6 '15 at 8:38
user24142user24142
3,0151116
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
$begingroup$
Lets begin by noticing that we only have to concern ourselves with $rin [0, 1]$: any number that can be so approximated will differ by an an integer from a number in that interval. Now, consider the sets
$$A_q = left[0, frac{C}{q^3} right) cup left( 1 - frac{C}{q^3}, 1 right]cup bigcup_{i=1}^{q-1} left( frac{i}{q} - frac{C}{q^3}, frac{i}{q} + frac{C}{q^3} right).$$
A point satisying the condition must be in an infinite number of sets of the form $A_q$. Put in terms of sets, this is saying that if$$ B_n = bigcup_{q=n}^infty A_q $$then the set of points satisfying the hypothesis in $[0, 1]$ is$$ X = bigcap_{n=1}^infty B_n.$$ To get the result, note that $lambda(A_q) leq 2C/q^2$, $lambda(B_n) leq 2C sum_{i=n}^infty 1/i^2$ and so $$lambda(X) < liminf_n B_n = 0$$ as required.
answered Aug 6 '15 at 8:38
user24142user24142
3,0151116
$endgroup$
add a comment |
$begingroup$
Lets begin by noticing that we only have to concern ourselves with $rin [0, 1]$: any number that can be so approximated will differ by an an integer from a number in that interval. Now, consider the sets
$$A_q = left[0, frac{C}{q^3} right) cup left( 1 - frac{C}{q^3}, 1 right]cup bigcup_{i=1}^{q-1} left( frac{i}{q} - frac{C}{q^3}, frac{i}{q} + frac{C}{q^3} right).$$
A point satisying the condition must be in an infinite number of sets of the form $A_q$. Put in terms of sets, this is saying that if$$ B_n = bigcup_{q=n}^infty A_q $$then the set of points satisfying the hypothesis in $[0, 1]$ is$$ X = bigcap_{n=1}^infty B_n.$$ To get the result, note that $lambda(A_q) leq 2C/q^2$, $lambda(B_n) leq 2C sum_{i=n}^infty 1/i^2$ and so $$lambda(X) < liminf_n B_n = 0$$ as required.
answered Aug 6 '15 at 8:38
user24142user24142
3,0151116
$endgroup$
add a comment |
$begingroup$
Lets begin by noticing that we only have to concern ourselves with $rin [0, 1]$: any number that can be so approximated will differ by an an integer from a number in that interval. Now, consider the sets
$$A_q = left[0, frac{C}{q^3} right) cup left( 1 - frac{C}{q^3}, 1 right]cup bigcup_{i=1}^{q-1} left( frac{i}{q} - frac{C}{q^3}, frac{i}{q} + frac{C}{q^3} right).$$
A point satisying the condition must be in an infinite number of sets of the form $A_q$. Put in terms of sets, this is saying that if$$ B_n = bigcup_{q=n}^infty A_q $$then the set of points satisfying the hypothesis in $[0, 1]$ is$$ X = bigcap_{n=1}^infty B_n.$$ To get the result, note that $lambda(A_q) leq 2C/q^2$, $lambda(B_n) leq 2C sum_{i=n}^infty 1/i^2$ and so $$lambda(X) < liminf_n B_n = 0$$ as required.
answered Aug 6 '15 at 8:38
user24142user24142
3,0151116
$endgroup$
Lets begin by noticing that we only have to concern ourselves with $rin [0, 1]$: any number that can be so approximated will differ by an an integer from a number in that interval. Now, consider the sets
$$A_q = left[0, frac{C}{q^3} right) cup left( 1 - frac{C}{q^3}, 1 right]cup bigcup_{i=1}^{q-1} left( frac{i}{q} - frac{C}{q^3}, frac{i}{q} + frac{C}{q^3} right).$$
A point satisying the condition must be in an infinite number of sets of the form $A_q$. Put in terms of sets, this is saying that if$$ B_n = bigcup_{q=n}^infty A_q $$then the set of points satisfying the hypothesis in $[0, 1]$ is$$ X = bigcap_{n=1}^infty B_n.$$ To get the result, note that $lambda(A_q) leq 2C/q^2$, $lambda(B_n) leq 2C sum_{i=n}^infty 1/i^2$ and so $$lambda(X) < liminf_n B_n = 0$$ as required.
answered Aug 6 '15 at 8:38
user24142user24142
3,0151116
edited Dec 18 '18 at 23:34
answered Aug 6 '15 at 8:38
user24142user24142
3,0151116
answered Aug 6 '15 at 8:38
user24142user24142
3,0151116
answered Aug 6 '15 at 8:38
user24142user24142
3,0151116
3,0151116
add a comment |
add a comment |
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