$I$ is a union of intervals such that there do no exist 2 points in $I$ with difference $1/12$.Prove the sum...
$begingroup$
Consider $I$ a union of disjoint intervals inlcuded in the interval $[0, 1]$ such that there do no exist 2 points in $I$ situated at distance $1/12$. Prove that the sum of the lengths of the intervals is at most $1/2$.
What I have shown is that this sum, is less than $11/12$, in the way I am going to describe below:
I used the notation: $I = [x_1, x_2] cup [x_3, x_4] cup ... cup [x_{2k + 1}, x_{2k + 2}]$, where $0 leq x_1 leq x_2 < x_3 leq x_4 < ... < x_{2k + 1} leq x_{2k + 2} leq 1$.
First of all, if we consider 2 points from the same interval $[x_{2i + 1}, x_{2i + 2}]$, from the condition of the problem we get that $x_{2i + 2} - x_{2i + 1} < 1/12$, which can be easily shown by proof by contradiction. So, each interval has length at most $1/12$.
Now consider 2 points situated in consecutive intervals: $a in [x_{2i + 1}, x_{2i + 2}]$, $b in [x_{2i + 3}, x_{2i + 4}]$. Because their difference cannot be equal to $1/12$, we have only 2 cases:
i) $x_{2i + 4} < x_{2i + 1} + 1/12$, which corresponds to having 2 intervals, together having a length less than 1/12;
ii) $x_{2i + 3} > x_{2i + 2} + 1/12$, which corresponds to having 2 intervals with length between them greater than 1/12;
Now, for the first case as the legth of the 2 intervals combined is less than $1/12$, we can reduce it to having only 1 interval with length less than $1/12$, so the entire problem reduces to having a number of intervals each of legth at most $1/12$ and with difference between them at least $1/12$.
Finally, to bound the number of such intervals, we use the fact that they are all included in $[0, 1]$, and because the distance between them is at least $1/12$ we have at most 11 intervals, thus the sum is $leq 11 cdot 1/12 = 11/12$
Do you have any suggestions on how I can reach $1/2$?
linear-algebra pigeonhole-principle interval-arithmetic
asked Dec 28 '18 at 18:50
SandelSandel
1816
$endgroup$
add a comment |
$begingroup$
Consider $I$ a union of disjoint intervals inlcuded in the interval $[0, 1]$ such that there do no exist 2 points in $I$ situated at distance $1/12$. Prove that the sum of the lengths of the intervals is at most $1/2$.
What I have shown is that this sum, is less than $11/12$, in the way I am going to describe below:
I used the notation: $I = [x_1, x_2] cup [x_3, x_4] cup ... cup [x_{2k + 1}, x_{2k + 2}]$, where $0 leq x_1 leq x_2 < x_3 leq x_4 < ... < x_{2k + 1} leq x_{2k + 2} leq 1$.
First of all, if we consider 2 points from the same interval $[x_{2i + 1}, x_{2i + 2}]$, from the condition of the problem we get that $x_{2i + 2} - x_{2i + 1} < 1/12$, which can be easily shown by proof by contradiction. So, each interval has length at most $1/12$.
Now consider 2 points situated in consecutive intervals: $a in [x_{2i + 1}, x_{2i + 2}]$, $b in [x_{2i + 3}, x_{2i + 4}]$. Because their difference cannot be equal to $1/12$, we have only 2 cases:
i) $x_{2i + 4} < x_{2i + 1} + 1/12$, which corresponds to having 2 intervals, together having a length less than 1/12;
ii) $x_{2i + 3} > x_{2i + 2} + 1/12$, which corresponds to having 2 intervals with length between them greater than 1/12;
Now, for the first case as the legth of the 2 intervals combined is less than $1/12$, we can reduce it to having only 1 interval with length less than $1/12$, so the entire problem reduces to having a number of intervals each of legth at most $1/12$ and with difference between them at least $1/12$.
Finally, to bound the number of such intervals, we use the fact that they are all included in $[0, 1]$, and because the distance between them is at least $1/12$ we have at most 11 intervals, thus the sum is $leq 11 cdot 1/12 = 11/12$
Do you have any suggestions on how I can reach $1/2$?
linear-algebra pigeonhole-principle interval-arithmetic
asked Dec 28 '18 at 18:50
SandelSandel
1816
$endgroup$
$begingroup$
You can show that the sum of lengths is at most $13/24$. This is because $I$ and $Icup(I+1/12)$ are disjoint, of equal measure and both subsets of $[0,13/12]$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 18:57
add a comment |
$begingroup$
Consider $I$ a union of disjoint intervals inlcuded in the interval $[0, 1]$ such that there do no exist 2 points in $I$ situated at distance $1/12$. Prove that the sum of the lengths of the intervals is at most $1/2$.
What I have shown is that this sum, is less than $11/12$, in the way I am going to describe below:
I used the notation: $I = [x_1, x_2] cup [x_3, x_4] cup ... cup [x_{2k + 1}, x_{2k + 2}]$, where $0 leq x_1 leq x_2 < x_3 leq x_4 < ... < x_{2k + 1} leq x_{2k + 2} leq 1$.
First of all, if we consider 2 points from the same interval $[x_{2i + 1}, x_{2i + 2}]$, from the condition of the problem we get that $x_{2i + 2} - x_{2i + 1} < 1/12$, which can be easily shown by proof by contradiction. So, each interval has length at most $1/12$.
Now consider 2 points situated in consecutive intervals: $a in [x_{2i + 1}, x_{2i + 2}]$, $b in [x_{2i + 3}, x_{2i + 4}]$. Because their difference cannot be equal to $1/12$, we have only 2 cases:
i) $x_{2i + 4} < x_{2i + 1} + 1/12$, which corresponds to having 2 intervals, together having a length less than 1/12;
ii) $x_{2i + 3} > x_{2i + 2} + 1/12$, which corresponds to having 2 intervals with length between them greater than 1/12;
Now, for the first case as the legth of the 2 intervals combined is less than $1/12$, we can reduce it to having only 1 interval with length less than $1/12$, so the entire problem reduces to having a number of intervals each of legth at most $1/12$ and with difference between them at least $1/12$.
Finally, to bound the number of such intervals, we use the fact that they are all included in $[0, 1]$, and because the distance between them is at least $1/12$ we have at most 11 intervals, thus the sum is $leq 11 cdot 1/12 = 11/12$
Do you have any suggestions on how I can reach $1/2$?
linear-algebra pigeonhole-principle interval-arithmetic
asked Dec 28 '18 at 18:50
SandelSandel
1816
$endgroup$
Consider $I$ a union of disjoint intervals inlcuded in the interval $[0, 1]$ such that there do no exist 2 points in $I$ situated at distance $1/12$. Prove that the sum of the lengths of the intervals is at most $1/2$.
What I have shown is that this sum, is less than $11/12$, in the way I am going to describe below:
I used the notation: $I = [x_1, x_2] cup [x_3, x_4] cup ... cup [x_{2k + 1}, x_{2k + 2}]$, where $0 leq x_1 leq x_2 < x_3 leq x_4 < ... < x_{2k + 1} leq x_{2k + 2} leq 1$.
First of all, if we consider 2 points from the same interval $[x_{2i + 1}, x_{2i + 2}]$, from the condition of the problem we get that $x_{2i + 2} - x_{2i + 1} < 1/12$, which can be easily shown by proof by contradiction. So, each interval has length at most $1/12$.
Now consider 2 points situated in consecutive intervals: $a in [x_{2i + 1}, x_{2i + 2}]$, $b in [x_{2i + 3}, x_{2i + 4}]$. Because their difference cannot be equal to $1/12$, we have only 2 cases:
i) $x_{2i + 4} < x_{2i + 1} + 1/12$, which corresponds to having 2 intervals, together having a length less than 1/12;
ii) $x_{2i + 3} > x_{2i + 2} + 1/12$, which corresponds to having 2 intervals with length between them greater than 1/12;
Now, for the first case as the legth of the 2 intervals combined is less than $1/12$, we can reduce it to having only 1 interval with length less than $1/12$, so the entire problem reduces to having a number of intervals each of legth at most $1/12$ and with difference between them at least $1/12$.
Finally, to bound the number of such intervals, we use the fact that they are all included in $[0, 1]$, and because the distance between them is at least $1/12$ we have at most 11 intervals, thus the sum is $leq 11 cdot 1/12 = 11/12$
Do you have any suggestions on how I can reach $1/2$?
linear-algebra pigeonhole-principle interval-arithmetic
linear-algebra pigeonhole-principle interval-arithmetic
asked Dec 28 '18 at 18:50
SandelSandel
1816
asked Dec 28 '18 at 18:50
SandelSandel
1816
edited Dec 28 '18 at 19:08
Sandel
asked Dec 28 '18 at 18:50
SandelSandel
1816
asked Dec 28 '18 at 18:50
SandelSandel
1816
asked Dec 28 '18 at 18:50
SandelSandel
1816
1816
$begingroup$
You can show that the sum of lengths is at most $13/24$. This is because $I$ and $Icup(I+1/12)$ are disjoint, of equal measure and both subsets of $[0,13/12]$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 18:57
add a comment |
$begingroup$
You can show that the sum of lengths is at most $13/24$. This is because $I$ and $Icup(I+1/12)$ are disjoint, of equal measure and both subsets of $[0,13/12]$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 18:57
$begingroup$
You can show that the sum of lengths is at most $13/24$. This is because $I$ and $Icup(I+1/12)$ are disjoint, of equal measure and both subsets of $[0,13/12]$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 18:57
$begingroup$
You can show that the sum of lengths is at most $13/24$. This is because $I$ and $Icup(I+1/12)$ are disjoint, of equal measure and both subsets of $[0,13/12]$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 18:57
add a comment |
1 Answer
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You can sort all the endpoint values modulo $1/12$ together with $0$ and $1/12$, let's call this sorted sequence ${s_i}$. Then for each $i$ we can observe the intervals $[s_i,s_{i+1}]$, $[s_i+1/12,s_{i+1}+1/12]$, ..., $[s_i+11/12,s_{i+1}+11/12]$. Call the union of these intervals $U_i$. For each $U_i$ at least half of the intervals must be disjoint from $I$. Hence the measure of $Icap U_i$ is at most half of the measure of $U_i$. Because the $U_i$ partition $[0,1]$, and because $Isubset[0,1]$ we find that the measure of $I$ is at most half the measure of $[0,1]$.
answered Dec 28 '18 at 19:24
SmileyCraftSmileyCraft
3,601517
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$begingroup$
You can sort all the endpoint values modulo $1/12$ together with $0$ and $1/12$, let's call this sorted sequence ${s_i}$. Then for each $i$ we can observe the intervals $[s_i,s_{i+1}]$, $[s_i+1/12,s_{i+1}+1/12]$, ..., $[s_i+11/12,s_{i+1}+11/12]$. Call the union of these intervals $U_i$. For each $U_i$ at least half of the intervals must be disjoint from $I$. Hence the measure of $Icap U_i$ is at most half of the measure of $U_i$. Because the $U_i$ partition $[0,1]$, and because $Isubset[0,1]$ we find that the measure of $I$ is at most half the measure of $[0,1]$.
answered Dec 28 '18 at 19:24
SmileyCraftSmileyCraft
3,601517
$endgroup$
add a comment |
$begingroup$
You can sort all the endpoint values modulo $1/12$ together with $0$ and $1/12$, let's call this sorted sequence ${s_i}$. Then for each $i$ we can observe the intervals $[s_i,s_{i+1}]$, $[s_i+1/12,s_{i+1}+1/12]$, ..., $[s_i+11/12,s_{i+1}+11/12]$. Call the union of these intervals $U_i$. For each $U_i$ at least half of the intervals must be disjoint from $I$. Hence the measure of $Icap U_i$ is at most half of the measure of $U_i$. Because the $U_i$ partition $[0,1]$, and because $Isubset[0,1]$ we find that the measure of $I$ is at most half the measure of $[0,1]$.
answered Dec 28 '18 at 19:24
SmileyCraftSmileyCraft
3,601517
$endgroup$
add a comment |
$begingroup$
You can sort all the endpoint values modulo $1/12$ together with $0$ and $1/12$, let's call this sorted sequence ${s_i}$. Then for each $i$ we can observe the intervals $[s_i,s_{i+1}]$, $[s_i+1/12,s_{i+1}+1/12]$, ..., $[s_i+11/12,s_{i+1}+11/12]$. Call the union of these intervals $U_i$. For each $U_i$ at least half of the intervals must be disjoint from $I$. Hence the measure of $Icap U_i$ is at most half of the measure of $U_i$. Because the $U_i$ partition $[0,1]$, and because $Isubset[0,1]$ we find that the measure of $I$ is at most half the measure of $[0,1]$.
answered Dec 28 '18 at 19:24
SmileyCraftSmileyCraft
3,601517
$endgroup$
You can sort all the endpoint values modulo $1/12$ together with $0$ and $1/12$, let's call this sorted sequence ${s_i}$. Then for each $i$ we can observe the intervals $[s_i,s_{i+1}]$, $[s_i+1/12,s_{i+1}+1/12]$, ..., $[s_i+11/12,s_{i+1}+11/12]$. Call the union of these intervals $U_i$. For each $U_i$ at least half of the intervals must be disjoint from $I$. Hence the measure of $Icap U_i$ is at most half of the measure of $U_i$. Because the $U_i$ partition $[0,1]$, and because $Isubset[0,1]$ we find that the measure of $I$ is at most half the measure of $[0,1]$.
answered Dec 28 '18 at 19:24
SmileyCraftSmileyCraft
3,601517
answered Dec 28 '18 at 19:24
SmileyCraftSmileyCraft
3,601517
answered Dec 28 '18 at 19:24
SmileyCraftSmileyCraft
3,601517
answered Dec 28 '18 at 19:24
SmileyCraftSmileyCraft
3,601517
3,601517
add a comment |
add a comment |
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$begingroup$
You can show that the sum of lengths is at most $13/24$. This is because $I$ and $Icup(I+1/12)$ are disjoint, of equal measure and both subsets of $[0,13/12]$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 18:57