It is true that a finite number of compact curves cannot cover an open disk?
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Im trying to prove that a finite number of compact curves cannot cover any open disk in $Bbb R^2$. I dont know if this is true but it seems.
If a curve is a $1$-dimensional submanifold then the result is clear because the dimension of a submanifold is unique, but I dont know how to approach this problem for any kind of compact curves.
Can someone show me if this is false or true? Preferably with the most elementary methods (no measure theory, by example).
geometry analysis curves
asked Mar 19 at 20:49
Masacroso
12.3k41746
add a comment |
up vote
0
down vote
favorite
Im trying to prove that a finite number of compact curves cannot cover any open disk in $Bbb R^2$. I dont know if this is true but it seems.
If a curve is a $1$-dimensional submanifold then the result is clear because the dimension of a submanifold is unique, but I dont know how to approach this problem for any kind of compact curves.
Can someone show me if this is false or true? Preferably with the most elementary methods (no measure theory, by example).
geometry analysis curves
asked Mar 19 at 20:49
Masacroso
12.3k41746
I see; I thought you wanted the curves to be contained within the disk.
– Clayton
Mar 19 at 20:52
1
You may be be interested in space filling curves](en.wikipedia.org/wiki/Space-filling_curve). (But they aren't rectifiable.) It would have been helpful if you'd included the word "rectifiable" in your question and not just the title.
– Rob Arthan
Mar 19 at 20:54
3
A rectifiable curve has Lebesgue measure zero.
– Lord Shark the Unknown
Mar 19 at 20:55
@Lord I know, but where this question appear there is no measure theory involved. This is why I need something more elementary.
– Masacroso
Mar 19 at 21:18
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Im trying to prove that a finite number of compact curves cannot cover any open disk in $Bbb R^2$. I dont know if this is true but it seems.
If a curve is a $1$-dimensional submanifold then the result is clear because the dimension of a submanifold is unique, but I dont know how to approach this problem for any kind of compact curves.
Can someone show me if this is false or true? Preferably with the most elementary methods (no measure theory, by example).
geometry analysis curves
asked Mar 19 at 20:49
Masacroso
12.3k41746
Im trying to prove that a finite number of compact curves cannot cover any open disk in $Bbb R^2$. I dont know if this is true but it seems.
If a curve is a $1$-dimensional submanifold then the result is clear because the dimension of a submanifold is unique, but I dont know how to approach this problem for any kind of compact curves.
Can someone show me if this is false or true? Preferably with the most elementary methods (no measure theory, by example).
geometry analysis curves
geometry analysis curves
asked Mar 19 at 20:49
Masacroso
12.3k41746
asked Mar 19 at 20:49
Masacroso
12.3k41746
edited Mar 20 at 14:12
asked Mar 19 at 20:49
Masacroso
12.3k41746
asked Mar 19 at 20:49
Masacroso
12.3k41746
asked Mar 19 at 20:49
Masacroso
12.3k41746
12.3k41746
I see; I thought you wanted the curves to be contained within the disk.
– Clayton
Mar 19 at 20:52
1
You may be be interested in space filling curves](en.wikipedia.org/wiki/Space-filling_curve). (But they aren't rectifiable.) It would have been helpful if you'd included the word "rectifiable" in your question and not just the title.
– Rob Arthan
Mar 19 at 20:54
3
A rectifiable curve has Lebesgue measure zero.
– Lord Shark the Unknown
Mar 19 at 20:55
@Lord I know, but where this question appear there is no measure theory involved. This is why I need something more elementary.
– Masacroso
Mar 19 at 21:18
add a comment |
I see; I thought you wanted the curves to be contained within the disk.
– Clayton
Mar 19 at 20:52
1
You may be be interested in space filling curves](en.wikipedia.org/wiki/Space-filling_curve). (But they aren't rectifiable.) It would have been helpful if you'd included the word "rectifiable" in your question and not just the title.
– Rob Arthan
Mar 19 at 20:54
3
A rectifiable curve has Lebesgue measure zero.
– Lord Shark the Unknown
Mar 19 at 20:55
@Lord I know, but where this question appear there is no measure theory involved. This is why I need something more elementary.
– Masacroso
Mar 19 at 21:18
I see; I thought you wanted the curves to be contained within the disk.
– Clayton
Mar 19 at 20:52
I see; I thought you wanted the curves to be contained within the disk.
– Clayton
Mar 19 at 20:52
1
1
You may be be interested in space filling curves](en.wikipedia.org/wiki/Space-filling_curve). (But they aren't rectifiable.) It would have been helpful if you'd included the word "rectifiable" in your question and not just the title.
– Rob Arthan
Mar 19 at 20:54
You may be be interested in space filling curves](en.wikipedia.org/wiki/Space-filling_curve). (But they aren't rectifiable.) It would have been helpful if you'd included the word "rectifiable" in your question and not just the title.
– Rob Arthan
Mar 19 at 20:54
3
3
A rectifiable curve has Lebesgue measure zero.
– Lord Shark the Unknown
Mar 19 at 20:55
A rectifiable curve has Lebesgue measure zero.
– Lord Shark the Unknown
Mar 19 at 20:55
@Lord I know, but where this question appear there is no measure theory involved. This is why I need something more elementary.
– Masacroso
Mar 19 at 21:18
@Lord I know, but where this question appear there is no measure theory involved. This is why I need something more elementary.
– Masacroso
Mar 19 at 21:18
add a comment |
1 Answer
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I found a very "simple" answer using elementary geometry.
Observe that for any path that connect $m$ points there is a minimum distance between the connection of any two different points. Hence the length of any path that connect $m$ points is bounded below by $L:=ellcdot (m-1)$, where $ell$ is the minimum distance between all pairs of different points and $m-1$ is the minimum number of segments required to connect all points.
Now consider a grid of $m$ points equally spaced in the plane by a distance $ell$, by example the vertex of a tessellation of the plane by isosceles triangles.
Then define a sequence of finer grids as follows: first consider the vertex of a isosceles triangle of side $ell$, this is the grid $G_0$. Now defines recursively $G_{n+1}$ by adding to $G_n$ the points in the middle of any side of an isosceles triangle defined by the points of $G_n$. Then if the minimum distance between points in $G_0$ is $ell$ then the minimum distance between points in $G_n$ is $ell/2^n$.
The number of points in $G_n$ can be counted using the triangular number $T_n:=binom{n+1}2$ for $n$ vertex in one side of the starting triangle (that is, the number of points in the segment that join any two points in $G_0$). The number of vertex $a_n$ in one side of the starting triangle is defined by the recurrence relation $a_0:=2$ and $a_{n+1}=2 a_n-1$. A brief calculation shows that $a_n=1+2^n$.
Thus the length of any path that connect the grid $G_n$ is bounded below by
$$
L_n=frac{ell}{2^n}cdot (T_{a_n}-1)=frac{ell}{2^n}cdotleft(frac{(a_n+1)a_n}{2}-1right)\=ellcdotleft(frac{(2+2^n)(1+2^n)}{2^{n+1}}-frac1{2^n}right)=ellcdot(2^{n-1}+3/2)
$$
and because $lim_{ntoinfty}L_n=infty$ then we can conclude there is no possible that a rectifiable curve cover an open disk.
answered Mar 19 at 23:09
Masacroso
12.3k41746
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I found a very "simple" answer using elementary geometry.
Observe that for any path that connect $m$ points there is a minimum distance between the connection of any two different points. Hence the length of any path that connect $m$ points is bounded below by $L:=ellcdot (m-1)$, where $ell$ is the minimum distance between all pairs of different points and $m-1$ is the minimum number of segments required to connect all points.
Now consider a grid of $m$ points equally spaced in the plane by a distance $ell$, by example the vertex of a tessellation of the plane by isosceles triangles.
Then define a sequence of finer grids as follows: first consider the vertex of a isosceles triangle of side $ell$, this is the grid $G_0$. Now defines recursively $G_{n+1}$ by adding to $G_n$ the points in the middle of any side of an isosceles triangle defined by the points of $G_n$. Then if the minimum distance between points in $G_0$ is $ell$ then the minimum distance between points in $G_n$ is $ell/2^n$.
The number of points in $G_n$ can be counted using the triangular number $T_n:=binom{n+1}2$ for $n$ vertex in one side of the starting triangle (that is, the number of points in the segment that join any two points in $G_0$). The number of vertex $a_n$ in one side of the starting triangle is defined by the recurrence relation $a_0:=2$ and $a_{n+1}=2 a_n-1$. A brief calculation shows that $a_n=1+2^n$.
Thus the length of any path that connect the grid $G_n$ is bounded below by
$$
L_n=frac{ell}{2^n}cdot (T_{a_n}-1)=frac{ell}{2^n}cdotleft(frac{(a_n+1)a_n}{2}-1right)\=ellcdotleft(frac{(2+2^n)(1+2^n)}{2^{n+1}}-frac1{2^n}right)=ellcdot(2^{n-1}+3/2)
$$
and because $lim_{ntoinfty}L_n=infty$ then we can conclude there is no possible that a rectifiable curve cover an open disk.
answered Mar 19 at 23:09
Masacroso
12.3k41746
add a comment |
up vote
0
down vote
accepted
I found a very "simple" answer using elementary geometry.
Observe that for any path that connect $m$ points there is a minimum distance between the connection of any two different points. Hence the length of any path that connect $m$ points is bounded below by $L:=ellcdot (m-1)$, where $ell$ is the minimum distance between all pairs of different points and $m-1$ is the minimum number of segments required to connect all points.
Now consider a grid of $m$ points equally spaced in the plane by a distance $ell$, by example the vertex of a tessellation of the plane by isosceles triangles.
Then define a sequence of finer grids as follows: first consider the vertex of a isosceles triangle of side $ell$, this is the grid $G_0$. Now defines recursively $G_{n+1}$ by adding to $G_n$ the points in the middle of any side of an isosceles triangle defined by the points of $G_n$. Then if the minimum distance between points in $G_0$ is $ell$ then the minimum distance between points in $G_n$ is $ell/2^n$.
The number of points in $G_n$ can be counted using the triangular number $T_n:=binom{n+1}2$ for $n$ vertex in one side of the starting triangle (that is, the number of points in the segment that join any two points in $G_0$). The number of vertex $a_n$ in one side of the starting triangle is defined by the recurrence relation $a_0:=2$ and $a_{n+1}=2 a_n-1$. A brief calculation shows that $a_n=1+2^n$.
Thus the length of any path that connect the grid $G_n$ is bounded below by
$$
L_n=frac{ell}{2^n}cdot (T_{a_n}-1)=frac{ell}{2^n}cdotleft(frac{(a_n+1)a_n}{2}-1right)\=ellcdotleft(frac{(2+2^n)(1+2^n)}{2^{n+1}}-frac1{2^n}right)=ellcdot(2^{n-1}+3/2)
$$
and because $lim_{ntoinfty}L_n=infty$ then we can conclude there is no possible that a rectifiable curve cover an open disk.
answered Mar 19 at 23:09
Masacroso
12.3k41746
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I found a very "simple" answer using elementary geometry.
Observe that for any path that connect $m$ points there is a minimum distance between the connection of any two different points. Hence the length of any path that connect $m$ points is bounded below by $L:=ellcdot (m-1)$, where $ell$ is the minimum distance between all pairs of different points and $m-1$ is the minimum number of segments required to connect all points.
Now consider a grid of $m$ points equally spaced in the plane by a distance $ell$, by example the vertex of a tessellation of the plane by isosceles triangles.
Then define a sequence of finer grids as follows: first consider the vertex of a isosceles triangle of side $ell$, this is the grid $G_0$. Now defines recursively $G_{n+1}$ by adding to $G_n$ the points in the middle of any side of an isosceles triangle defined by the points of $G_n$. Then if the minimum distance between points in $G_0$ is $ell$ then the minimum distance between points in $G_n$ is $ell/2^n$.
The number of points in $G_n$ can be counted using the triangular number $T_n:=binom{n+1}2$ for $n$ vertex in one side of the starting triangle (that is, the number of points in the segment that join any two points in $G_0$). The number of vertex $a_n$ in one side of the starting triangle is defined by the recurrence relation $a_0:=2$ and $a_{n+1}=2 a_n-1$. A brief calculation shows that $a_n=1+2^n$.
Thus the length of any path that connect the grid $G_n$ is bounded below by
$$
L_n=frac{ell}{2^n}cdot (T_{a_n}-1)=frac{ell}{2^n}cdotleft(frac{(a_n+1)a_n}{2}-1right)\=ellcdotleft(frac{(2+2^n)(1+2^n)}{2^{n+1}}-frac1{2^n}right)=ellcdot(2^{n-1}+3/2)
$$
and because $lim_{ntoinfty}L_n=infty$ then we can conclude there is no possible that a rectifiable curve cover an open disk.
answered Mar 19 at 23:09
Masacroso
12.3k41746
I found a very "simple" answer using elementary geometry.
Observe that for any path that connect $m$ points there is a minimum distance between the connection of any two different points. Hence the length of any path that connect $m$ points is bounded below by $L:=ellcdot (m-1)$, where $ell$ is the minimum distance between all pairs of different points and $m-1$ is the minimum number of segments required to connect all points.
Now consider a grid of $m$ points equally spaced in the plane by a distance $ell$, by example the vertex of a tessellation of the plane by isosceles triangles.
Then define a sequence of finer grids as follows: first consider the vertex of a isosceles triangle of side $ell$, this is the grid $G_0$. Now defines recursively $G_{n+1}$ by adding to $G_n$ the points in the middle of any side of an isosceles triangle defined by the points of $G_n$. Then if the minimum distance between points in $G_0$ is $ell$ then the minimum distance between points in $G_n$ is $ell/2^n$.
The number of points in $G_n$ can be counted using the triangular number $T_n:=binom{n+1}2$ for $n$ vertex in one side of the starting triangle (that is, the number of points in the segment that join any two points in $G_0$). The number of vertex $a_n$ in one side of the starting triangle is defined by the recurrence relation $a_0:=2$ and $a_{n+1}=2 a_n-1$. A brief calculation shows that $a_n=1+2^n$.
Thus the length of any path that connect the grid $G_n$ is bounded below by
$$
L_n=frac{ell}{2^n}cdot (T_{a_n}-1)=frac{ell}{2^n}cdotleft(frac{(a_n+1)a_n}{2}-1right)\=ellcdotleft(frac{(2+2^n)(1+2^n)}{2^{n+1}}-frac1{2^n}right)=ellcdot(2^{n-1}+3/2)
$$
and because $lim_{ntoinfty}L_n=infty$ then we can conclude there is no possible that a rectifiable curve cover an open disk.
answered Mar 19 at 23:09
Masacroso
12.3k41746
edited 2 days ago
answered Mar 19 at 23:09
Masacroso
12.3k41746
answered Mar 19 at 23:09
Masacroso
12.3k41746
answered Mar 19 at 23:09
Masacroso
12.3k41746
12.3k41746
add a comment |
add a comment |
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I see; I thought you wanted the curves to be contained within the disk.
– Clayton
Mar 19 at 20:52
1
You may be be interested in space filling curves](en.wikipedia.org/wiki/Space-filling_curve). (But they aren't rectifiable.) It would have been helpful if you'd included the word "rectifiable" in your question and not just the title.
– Rob Arthan
Mar 19 at 20:54
3
A rectifiable curve has Lebesgue measure zero.
– Lord Shark the Unknown
Mar 19 at 20:55
@Lord I know, but where this question appear there is no measure theory involved. This is why I need something more elementary.
– Masacroso
Mar 19 at 21:18