Mixing
Koch curve from Cantor sets (paradox)
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The Koch curve is normally constructed by taking a line segment, replacing the middle third with two copies of itself forming legs in an equilateral triangle, and then repeating this recursively for every subsegment. See image below. At every step, the length of the curve is multiplied by $4/3$ so the final length is infinite.
Notice that every line segment undergoes the construction of the Cantor set:
Therefore one could consider replacing every line segment with Cantor sets already from the beginning. Thus, start with the Cantor set, take two smaller copies of the Cantor set and make a $wedge$ in the middle opening, and then repeat recursively. In this case the final length will be zero since the Cantor set has length zero.
Question: How is this paradox resolved? Will the limit set of my construction be different than the ordinary Koch curve? If so, what points are missing?
fractals paradoxes cantor-set
asked 2 days ago
md2perpe
7,24611027
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show 5 more comments
up vote
4
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The Koch curve is normally constructed by taking a line segment, replacing the middle third with two copies of itself forming legs in an equilateral triangle, and then repeating this recursively for every subsegment. See image below. At every step, the length of the curve is multiplied by $4/3$ so the final length is infinite.
Notice that every line segment undergoes the construction of the Cantor set:
Therefore one could consider replacing every line segment with Cantor sets already from the beginning. Thus, start with the Cantor set, take two smaller copies of the Cantor set and make a $wedge$ in the middle opening, and then repeat recursively. In this case the final length will be zero since the Cantor set has length zero.
Question: How is this paradox resolved? Will the limit set of my construction be different than the ordinary Koch curve? If so, what points are missing?
fractals paradoxes cantor-set
asked 2 days ago
md2perpe
7,24611027
Aren't there a countably infinite number of Cantor sets in the Koch snowflake if you look at it this way?
– Michael Seifert
2 days ago
@MichaelSeifert. Yes, but a countable union of sets with zero length (Lebesgue measure) also has zero length.
– md2perpe
2 days ago
I'm not entirely sure what the paradox is here. You're just taking something with $0$ length and "adding" it together countably infinite times, right? I'm not sure why anyone would expect it to have positive length, in that case. Am I missing something?
– Clayton
2 days ago
1
You've not argued that the final Koch snowflake is in fact the union of countably many Cantor sets (and indeed it isn't), so there's no contradiction.
– Noah Schweber
2 days ago
1
The MSE question 12906 "The staircase paradox, or why $pine 4$" seems similar. Or else MSE question 1525410 "Length of diagonal compared to the limit of lengths of stair-shaped curves converging to it".
– Somos
2 days ago
|
show 5 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The Koch curve is normally constructed by taking a line segment, replacing the middle third with two copies of itself forming legs in an equilateral triangle, and then repeating this recursively for every subsegment. See image below. At every step, the length of the curve is multiplied by $4/3$ so the final length is infinite.
Notice that every line segment undergoes the construction of the Cantor set:
Therefore one could consider replacing every line segment with Cantor sets already from the beginning. Thus, start with the Cantor set, take two smaller copies of the Cantor set and make a $wedge$ in the middle opening, and then repeat recursively. In this case the final length will be zero since the Cantor set has length zero.
Question: How is this paradox resolved? Will the limit set of my construction be different than the ordinary Koch curve? If so, what points are missing?
fractals paradoxes cantor-set
asked 2 days ago
md2perpe
7,24611027
The Koch curve is normally constructed by taking a line segment, replacing the middle third with two copies of itself forming legs in an equilateral triangle, and then repeating this recursively for every subsegment. See image below. At every step, the length of the curve is multiplied by $4/3$ so the final length is infinite.
Notice that every line segment undergoes the construction of the Cantor set:
Therefore one could consider replacing every line segment with Cantor sets already from the beginning. Thus, start with the Cantor set, take two smaller copies of the Cantor set and make a $wedge$ in the middle opening, and then repeat recursively. In this case the final length will be zero since the Cantor set has length zero.
Question: How is this paradox resolved? Will the limit set of my construction be different than the ordinary Koch curve? If so, what points are missing?
fractals paradoxes cantor-set
fractals paradoxes cantor-set
asked 2 days ago
md2perpe
7,24611027
asked 2 days ago
md2perpe
7,24611027
asked 2 days ago
md2perpe
7,24611027
asked 2 days ago
md2perpe
7,24611027
asked 2 days ago
md2perpe
7,24611027
7,24611027
Aren't there a countably infinite number of Cantor sets in the Koch snowflake if you look at it this way?
– Michael Seifert
2 days ago
@MichaelSeifert. Yes, but a countable union of sets with zero length (Lebesgue measure) also has zero length.
– md2perpe
2 days ago
I'm not entirely sure what the paradox is here. You're just taking something with $0$ length and "adding" it together countably infinite times, right? I'm not sure why anyone would expect it to have positive length, in that case. Am I missing something?
– Clayton
2 days ago
1
You've not argued that the final Koch snowflake is in fact the union of countably many Cantor sets (and indeed it isn't), so there's no contradiction.
– Noah Schweber
2 days ago
1
The MSE question 12906 "The staircase paradox, or why $pine 4$" seems similar. Or else MSE question 1525410 "Length of diagonal compared to the limit of lengths of stair-shaped curves converging to it".
– Somos
2 days ago
|
show 5 more comments
Aren't there a countably infinite number of Cantor sets in the Koch snowflake if you look at it this way?
– Michael Seifert
2 days ago
@MichaelSeifert. Yes, but a countable union of sets with zero length (Lebesgue measure) also has zero length.
– md2perpe
2 days ago
I'm not entirely sure what the paradox is here. You're just taking something with $0$ length and "adding" it together countably infinite times, right? I'm not sure why anyone would expect it to have positive length, in that case. Am I missing something?
– Clayton
2 days ago
1
You've not argued that the final Koch snowflake is in fact the union of countably many Cantor sets (and indeed it isn't), so there's no contradiction.
– Noah Schweber
2 days ago
1
The MSE question 12906 "The staircase paradox, or why $pine 4$" seems similar. Or else MSE question 1525410 "Length of diagonal compared to the limit of lengths of stair-shaped curves converging to it".
– Somos
2 days ago
Aren't there a countably infinite number of Cantor sets in the Koch snowflake if you look at it this way?
– Michael Seifert
2 days ago
Aren't there a countably infinite number of Cantor sets in the Koch snowflake if you look at it this way?
– Michael Seifert
2 days ago
@MichaelSeifert. Yes, but a countable union of sets with zero length (Lebesgue measure) also has zero length.
– md2perpe
2 days ago
@MichaelSeifert. Yes, but a countable union of sets with zero length (Lebesgue measure) also has zero length.
– md2perpe
2 days ago
I'm not entirely sure what the paradox is here. You're just taking something with $0$ length and "adding" it together countably infinite times, right? I'm not sure why anyone would expect it to have positive length, in that case. Am I missing something?
– Clayton
2 days ago
I'm not entirely sure what the paradox is here. You're just taking something with $0$ length and "adding" it together countably infinite times, right? I'm not sure why anyone would expect it to have positive length, in that case. Am I missing something?
– Clayton
2 days ago
1
1
You've not argued that the final Koch snowflake is in fact the union of countably many Cantor sets (and indeed it isn't), so there's no contradiction.
– Noah Schweber
2 days ago
You've not argued that the final Koch snowflake is in fact the union of countably many Cantor sets (and indeed it isn't), so there's no contradiction.
– Noah Schweber
2 days ago
1
1
The MSE question 12906 "The staircase paradox, or why $pine 4$" seems similar. Or else MSE question 1525410 "Length of diagonal compared to the limit of lengths of stair-shaped curves converging to it".
– Somos
2 days ago
The MSE question 12906 "The staircase paradox, or why $pine 4$" seems similar. Or else MSE question 1525410 "Length of diagonal compared to the limit of lengths of stair-shaped curves converging to it".
– Somos
2 days ago
|
show 5 more comments
2 Answers
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up vote
3
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Convergence in fractal geometry is typically defined in terms of the Hausdorff metric. Roughly, two sets are "close" with respect to the Hausdorff metric, if every point in one is close to some point of the other.
Your collection of Cantor sets is indeed dense in the Koch curve with respect to the Hausdorff metric. The Hausdorff metric, however, doesn't respect length. That is, two sets can be close in the Hausdorff metric, yet their lengths can be very far apart. You've found one example illustrating this but there are others.
For example, if
$$
Q_n = {k/n:0leq k leq n},
$$
then the Hausdorff distance between $Q_n$ and the unit interval is less than $1/n$. $Q_n$ is finite, yet the sequence of $Q_n$s converges to a set of positive length. Similarly, you could strengthen your example by using the set of endpoints of the intervals that approximate the Koch curve.
Here is a strategy to find points in the Koch curve that do not lie on any of your Cantor sets. First, note that the Koch curve is invariant set of the iterated function system:
$$begin{align}
T_1(x,y) &= left(frac{x}{3},frac{y}{3}right) \
T_2(x,y) &= left(frac{1}{6} left(x-sqrt{3}
y+2right),frac{1}{6} left(sqrt{3}
x+yright)right) \
T_3(x,y) &= left(frac{1}{6} left(x+sqrt{3}
y+3right),frac{1}{6} left(-sqrt{3}
x+y+sqrt{3}right)right) \
T_4(x,y) &= left( frac{x}{3},frac{y+2}{3} right).
end{align}$$
These functions map the Koch curve onto the four sub-parts below
Now, any point On the Koch curve can be realized as the limit of a sequence
$$begin{align}
& T_{i_1}(0,0) \
& T_{i_1} circ T_{i_2}(0,0)\
& vdots \
& T_{i_1} circ T_{i_2} circ cdots T_{i_n}(0,0) \
& vdots
end{align}$$
where $(i_1,i_2,i_3,ldots)$ is a sequence in ${1,2,3,4}$. The point lies on one of your Cantor sets precisely when the sequence contains only finitely many 2s and 3s, so that it ends in a string of only 1s and 4s.
If, for example, the sequence has only 1s and 4s, and no 2s or 3s, then we get a point in Cantor's ternary set lying on the unit interval. If the sequence start with a 2 and then contains only 1s and 4s, we generate a point in the red Cantor set shown below; this is exactly the image of the ternary Cantor set under the function $T_2$. If the sequence starts 3, then 2, then contains only 1s and 4s, we generate a point in the blue Cantor set below; this is exactly the the image of the ternary Cantor set under the function $T_3 circ T_2$.
Finally, if we have any other sequence of 1s, 2s, 3s, and 4s, then we generate some other point on the Koch curve that is not in any of your Cantor sets. There are uncountably many such points. I suppose the simplest one to find explicitly corresponds to the sequence containing only 2s, which is exactly the fixed point of $T_2$. To find it, we need only solve
$$T_2(x,y) = (x,y),$$
which yields $(5/14,sqrt{3}/14)$. That point is shown in red in the sequence of approximations below.
If we zoom into the last picture on the red point so that it's centered in a square of side length 0.04, we get the following:
Thus, the edges keep jutting out closer to the point but never actually hit it. It's in the limit but not on any of the edges.
answered 2 days ago
Mark McClure
23.1k34170
Could you give me some points that will be in the Koch curve but not in my construction?
– md2perpe
yesterday
@md2perpe Sure - the point $(5/14,sqrt{3}/14)$ (along with many other points) is in the Koch curve but not in your construction, as explained in my edit.
– Mark McClure
yesterday
So that point is neither in any step of the construction of the Koch curve nor in any step of my construction. Therefore it's not in the $limsup$ limit of neither sequence. But it is in the Hausdorff limit of both?
– md2perpe
yesterday
That is correct. If you were thinking in terms of the limsup, that might explain your confusion.
– Mark McClure
yesterday
add a comment |
up vote
1
down vote
Not every point in the Koch curve is in one of your Cantor sets! Indeed, this follows immediately from the Baire category theorem: each of your Cantor sets is a closed subset of the Koch curve with empty interior, and the union of countably many such sets will still have empty interior and thus cannot be the whole Koch curve.
The proof of the Baire category theorem also gives a recipe for finding points that are not in any of the Cantor sets. Let's say the $n$th Cantor set is $C_n$ and the Koch curve is $K$. Start with a point $x_0in Ksetminus C_0$; then there is some closed ball $B_0$ around $x_0$ which does not intersect $C_0$. Then pick a point $x_1in B_0setminus C_1$. There is again some closed ball $B_1$ around $x_1$ which does not intersect $C_1$. Then pick a point $x_2in B_0cap B_1setminus C_2$, and so on.
This procedure gives a sequence of points $(x_n)$ in $K$ and balls $B_n$ such that $x_min B_n$ for all $mgeq n$ and $B_ncap C_n=emptyset$. Since $K$ is compact, this sequence $(x_n)$ must accumulate at some point $xin K$ (in fact, we can choose the balls $B_n$ to have radius converging to $0$ and then $(x_n)$ will be Cauchy and thus will converge). Then $xin B_n$ for all $n$, so $x$ cannot be in $C_n$ for any $n$.
Another way to think of it is that the Koch curve is parametrized by a continuous function $f:[0,1]to mathbb{R}^2$ which is the limit of functions $f_n:[0,1]tomathbb{R}^2$ parametrizing the approximations to the Koch curve. Each of your Cantor sets corresponds to the image of some Cantor set in $[0,1]$ under $f$, and that Cantor set in $[0,1]$ will have Lebesgue measure $0$ if you've chosen your parametrization reasonably. So, the union of all those Cantor sets in $[0,1]$ is still a null set (and a set of first category), so most points in $[0,1]$ aren't in any of those Cantor sets.
Note that you can also think of a point $xin[0,1]$ that is not in any of those Cantor sets as a point such that the sequence $(f_n(x))$ never stabilizes but instead keeps moving infinitely often as you build the successive approximations. The value of $f(x)$ will then be the limit of this sequence, a point of the Koch curve that was not in any of the approximations but instead is a limit of points in the approximations.
answered 2 days ago
Eric Wofsey
175k12202326
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2 Answers
2
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Convergence in fractal geometry is typically defined in terms of the Hausdorff metric. Roughly, two sets are "close" with respect to the Hausdorff metric, if every point in one is close to some point of the other.
Your collection of Cantor sets is indeed dense in the Koch curve with respect to the Hausdorff metric. The Hausdorff metric, however, doesn't respect length. That is, two sets can be close in the Hausdorff metric, yet their lengths can be very far apart. You've found one example illustrating this but there are others.
For example, if
$$
Q_n = {k/n:0leq k leq n},
$$
then the Hausdorff distance between $Q_n$ and the unit interval is less than $1/n$. $Q_n$ is finite, yet the sequence of $Q_n$s converges to a set of positive length. Similarly, you could strengthen your example by using the set of endpoints of the intervals that approximate the Koch curve.
Here is a strategy to find points in the Koch curve that do not lie on any of your Cantor sets. First, note that the Koch curve is invariant set of the iterated function system:
$$begin{align}
T_1(x,y) &= left(frac{x}{3},frac{y}{3}right) \
T_2(x,y) &= left(frac{1}{6} left(x-sqrt{3}
y+2right),frac{1}{6} left(sqrt{3}
x+yright)right) \
T_3(x,y) &= left(frac{1}{6} left(x+sqrt{3}
y+3right),frac{1}{6} left(-sqrt{3}
x+y+sqrt{3}right)right) \
T_4(x,y) &= left( frac{x}{3},frac{y+2}{3} right).
end{align}$$
These functions map the Koch curve onto the four sub-parts below
Now, any point On the Koch curve can be realized as the limit of a sequence
$$begin{align}
& T_{i_1}(0,0) \
& T_{i_1} circ T_{i_2}(0,0)\
& vdots \
& T_{i_1} circ T_{i_2} circ cdots T_{i_n}(0,0) \
& vdots
end{align}$$
where $(i_1,i_2,i_3,ldots)$ is a sequence in ${1,2,3,4}$. The point lies on one of your Cantor sets precisely when the sequence contains only finitely many 2s and 3s, so that it ends in a string of only 1s and 4s.
If, for example, the sequence has only 1s and 4s, and no 2s or 3s, then we get a point in Cantor's ternary set lying on the unit interval. If the sequence start with a 2 and then contains only 1s and 4s, we generate a point in the red Cantor set shown below; this is exactly the image of the ternary Cantor set under the function $T_2$. If the sequence starts 3, then 2, then contains only 1s and 4s, we generate a point in the blue Cantor set below; this is exactly the the image of the ternary Cantor set under the function $T_3 circ T_2$.
Finally, if we have any other sequence of 1s, 2s, 3s, and 4s, then we generate some other point on the Koch curve that is not in any of your Cantor sets. There are uncountably many such points. I suppose the simplest one to find explicitly corresponds to the sequence containing only 2s, which is exactly the fixed point of $T_2$. To find it, we need only solve
$$T_2(x,y) = (x,y),$$
which yields $(5/14,sqrt{3}/14)$. That point is shown in red in the sequence of approximations below.
If we zoom into the last picture on the red point so that it's centered in a square of side length 0.04, we get the following:
Thus, the edges keep jutting out closer to the point but never actually hit it. It's in the limit but not on any of the edges.
answered 2 days ago
Mark McClure
23.1k34170
Could you give me some points that will be in the Koch curve but not in my construction?
– md2perpe
yesterday
@md2perpe Sure - the point $(5/14,sqrt{3}/14)$ (along with many other points) is in the Koch curve but not in your construction, as explained in my edit.
– Mark McClure
yesterday
So that point is neither in any step of the construction of the Koch curve nor in any step of my construction. Therefore it's not in the $limsup$ limit of neither sequence. But it is in the Hausdorff limit of both?
– md2perpe
yesterday
That is correct. If you were thinking in terms of the limsup, that might explain your confusion.
– Mark McClure
yesterday
add a comment |
up vote
3
down vote
Convergence in fractal geometry is typically defined in terms of the Hausdorff metric. Roughly, two sets are "close" with respect to the Hausdorff metric, if every point in one is close to some point of the other.
Your collection of Cantor sets is indeed dense in the Koch curve with respect to the Hausdorff metric. The Hausdorff metric, however, doesn't respect length. That is, two sets can be close in the Hausdorff metric, yet their lengths can be very far apart. You've found one example illustrating this but there are others.
For example, if
$$
Q_n = {k/n:0leq k leq n},
$$
then the Hausdorff distance between $Q_n$ and the unit interval is less than $1/n$. $Q_n$ is finite, yet the sequence of $Q_n$s converges to a set of positive length. Similarly, you could strengthen your example by using the set of endpoints of the intervals that approximate the Koch curve.
Here is a strategy to find points in the Koch curve that do not lie on any of your Cantor sets. First, note that the Koch curve is invariant set of the iterated function system:
$$begin{align}
T_1(x,y) &= left(frac{x}{3},frac{y}{3}right) \
T_2(x,y) &= left(frac{1}{6} left(x-sqrt{3}
y+2right),frac{1}{6} left(sqrt{3}
x+yright)right) \
T_3(x,y) &= left(frac{1}{6} left(x+sqrt{3}
y+3right),frac{1}{6} left(-sqrt{3}
x+y+sqrt{3}right)right) \
T_4(x,y) &= left( frac{x}{3},frac{y+2}{3} right).
end{align}$$
These functions map the Koch curve onto the four sub-parts below
Now, any point On the Koch curve can be realized as the limit of a sequence
$$begin{align}
& T_{i_1}(0,0) \
& T_{i_1} circ T_{i_2}(0,0)\
& vdots \
& T_{i_1} circ T_{i_2} circ cdots T_{i_n}(0,0) \
& vdots
end{align}$$
where $(i_1,i_2,i_3,ldots)$ is a sequence in ${1,2,3,4}$. The point lies on one of your Cantor sets precisely when the sequence contains only finitely many 2s and 3s, so that it ends in a string of only 1s and 4s.
If, for example, the sequence has only 1s and 4s, and no 2s or 3s, then we get a point in Cantor's ternary set lying on the unit interval. If the sequence start with a 2 and then contains only 1s and 4s, we generate a point in the red Cantor set shown below; this is exactly the image of the ternary Cantor set under the function $T_2$. If the sequence starts 3, then 2, then contains only 1s and 4s, we generate a point in the blue Cantor set below; this is exactly the the image of the ternary Cantor set under the function $T_3 circ T_2$.
Finally, if we have any other sequence of 1s, 2s, 3s, and 4s, then we generate some other point on the Koch curve that is not in any of your Cantor sets. There are uncountably many such points. I suppose the simplest one to find explicitly corresponds to the sequence containing only 2s, which is exactly the fixed point of $T_2$. To find it, we need only solve
$$T_2(x,y) = (x,y),$$
which yields $(5/14,sqrt{3}/14)$. That point is shown in red in the sequence of approximations below.
If we zoom into the last picture on the red point so that it's centered in a square of side length 0.04, we get the following:
Thus, the edges keep jutting out closer to the point but never actually hit it. It's in the limit but not on any of the edges.
answered 2 days ago
Mark McClure
23.1k34170
Could you give me some points that will be in the Koch curve but not in my construction?
– md2perpe
yesterday
@md2perpe Sure - the point $(5/14,sqrt{3}/14)$ (along with many other points) is in the Koch curve but not in your construction, as explained in my edit.
– Mark McClure
yesterday
So that point is neither in any step of the construction of the Koch curve nor in any step of my construction. Therefore it's not in the $limsup$ limit of neither sequence. But it is in the Hausdorff limit of both?
– md2perpe
yesterday
That is correct. If you were thinking in terms of the limsup, that might explain your confusion.
– Mark McClure
yesterday
add a comment |
up vote
3
down vote
up vote
3
down vote
Convergence in fractal geometry is typically defined in terms of the Hausdorff metric. Roughly, two sets are "close" with respect to the Hausdorff metric, if every point in one is close to some point of the other.
Your collection of Cantor sets is indeed dense in the Koch curve with respect to the Hausdorff metric. The Hausdorff metric, however, doesn't respect length. That is, two sets can be close in the Hausdorff metric, yet their lengths can be very far apart. You've found one example illustrating this but there are others.
For example, if
$$
Q_n = {k/n:0leq k leq n},
$$
then the Hausdorff distance between $Q_n$ and the unit interval is less than $1/n$. $Q_n$ is finite, yet the sequence of $Q_n$s converges to a set of positive length. Similarly, you could strengthen your example by using the set of endpoints of the intervals that approximate the Koch curve.
Here is a strategy to find points in the Koch curve that do not lie on any of your Cantor sets. First, note that the Koch curve is invariant set of the iterated function system:
$$begin{align}
T_1(x,y) &= left(frac{x}{3},frac{y}{3}right) \
T_2(x,y) &= left(frac{1}{6} left(x-sqrt{3}
y+2right),frac{1}{6} left(sqrt{3}
x+yright)right) \
T_3(x,y) &= left(frac{1}{6} left(x+sqrt{3}
y+3right),frac{1}{6} left(-sqrt{3}
x+y+sqrt{3}right)right) \
T_4(x,y) &= left( frac{x}{3},frac{y+2}{3} right).
end{align}$$
These functions map the Koch curve onto the four sub-parts below
Now, any point On the Koch curve can be realized as the limit of a sequence
$$begin{align}
& T_{i_1}(0,0) \
& T_{i_1} circ T_{i_2}(0,0)\
& vdots \
& T_{i_1} circ T_{i_2} circ cdots T_{i_n}(0,0) \
& vdots
end{align}$$
where $(i_1,i_2,i_3,ldots)$ is a sequence in ${1,2,3,4}$. The point lies on one of your Cantor sets precisely when the sequence contains only finitely many 2s and 3s, so that it ends in a string of only 1s and 4s.
If, for example, the sequence has only 1s and 4s, and no 2s or 3s, then we get a point in Cantor's ternary set lying on the unit interval. If the sequence start with a 2 and then contains only 1s and 4s, we generate a point in the red Cantor set shown below; this is exactly the image of the ternary Cantor set under the function $T_2$. If the sequence starts 3, then 2, then contains only 1s and 4s, we generate a point in the blue Cantor set below; this is exactly the the image of the ternary Cantor set under the function $T_3 circ T_2$.
Finally, if we have any other sequence of 1s, 2s, 3s, and 4s, then we generate some other point on the Koch curve that is not in any of your Cantor sets. There are uncountably many such points. I suppose the simplest one to find explicitly corresponds to the sequence containing only 2s, which is exactly the fixed point of $T_2$. To find it, we need only solve
$$T_2(x,y) = (x,y),$$
which yields $(5/14,sqrt{3}/14)$. That point is shown in red in the sequence of approximations below.
If we zoom into the last picture on the red point so that it's centered in a square of side length 0.04, we get the following:
Thus, the edges keep jutting out closer to the point but never actually hit it. It's in the limit but not on any of the edges.
answered 2 days ago
Mark McClure
23.1k34170
Convergence in fractal geometry is typically defined in terms of the Hausdorff metric. Roughly, two sets are "close" with respect to the Hausdorff metric, if every point in one is close to some point of the other.
Your collection of Cantor sets is indeed dense in the Koch curve with respect to the Hausdorff metric. The Hausdorff metric, however, doesn't respect length. That is, two sets can be close in the Hausdorff metric, yet their lengths can be very far apart. You've found one example illustrating this but there are others.
For example, if
$$
Q_n = {k/n:0leq k leq n},
$$
then the Hausdorff distance between $Q_n$ and the unit interval is less than $1/n$. $Q_n$ is finite, yet the sequence of $Q_n$s converges to a set of positive length. Similarly, you could strengthen your example by using the set of endpoints of the intervals that approximate the Koch curve.
Here is a strategy to find points in the Koch curve that do not lie on any of your Cantor sets. First, note that the Koch curve is invariant set of the iterated function system:
$$begin{align}
T_1(x,y) &= left(frac{x}{3},frac{y}{3}right) \
T_2(x,y) &= left(frac{1}{6} left(x-sqrt{3}
y+2right),frac{1}{6} left(sqrt{3}
x+yright)right) \
T_3(x,y) &= left(frac{1}{6} left(x+sqrt{3}
y+3right),frac{1}{6} left(-sqrt{3}
x+y+sqrt{3}right)right) \
T_4(x,y) &= left( frac{x}{3},frac{y+2}{3} right).
end{align}$$
These functions map the Koch curve onto the four sub-parts below
Now, any point On the Koch curve can be realized as the limit of a sequence
$$begin{align}
& T_{i_1}(0,0) \
& T_{i_1} circ T_{i_2}(0,0)\
& vdots \
& T_{i_1} circ T_{i_2} circ cdots T_{i_n}(0,0) \
& vdots
end{align}$$
where $(i_1,i_2,i_3,ldots)$ is a sequence in ${1,2,3,4}$. The point lies on one of your Cantor sets precisely when the sequence contains only finitely many 2s and 3s, so that it ends in a string of only 1s and 4s.
If, for example, the sequence has only 1s and 4s, and no 2s or 3s, then we get a point in Cantor's ternary set lying on the unit interval. If the sequence start with a 2 and then contains only 1s and 4s, we generate a point in the red Cantor set shown below; this is exactly the image of the ternary Cantor set under the function $T_2$. If the sequence starts 3, then 2, then contains only 1s and 4s, we generate a point in the blue Cantor set below; this is exactly the the image of the ternary Cantor set under the function $T_3 circ T_2$.
Finally, if we have any other sequence of 1s, 2s, 3s, and 4s, then we generate some other point on the Koch curve that is not in any of your Cantor sets. There are uncountably many such points. I suppose the simplest one to find explicitly corresponds to the sequence containing only 2s, which is exactly the fixed point of $T_2$. To find it, we need only solve
$$T_2(x,y) = (x,y),$$
which yields $(5/14,sqrt{3}/14)$. That point is shown in red in the sequence of approximations below.
If we zoom into the last picture on the red point so that it's centered in a square of side length 0.04, we get the following:
Thus, the edges keep jutting out closer to the point but never actually hit it. It's in the limit but not on any of the edges.
answered 2 days ago
Mark McClure
23.1k34170
edited yesterday
answered 2 days ago
Mark McClure
23.1k34170
answered 2 days ago
Mark McClure
23.1k34170
answered 2 days ago
Mark McClure
23.1k34170
23.1k34170
Could you give me some points that will be in the Koch curve but not in my construction?
– md2perpe
yesterday
@md2perpe Sure - the point $(5/14,sqrt{3}/14)$ (along with many other points) is in the Koch curve but not in your construction, as explained in my edit.
– Mark McClure
yesterday
So that point is neither in any step of the construction of the Koch curve nor in any step of my construction. Therefore it's not in the $limsup$ limit of neither sequence. But it is in the Hausdorff limit of both?
– md2perpe
yesterday
That is correct. If you were thinking in terms of the limsup, that might explain your confusion.
– Mark McClure
yesterday
add a comment |
Could you give me some points that will be in the Koch curve but not in my construction?
– md2perpe
yesterday
@md2perpe Sure - the point $(5/14,sqrt{3}/14)$ (along with many other points) is in the Koch curve but not in your construction, as explained in my edit.
– Mark McClure
yesterday
So that point is neither in any step of the construction of the Koch curve nor in any step of my construction. Therefore it's not in the $limsup$ limit of neither sequence. But it is in the Hausdorff limit of both?
– md2perpe
yesterday
That is correct. If you were thinking in terms of the limsup, that might explain your confusion.
– Mark McClure
yesterday
Could you give me some points that will be in the Koch curve but not in my construction?
– md2perpe
yesterday
Could you give me some points that will be in the Koch curve but not in my construction?
– md2perpe
yesterday
@md2perpe Sure - the point $(5/14,sqrt{3}/14)$ (along with many other points) is in the Koch curve but not in your construction, as explained in my edit.
– Mark McClure
yesterday
@md2perpe Sure - the point $(5/14,sqrt{3}/14)$ (along with many other points) is in the Koch curve but not in your construction, as explained in my edit.
– Mark McClure
yesterday
So that point is neither in any step of the construction of the Koch curve nor in any step of my construction. Therefore it's not in the $limsup$ limit of neither sequence. But it is in the Hausdorff limit of both?
– md2perpe
yesterday
So that point is neither in any step of the construction of the Koch curve nor in any step of my construction. Therefore it's not in the $limsup$ limit of neither sequence. But it is in the Hausdorff limit of both?
– md2perpe
yesterday
That is correct. If you were thinking in terms of the limsup, that might explain your confusion.
– Mark McClure
yesterday
That is correct. If you were thinking in terms of the limsup, that might explain your confusion.
– Mark McClure
yesterday
add a comment |
up vote
1
down vote
Not every point in the Koch curve is in one of your Cantor sets! Indeed, this follows immediately from the Baire category theorem: each of your Cantor sets is a closed subset of the Koch curve with empty interior, and the union of countably many such sets will still have empty interior and thus cannot be the whole Koch curve.
The proof of the Baire category theorem also gives a recipe for finding points that are not in any of the Cantor sets. Let's say the $n$th Cantor set is $C_n$ and the Koch curve is $K$. Start with a point $x_0in Ksetminus C_0$; then there is some closed ball $B_0$ around $x_0$ which does not intersect $C_0$. Then pick a point $x_1in B_0setminus C_1$. There is again some closed ball $B_1$ around $x_1$ which does not intersect $C_1$. Then pick a point $x_2in B_0cap B_1setminus C_2$, and so on.
This procedure gives a sequence of points $(x_n)$ in $K$ and balls $B_n$ such that $x_min B_n$ for all $mgeq n$ and $B_ncap C_n=emptyset$. Since $K$ is compact, this sequence $(x_n)$ must accumulate at some point $xin K$ (in fact, we can choose the balls $B_n$ to have radius converging to $0$ and then $(x_n)$ will be Cauchy and thus will converge). Then $xin B_n$ for all $n$, so $x$ cannot be in $C_n$ for any $n$.
Another way to think of it is that the Koch curve is parametrized by a continuous function $f:[0,1]to mathbb{R}^2$ which is the limit of functions $f_n:[0,1]tomathbb{R}^2$ parametrizing the approximations to the Koch curve. Each of your Cantor sets corresponds to the image of some Cantor set in $[0,1]$ under $f$, and that Cantor set in $[0,1]$ will have Lebesgue measure $0$ if you've chosen your parametrization reasonably. So, the union of all those Cantor sets in $[0,1]$ is still a null set (and a set of first category), so most points in $[0,1]$ aren't in any of those Cantor sets.
Note that you can also think of a point $xin[0,1]$ that is not in any of those Cantor sets as a point such that the sequence $(f_n(x))$ never stabilizes but instead keeps moving infinitely often as you build the successive approximations. The value of $f(x)$ will then be the limit of this sequence, a point of the Koch curve that was not in any of the approximations but instead is a limit of points in the approximations.
answered 2 days ago
Eric Wofsey
175k12202326
add a comment |
up vote
1
down vote
Not every point in the Koch curve is in one of your Cantor sets! Indeed, this follows immediately from the Baire category theorem: each of your Cantor sets is a closed subset of the Koch curve with empty interior, and the union of countably many such sets will still have empty interior and thus cannot be the whole Koch curve.
The proof of the Baire category theorem also gives a recipe for finding points that are not in any of the Cantor sets. Let's say the $n$th Cantor set is $C_n$ and the Koch curve is $K$. Start with a point $x_0in Ksetminus C_0$; then there is some closed ball $B_0$ around $x_0$ which does not intersect $C_0$. Then pick a point $x_1in B_0setminus C_1$. There is again some closed ball $B_1$ around $x_1$ which does not intersect $C_1$. Then pick a point $x_2in B_0cap B_1setminus C_2$, and so on.
This procedure gives a sequence of points $(x_n)$ in $K$ and balls $B_n$ such that $x_min B_n$ for all $mgeq n$ and $B_ncap C_n=emptyset$. Since $K$ is compact, this sequence $(x_n)$ must accumulate at some point $xin K$ (in fact, we can choose the balls $B_n$ to have radius converging to $0$ and then $(x_n)$ will be Cauchy and thus will converge). Then $xin B_n$ for all $n$, so $x$ cannot be in $C_n$ for any $n$.
Another way to think of it is that the Koch curve is parametrized by a continuous function $f:[0,1]to mathbb{R}^2$ which is the limit of functions $f_n:[0,1]tomathbb{R}^2$ parametrizing the approximations to the Koch curve. Each of your Cantor sets corresponds to the image of some Cantor set in $[0,1]$ under $f$, and that Cantor set in $[0,1]$ will have Lebesgue measure $0$ if you've chosen your parametrization reasonably. So, the union of all those Cantor sets in $[0,1]$ is still a null set (and a set of first category), so most points in $[0,1]$ aren't in any of those Cantor sets.
Note that you can also think of a point $xin[0,1]$ that is not in any of those Cantor sets as a point such that the sequence $(f_n(x))$ never stabilizes but instead keeps moving infinitely often as you build the successive approximations. The value of $f(x)$ will then be the limit of this sequence, a point of the Koch curve that was not in any of the approximations but instead is a limit of points in the approximations.
answered 2 days ago
Eric Wofsey
175k12202326
add a comment |
up vote
1
down vote
up vote
1
down vote
Not every point in the Koch curve is in one of your Cantor sets! Indeed, this follows immediately from the Baire category theorem: each of your Cantor sets is a closed subset of the Koch curve with empty interior, and the union of countably many such sets will still have empty interior and thus cannot be the whole Koch curve.
The proof of the Baire category theorem also gives a recipe for finding points that are not in any of the Cantor sets. Let's say the $n$th Cantor set is $C_n$ and the Koch curve is $K$. Start with a point $x_0in Ksetminus C_0$; then there is some closed ball $B_0$ around $x_0$ which does not intersect $C_0$. Then pick a point $x_1in B_0setminus C_1$. There is again some closed ball $B_1$ around $x_1$ which does not intersect $C_1$. Then pick a point $x_2in B_0cap B_1setminus C_2$, and so on.
This procedure gives a sequence of points $(x_n)$ in $K$ and balls $B_n$ such that $x_min B_n$ for all $mgeq n$ and $B_ncap C_n=emptyset$. Since $K$ is compact, this sequence $(x_n)$ must accumulate at some point $xin K$ (in fact, we can choose the balls $B_n$ to have radius converging to $0$ and then $(x_n)$ will be Cauchy and thus will converge). Then $xin B_n$ for all $n$, so $x$ cannot be in $C_n$ for any $n$.
Another way to think of it is that the Koch curve is parametrized by a continuous function $f:[0,1]to mathbb{R}^2$ which is the limit of functions $f_n:[0,1]tomathbb{R}^2$ parametrizing the approximations to the Koch curve. Each of your Cantor sets corresponds to the image of some Cantor set in $[0,1]$ under $f$, and that Cantor set in $[0,1]$ will have Lebesgue measure $0$ if you've chosen your parametrization reasonably. So, the union of all those Cantor sets in $[0,1]$ is still a null set (and a set of first category), so most points in $[0,1]$ aren't in any of those Cantor sets.
Note that you can also think of a point $xin[0,1]$ that is not in any of those Cantor sets as a point such that the sequence $(f_n(x))$ never stabilizes but instead keeps moving infinitely often as you build the successive approximations. The value of $f(x)$ will then be the limit of this sequence, a point of the Koch curve that was not in any of the approximations but instead is a limit of points in the approximations.
answered 2 days ago
Eric Wofsey
175k12202326
Not every point in the Koch curve is in one of your Cantor sets! Indeed, this follows immediately from the Baire category theorem: each of your Cantor sets is a closed subset of the Koch curve with empty interior, and the union of countably many such sets will still have empty interior and thus cannot be the whole Koch curve.
The proof of the Baire category theorem also gives a recipe for finding points that are not in any of the Cantor sets. Let's say the $n$th Cantor set is $C_n$ and the Koch curve is $K$. Start with a point $x_0in Ksetminus C_0$; then there is some closed ball $B_0$ around $x_0$ which does not intersect $C_0$. Then pick a point $x_1in B_0setminus C_1$. There is again some closed ball $B_1$ around $x_1$ which does not intersect $C_1$. Then pick a point $x_2in B_0cap B_1setminus C_2$, and so on.
This procedure gives a sequence of points $(x_n)$ in $K$ and balls $B_n$ such that $x_min B_n$ for all $mgeq n$ and $B_ncap C_n=emptyset$. Since $K$ is compact, this sequence $(x_n)$ must accumulate at some point $xin K$ (in fact, we can choose the balls $B_n$ to have radius converging to $0$ and then $(x_n)$ will be Cauchy and thus will converge). Then $xin B_n$ for all $n$, so $x$ cannot be in $C_n$ for any $n$.
Another way to think of it is that the Koch curve is parametrized by a continuous function $f:[0,1]to mathbb{R}^2$ which is the limit of functions $f_n:[0,1]tomathbb{R}^2$ parametrizing the approximations to the Koch curve. Each of your Cantor sets corresponds to the image of some Cantor set in $[0,1]$ under $f$, and that Cantor set in $[0,1]$ will have Lebesgue measure $0$ if you've chosen your parametrization reasonably. So, the union of all those Cantor sets in $[0,1]$ is still a null set (and a set of first category), so most points in $[0,1]$ aren't in any of those Cantor sets.
Note that you can also think of a point $xin[0,1]$ that is not in any of those Cantor sets as a point such that the sequence $(f_n(x))$ never stabilizes but instead keeps moving infinitely often as you build the successive approximations. The value of $f(x)$ will then be the limit of this sequence, a point of the Koch curve that was not in any of the approximations but instead is a limit of points in the approximations.
answered 2 days ago
Eric Wofsey
175k12202326
edited 2 days ago
answered 2 days ago
Eric Wofsey
175k12202326
answered 2 days ago
Eric Wofsey
175k12202326
answered 2 days ago
Eric Wofsey
175k12202326
175k12202326
add a comment |
add a comment |
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Aren't there a countably infinite number of Cantor sets in the Koch snowflake if you look at it this way?
– Michael Seifert
2 days ago
@MichaelSeifert. Yes, but a countable union of sets with zero length (Lebesgue measure) also has zero length.
– md2perpe
2 days ago
I'm not entirely sure what the paradox is here. You're just taking something with $0$ length and "adding" it together countably infinite times, right? I'm not sure why anyone would expect it to have positive length, in that case. Am I missing something?
– Clayton
2 days ago
1
You've not argued that the final Koch snowflake is in fact the union of countably many Cantor sets (and indeed it isn't), so there's no contradiction.
– Noah Schweber
2 days ago
1
The MSE question 12906 "The staircase paradox, or why $pine 4$" seems similar. Or else MSE question 1525410 "Length of diagonal compared to the limit of lengths of stair-shaped curves converging to it".
– Somos
2 days ago