Mapping a square onto the real number line or a line segment
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A square can be filled by Hilbert's curve, a one dimensional object. Is there an equation (a continuous bijection) that takes the x-y coordinates of a point in a finite square and maps them onto a line segment or the real number line ? If there is, what is it and how is it derived ?
geometry
asked Aug 27 at 18:31
Xaovnx
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A square can be filled by Hilbert's curve, a one dimensional object. Is there an equation (a continuous bijection) that takes the x-y coordinates of a point in a finite square and maps them onto a line segment or the real number line ? If there is, what is it and how is it derived ?
geometry
asked Aug 27 at 18:31
Xaovnx
12
1
Do you want a surjective map? Or an injection? There are several maps from a square to the real line, like $(x,y) mapsto x$...
– Babelfish
Aug 27 at 18:36
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A square can be filled by Hilbert's curve, a one dimensional object. Is there an equation (a continuous bijection) that takes the x-y coordinates of a point in a finite square and maps them onto a line segment or the real number line ? If there is, what is it and how is it derived ?
geometry
asked Aug 27 at 18:31
Xaovnx
12
A square can be filled by Hilbert's curve, a one dimensional object. Is there an equation (a continuous bijection) that takes the x-y coordinates of a point in a finite square and maps them onto a line segment or the real number line ? If there is, what is it and how is it derived ?
geometry
asked Aug 27 at 18:31
Xaovnx
12
edited Aug 27 at 22:01
asked Aug 27 at 18:31
Xaovnx
12
asked Aug 27 at 18:31
Xaovnx
12
asked Aug 27 at 18:31
Xaovnx
12
12
1
Do you want a surjective map? Or an injection? There are several maps from a square to the real line, like $(x,y) mapsto x$...
– Babelfish
Aug 27 at 18:36
add a comment |Â
1
Do you want a surjective map? Or an injection? There are several maps from a square to the real line, like $(x,y) mapsto x$...
– Babelfish
Aug 27 at 18:36
1
1
Do you want a surjective map? Or an injection? There are several maps from a square to the real line, like $(x,y) mapsto x$...
– Babelfish
Aug 27 at 18:36
Do you want a surjective map? Or an injection? There are several maps from a square to the real line, like $(x,y) mapsto x$...
– Babelfish
Aug 27 at 18:36
add a comment |Â
2 Answers
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Take for example $f(x, y) =x$ $(x, y) in[0,1]×[0,1]$
answered Aug 27 at 18:47
dmtri
810317
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If you view the coordinates as digit strings it is easy to get close. Let $x=0.x_1x_2x_3x_4ldots$ and $y=0.y_1y_2y_3y_4ldots$. Map it to $0x_1y_1x_2y_2x_3y_3ldots$. This is a bijection between pairs of digit strings and single digit strings.
The problem is that some reals, the ones with terminating decimals, have two representations. We can just choose one of the representations for the single number side, say the one that ends in all $0$s. Unfortunately you generate both versions when you unpack the single number into two. Your intuition should tell you that as this is only a countable infinity of numbers we can sweep it under the rug. The secret is to get a bijection between the digit strings and the reals. If you have two digit strings that represent $frac a10^n$ send the one ending in all $0$s to $frac a10^2n-1$ and the one ending in $9$s to $frac a10^2n$. Al the other digit strings can just be viewed as decimals and all is well
answered Aug 27 at 19:21
Ross Millikan
279k22188355
I came across this solution before, there are many disadvantages including the fact that the mapping changes when the base is changed and the fact that it isn't continuous. I am looking for an equation that uses standard operations and functions to make the mapping. Originally my idea was to "pull" on both ends of Hilbert's curve to get the real number line.
– Xaovnx
Aug 27 at 21:57
There is no continuous mapping. That can be demonstrated by the topological dimension of the two spaces. It is important not to think of functions as limited to those that have nice formulas-there are many more functions than that.
– Ross Millikan
Aug 28 at 0:45
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Take for example $f(x, y) =x$ $(x, y) in[0,1]×[0,1]$
answered Aug 27 at 18:47
dmtri
810317
add a comment |Â
up vote
1
down vote
Take for example $f(x, y) =x$ $(x, y) in[0,1]×[0,1]$
answered Aug 27 at 18:47
dmtri
810317
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Take for example $f(x, y) =x$ $(x, y) in[0,1]×[0,1]$
answered Aug 27 at 18:47
dmtri
810317
Take for example $f(x, y) =x$ $(x, y) in[0,1]×[0,1]$
answered Aug 27 at 18:47
dmtri
810317
answered Aug 27 at 18:47
dmtri
810317
answered Aug 27 at 18:47
dmtri
810317
answered Aug 27 at 18:47
dmtri
810317
810317
add a comment |Â
add a comment |Â
up vote
1
down vote
If you view the coordinates as digit strings it is easy to get close. Let $x=0.x_1x_2x_3x_4ldots$ and $y=0.y_1y_2y_3y_4ldots$. Map it to $0x_1y_1x_2y_2x_3y_3ldots$. This is a bijection between pairs of digit strings and single digit strings.
The problem is that some reals, the ones with terminating decimals, have two representations. We can just choose one of the representations for the single number side, say the one that ends in all $0$s. Unfortunately you generate both versions when you unpack the single number into two. Your intuition should tell you that as this is only a countable infinity of numbers we can sweep it under the rug. The secret is to get a bijection between the digit strings and the reals. If you have two digit strings that represent $frac a10^n$ send the one ending in all $0$s to $frac a10^2n-1$ and the one ending in $9$s to $frac a10^2n$. Al the other digit strings can just be viewed as decimals and all is well
answered Aug 27 at 19:21
Ross Millikan
279k22188355
I came across this solution before, there are many disadvantages including the fact that the mapping changes when the base is changed and the fact that it isn't continuous. I am looking for an equation that uses standard operations and functions to make the mapping. Originally my idea was to "pull" on both ends of Hilbert's curve to get the real number line.
– Xaovnx
Aug 27 at 21:57
There is no continuous mapping. That can be demonstrated by the topological dimension of the two spaces. It is important not to think of functions as limited to those that have nice formulas-there are many more functions than that.
– Ross Millikan
Aug 28 at 0:45
add a comment |Â
up vote
1
down vote
If you view the coordinates as digit strings it is easy to get close. Let $x=0.x_1x_2x_3x_4ldots$ and $y=0.y_1y_2y_3y_4ldots$. Map it to $0x_1y_1x_2y_2x_3y_3ldots$. This is a bijection between pairs of digit strings and single digit strings.
The problem is that some reals, the ones with terminating decimals, have two representations. We can just choose one of the representations for the single number side, say the one that ends in all $0$s. Unfortunately you generate both versions when you unpack the single number into two. Your intuition should tell you that as this is only a countable infinity of numbers we can sweep it under the rug. The secret is to get a bijection between the digit strings and the reals. If you have two digit strings that represent $frac a10^n$ send the one ending in all $0$s to $frac a10^2n-1$ and the one ending in $9$s to $frac a10^2n$. Al the other digit strings can just be viewed as decimals and all is well
answered Aug 27 at 19:21
Ross Millikan
279k22188355
I came across this solution before, there are many disadvantages including the fact that the mapping changes when the base is changed and the fact that it isn't continuous. I am looking for an equation that uses standard operations and functions to make the mapping. Originally my idea was to "pull" on both ends of Hilbert's curve to get the real number line.
– Xaovnx
Aug 27 at 21:57
There is no continuous mapping. That can be demonstrated by the topological dimension of the two spaces. It is important not to think of functions as limited to those that have nice formulas-there are many more functions than that.
– Ross Millikan
Aug 28 at 0:45
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you view the coordinates as digit strings it is easy to get close. Let $x=0.x_1x_2x_3x_4ldots$ and $y=0.y_1y_2y_3y_4ldots$. Map it to $0x_1y_1x_2y_2x_3y_3ldots$. This is a bijection between pairs of digit strings and single digit strings.
The problem is that some reals, the ones with terminating decimals, have two representations. We can just choose one of the representations for the single number side, say the one that ends in all $0$s. Unfortunately you generate both versions when you unpack the single number into two. Your intuition should tell you that as this is only a countable infinity of numbers we can sweep it under the rug. The secret is to get a bijection between the digit strings and the reals. If you have two digit strings that represent $frac a10^n$ send the one ending in all $0$s to $frac a10^2n-1$ and the one ending in $9$s to $frac a10^2n$. Al the other digit strings can just be viewed as decimals and all is well
answered Aug 27 at 19:21
Ross Millikan
279k22188355
If you view the coordinates as digit strings it is easy to get close. Let $x=0.x_1x_2x_3x_4ldots$ and $y=0.y_1y_2y_3y_4ldots$. Map it to $0x_1y_1x_2y_2x_3y_3ldots$. This is a bijection between pairs of digit strings and single digit strings.
The problem is that some reals, the ones with terminating decimals, have two representations. We can just choose one of the representations for the single number side, say the one that ends in all $0$s. Unfortunately you generate both versions when you unpack the single number into two. Your intuition should tell you that as this is only a countable infinity of numbers we can sweep it under the rug. The secret is to get a bijection between the digit strings and the reals. If you have two digit strings that represent $frac a10^n$ send the one ending in all $0$s to $frac a10^2n-1$ and the one ending in $9$s to $frac a10^2n$. Al the other digit strings can just be viewed as decimals and all is well
answered Aug 27 at 19:21
Ross Millikan
279k22188355
answered Aug 27 at 19:21
Ross Millikan
279k22188355
answered Aug 27 at 19:21
Ross Millikan
279k22188355
answered Aug 27 at 19:21
Ross Millikan
279k22188355
279k22188355
I came across this solution before, there are many disadvantages including the fact that the mapping changes when the base is changed and the fact that it isn't continuous. I am looking for an equation that uses standard operations and functions to make the mapping. Originally my idea was to "pull" on both ends of Hilbert's curve to get the real number line.
– Xaovnx
Aug 27 at 21:57
There is no continuous mapping. That can be demonstrated by the topological dimension of the two spaces. It is important not to think of functions as limited to those that have nice formulas-there are many more functions than that.
– Ross Millikan
Aug 28 at 0:45
add a comment |Â
I came across this solution before, there are many disadvantages including the fact that the mapping changes when the base is changed and the fact that it isn't continuous. I am looking for an equation that uses standard operations and functions to make the mapping. Originally my idea was to "pull" on both ends of Hilbert's curve to get the real number line.
– Xaovnx
Aug 27 at 21:57
There is no continuous mapping. That can be demonstrated by the topological dimension of the two spaces. It is important not to think of functions as limited to those that have nice formulas-there are many more functions than that.
– Ross Millikan
Aug 28 at 0:45
I came across this solution before, there are many disadvantages including the fact that the mapping changes when the base is changed and the fact that it isn't continuous. I am looking for an equation that uses standard operations and functions to make the mapping. Originally my idea was to "pull" on both ends of Hilbert's curve to get the real number line.
– Xaovnx
Aug 27 at 21:57
I came across this solution before, there are many disadvantages including the fact that the mapping changes when the base is changed and the fact that it isn't continuous. I am looking for an equation that uses standard operations and functions to make the mapping. Originally my idea was to "pull" on both ends of Hilbert's curve to get the real number line.
– Xaovnx
Aug 27 at 21:57
There is no continuous mapping. That can be demonstrated by the topological dimension of the two spaces. It is important not to think of functions as limited to those that have nice formulas-there are many more functions than that.
– Ross Millikan
Aug 28 at 0:45
There is no continuous mapping. That can be demonstrated by the topological dimension of the two spaces. It is important not to think of functions as limited to those that have nice formulas-there are many more functions than that.
– Ross Millikan
Aug 28 at 0:45
add a comment |Â
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1
Do you want a surjective map? Or an injection? There are several maps from a square to the real line, like $(x,y) mapsto x$...
– Babelfish
Aug 27 at 18:36