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Constructing a 5-dimensional injective function
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I am trying to find a 5-dimensional injective function, any suggestions or any ideas as to how I can construct one? Formally I am looking for a function $f: mathbbR^5 rightarrow mathbbR$ which is injective.
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asked Aug 1 at 16:56
Fabric
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up vote
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I am trying to find a 5-dimensional injective function, any suggestions or any ideas as to how I can construct one? Formally I am looking for a function $f: mathbbR^5 rightarrow mathbbR$ which is injective.
functions
asked Aug 1 at 16:56
Fabric
377
1
Iterlacing the digits will give you an bijective function from $mathbb R_+^5 to mathbb R_+$ And you can compose with bijective functions from $mathbb Rto mathbb R_+$ and back again.
– Doug M
Aug 1 at 17:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to find a 5-dimensional injective function, any suggestions or any ideas as to how I can construct one? Formally I am looking for a function $f: mathbbR^5 rightarrow mathbbR$ which is injective.
functions
asked Aug 1 at 16:56
Fabric
377
I am trying to find a 5-dimensional injective function, any suggestions or any ideas as to how I can construct one? Formally I am looking for a function $f: mathbbR^5 rightarrow mathbbR$ which is injective.
functions
asked Aug 1 at 16:56
Fabric
377
asked Aug 1 at 16:56
Fabric
377
asked Aug 1 at 16:56
Fabric
377
asked Aug 1 at 16:56
Fabric
377
377
1
Iterlacing the digits will give you an bijective function from $mathbb R_+^5 to mathbb R_+$ And you can compose with bijective functions from $mathbb Rto mathbb R_+$ and back again.
– Doug M
Aug 1 at 17:07
add a comment |Â
1
Iterlacing the digits will give you an bijective function from $mathbb R_+^5 to mathbb R_+$ And you can compose with bijective functions from $mathbb Rto mathbb R_+$ and back again.
– Doug M
Aug 1 at 17:07
1
1
Iterlacing the digits will give you an bijective function from $mathbb R_+^5 to mathbb R_+$ And you can compose with bijective functions from $mathbb Rto mathbb R_+$ and back again.
– Doug M
Aug 1 at 17:07
Iterlacing the digits will give you an bijective function from $mathbb R_+^5 to mathbb R_+$ And you can compose with bijective functions from $mathbb Rto mathbb R_+$ and back again.
– Doug M
Aug 1 at 17:07
add a comment |Â
1 Answer
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First we construct an injective function $mathbbR^2rightarrow mathbbR.$ To simplify matters, it suffices to construct an injective function $(0,1)rightarrow mathbbR$ since $tan(pi (x-frac12))$ is a bijection $(0,1)rightarrow mathbbR.$
Given a pair of real numbers $(a,b)in(0,1)^2$ they have unique binary representations $a = sum_i=0^inftyfraca_i2^i$ and $b = sum_i=0^inftyfracb_i2^i$ with neither $(a_i)$ nor $(b_i)$ ending in an infinite tail of $1$'s. Then it's not hard to see that the map sending $(a,b)$ to $0.a_1 b_1 a_2 b_2 ldots$ (written in binary) is an injective map $(0,1)^2rightarrow (0,1).$ Notice that this decimal will not end in infinite sequence of $1$'s by construction.
Now by identifying $mathbbR^5$ with $mathbbR^2times mathbbR^3,$ we obtain an injective map $mathbbR^5rightarrow mathbbRtimes mathbbR^3 = mathbbR^4.$ Similarly, we have injective maps $mathbbR^4rightarrow mathbbR^3$, $mathbbR^3rightarrow mathbbR^2$, and $mathbbR^2rightarrow mathbbR.$ By composing these maps we get an injection $mathbbR^5rightarrow mathbbR$ as desired.
answered Aug 1 at 17:14
William Swartworth
398210
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First we construct an injective function $mathbbR^2rightarrow mathbbR.$ To simplify matters, it suffices to construct an injective function $(0,1)rightarrow mathbbR$ since $tan(pi (x-frac12))$ is a bijection $(0,1)rightarrow mathbbR.$
Given a pair of real numbers $(a,b)in(0,1)^2$ they have unique binary representations $a = sum_i=0^inftyfraca_i2^i$ and $b = sum_i=0^inftyfracb_i2^i$ with neither $(a_i)$ nor $(b_i)$ ending in an infinite tail of $1$'s. Then it's not hard to see that the map sending $(a,b)$ to $0.a_1 b_1 a_2 b_2 ldots$ (written in binary) is an injective map $(0,1)^2rightarrow (0,1).$ Notice that this decimal will not end in infinite sequence of $1$'s by construction.
Now by identifying $mathbbR^5$ with $mathbbR^2times mathbbR^3,$ we obtain an injective map $mathbbR^5rightarrow mathbbRtimes mathbbR^3 = mathbbR^4.$ Similarly, we have injective maps $mathbbR^4rightarrow mathbbR^3$, $mathbbR^3rightarrow mathbbR^2$, and $mathbbR^2rightarrow mathbbR.$ By composing these maps we get an injection $mathbbR^5rightarrow mathbbR$ as desired.
answered Aug 1 at 17:14
William Swartworth
398210
add a comment |Â
up vote
2
down vote
accepted
First we construct an injective function $mathbbR^2rightarrow mathbbR.$ To simplify matters, it suffices to construct an injective function $(0,1)rightarrow mathbbR$ since $tan(pi (x-frac12))$ is a bijection $(0,1)rightarrow mathbbR.$
Given a pair of real numbers $(a,b)in(0,1)^2$ they have unique binary representations $a = sum_i=0^inftyfraca_i2^i$ and $b = sum_i=0^inftyfracb_i2^i$ with neither $(a_i)$ nor $(b_i)$ ending in an infinite tail of $1$'s. Then it's not hard to see that the map sending $(a,b)$ to $0.a_1 b_1 a_2 b_2 ldots$ (written in binary) is an injective map $(0,1)^2rightarrow (0,1).$ Notice that this decimal will not end in infinite sequence of $1$'s by construction.
Now by identifying $mathbbR^5$ with $mathbbR^2times mathbbR^3,$ we obtain an injective map $mathbbR^5rightarrow mathbbRtimes mathbbR^3 = mathbbR^4.$ Similarly, we have injective maps $mathbbR^4rightarrow mathbbR^3$, $mathbbR^3rightarrow mathbbR^2$, and $mathbbR^2rightarrow mathbbR.$ By composing these maps we get an injection $mathbbR^5rightarrow mathbbR$ as desired.
answered Aug 1 at 17:14
William Swartworth
398210
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First we construct an injective function $mathbbR^2rightarrow mathbbR.$ To simplify matters, it suffices to construct an injective function $(0,1)rightarrow mathbbR$ since $tan(pi (x-frac12))$ is a bijection $(0,1)rightarrow mathbbR.$
Given a pair of real numbers $(a,b)in(0,1)^2$ they have unique binary representations $a = sum_i=0^inftyfraca_i2^i$ and $b = sum_i=0^inftyfracb_i2^i$ with neither $(a_i)$ nor $(b_i)$ ending in an infinite tail of $1$'s. Then it's not hard to see that the map sending $(a,b)$ to $0.a_1 b_1 a_2 b_2 ldots$ (written in binary) is an injective map $(0,1)^2rightarrow (0,1).$ Notice that this decimal will not end in infinite sequence of $1$'s by construction.
Now by identifying $mathbbR^5$ with $mathbbR^2times mathbbR^3,$ we obtain an injective map $mathbbR^5rightarrow mathbbRtimes mathbbR^3 = mathbbR^4.$ Similarly, we have injective maps $mathbbR^4rightarrow mathbbR^3$, $mathbbR^3rightarrow mathbbR^2$, and $mathbbR^2rightarrow mathbbR.$ By composing these maps we get an injection $mathbbR^5rightarrow mathbbR$ as desired.
answered Aug 1 at 17:14
William Swartworth
398210
First we construct an injective function $mathbbR^2rightarrow mathbbR.$ To simplify matters, it suffices to construct an injective function $(0,1)rightarrow mathbbR$ since $tan(pi (x-frac12))$ is a bijection $(0,1)rightarrow mathbbR.$
Given a pair of real numbers $(a,b)in(0,1)^2$ they have unique binary representations $a = sum_i=0^inftyfraca_i2^i$ and $b = sum_i=0^inftyfracb_i2^i$ with neither $(a_i)$ nor $(b_i)$ ending in an infinite tail of $1$'s. Then it's not hard to see that the map sending $(a,b)$ to $0.a_1 b_1 a_2 b_2 ldots$ (written in binary) is an injective map $(0,1)^2rightarrow (0,1).$ Notice that this decimal will not end in infinite sequence of $1$'s by construction.
Now by identifying $mathbbR^5$ with $mathbbR^2times mathbbR^3,$ we obtain an injective map $mathbbR^5rightarrow mathbbRtimes mathbbR^3 = mathbbR^4.$ Similarly, we have injective maps $mathbbR^4rightarrow mathbbR^3$, $mathbbR^3rightarrow mathbbR^2$, and $mathbbR^2rightarrow mathbbR.$ By composing these maps we get an injection $mathbbR^5rightarrow mathbbR$ as desired.
answered Aug 1 at 17:14
William Swartworth
398210
edited Aug 1 at 17:22
answered Aug 1 at 17:14
William Swartworth
398210
answered Aug 1 at 17:14
William Swartworth
398210
answered Aug 1 at 17:14
William Swartworth
398210
398210
add a comment |Â
add a comment |Â
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1
Iterlacing the digits will give you an bijective function from $mathbb R_+^5 to mathbb R_+$ And you can compose with bijective functions from $mathbb Rto mathbb R_+$ and back again.
– Doug M
Aug 1 at 17:07