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Give a Bijection between $mathbbRsetminusmathbbZ$ and $mathbbRsetminusmathbbN$ [duplicate]
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Proof there is a 1-1 correspondence between an uncountable set and itself minus a countable part of it
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Give a Bijection between $mathbbRsetminusmathbbZ$ and $mathbbRsetminusmathbbN$
I got a bijection between $mathbbN$ and $mathbbZ$
Given by
$phi(1)=0$
$phi(2)=-1$
$phi(3)=1$
$phi(4)=-2$
$phi(5)=2$
$phi(6)=-3$
But I cannot figure how to find a bijection between $mathbbRsetminusmathbbZ$ and $mathbbRsetminusmathbbN$
discrete-mathematics elementary-set-theory
edited Jul 29 at 17:41
José Carlos Santos
112k1696173
asked Jul 29 at 17:22
Rakesh Bhatt
553112
marked as duplicate by Jyrki Lahtonen, Arnaud Mortier, amWhy discrete-mathematics
Users with the  discrete-mathematics badge can single-handedly close discrete-mathematics questions as duplicates and reopen them as needed.
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Jul 30 at 10:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
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down vote
favorite
This question already has an answer here:
Proof there is a 1-1 correspondence between an uncountable set and itself minus a countable part of it
1 answer
Give a Bijection between $mathbbRsetminusmathbbZ$ and $mathbbRsetminusmathbbN$
I got a bijection between $mathbbN$ and $mathbbZ$
Given by
$phi(1)=0$
$phi(2)=-1$
$phi(3)=1$
$phi(4)=-2$
$phi(5)=2$
$phi(6)=-3$
But I cannot figure how to find a bijection between $mathbbRsetminusmathbbZ$ and $mathbbRsetminusmathbbN$
discrete-mathematics elementary-set-theory
edited Jul 29 at 17:41
José Carlos Santos
112k1696173
asked Jul 29 at 17:22
Rakesh Bhatt
553112
marked as duplicate by Jyrki Lahtonen, Arnaud Mortier, amWhy discrete-mathematics
Users with the  discrete-mathematics badge can single-handedly close discrete-mathematics questions as duplicates and reopen them as needed.
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
5
Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way
– Rocket Man
Jul 29 at 17:27
1
Start with an identity map from reals to reals. Modify this map on the integers. Now drop the integers from the domain.
– Steve D
Jul 29 at 17:30
2
I don't know who you are addressing your comment to, but @RocketMan 's comment gives you the solution, pure and simple.
– Arnaud Mortier
Jul 29 at 17:31
6
Please don’t ask for help and then “Bro†prospective helpers. Thanks.
– Brian Tung
Jul 29 at 17:39
See here for a more general result.
– Jyrki Lahtonen
Jul 29 at 18:10
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
Proof there is a 1-1 correspondence between an uncountable set and itself minus a countable part of it
1 answer
Give a Bijection between $mathbbRsetminusmathbbZ$ and $mathbbRsetminusmathbbN$
I got a bijection between $mathbbN$ and $mathbbZ$
Given by
$phi(1)=0$
$phi(2)=-1$
$phi(3)=1$
$phi(4)=-2$
$phi(5)=2$
$phi(6)=-3$
But I cannot figure how to find a bijection between $mathbbRsetminusmathbbZ$ and $mathbbRsetminusmathbbN$
discrete-mathematics elementary-set-theory
edited Jul 29 at 17:41
José Carlos Santos
112k1696173
asked Jul 29 at 17:22
Rakesh Bhatt
553112
This question already has an answer here:
Proof there is a 1-1 correspondence between an uncountable set and itself minus a countable part of it
1 answer
Give a Bijection between $mathbbRsetminusmathbbZ$ and $mathbbRsetminusmathbbN$
I got a bijection between $mathbbN$ and $mathbbZ$
Given by
$phi(1)=0$
$phi(2)=-1$
$phi(3)=1$
$phi(4)=-2$
$phi(5)=2$
$phi(6)=-3$
But I cannot figure how to find a bijection between $mathbbRsetminusmathbbZ$ and $mathbbRsetminusmathbbN$
This question already has an answer here:
Proof there is a 1-1 correspondence between an uncountable set and itself minus a countable part of it
1 answer
discrete-mathematics elementary-set-theory
edited Jul 29 at 17:41
José Carlos Santos
112k1696173
asked Jul 29 at 17:22
Rakesh Bhatt
553112
edited Jul 29 at 17:41
José Carlos Santos
112k1696173
edited Jul 29 at 17:41
José Carlos Santos
112k1696173
edited Jul 29 at 17:41
José Carlos Santos
112k1696173
112k1696173
asked Jul 29 at 17:22
Rakesh Bhatt
553112
asked Jul 29 at 17:22
Rakesh Bhatt
553112
asked Jul 29 at 17:22
Rakesh Bhatt
553112
553112
marked as duplicate by Jyrki Lahtonen, Arnaud Mortier, amWhy discrete-mathematics
Users with the  discrete-mathematics badge can single-handedly close discrete-mathematics questions as duplicates and reopen them as needed.
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
5
Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way
– Rocket Man
Jul 29 at 17:27
1
Start with an identity map from reals to reals. Modify this map on the integers. Now drop the integers from the domain.
– Steve D
Jul 29 at 17:30
2
I don't know who you are addressing your comment to, but @RocketMan 's comment gives you the solution, pure and simple.
– Arnaud Mortier
Jul 29 at 17:31
6
Please don’t ask for help and then “Bro†prospective helpers. Thanks.
– Brian Tung
Jul 29 at 17:39
See here for a more general result.
– Jyrki Lahtonen
Jul 29 at 18:10
add a comment |Â
5
Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way
– Rocket Man
Jul 29 at 17:27
1
Start with an identity map from reals to reals. Modify this map on the integers. Now drop the integers from the domain.
– Steve D
Jul 29 at 17:30
2
I don't know who you are addressing your comment to, but @RocketMan 's comment gives you the solution, pure and simple.
– Arnaud Mortier
Jul 29 at 17:31
6
Please don’t ask for help and then “Bro†prospective helpers. Thanks.
– Brian Tung
Jul 29 at 17:39
See here for a more general result.
– Jyrki Lahtonen
Jul 29 at 18:10
5
5
Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way
– Rocket Man
Jul 29 at 17:27
Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way
– Rocket Man
Jul 29 at 17:27
1
1
Start with an identity map from reals to reals. Modify this map on the integers. Now drop the integers from the domain.
– Steve D
Jul 29 at 17:30
Start with an identity map from reals to reals. Modify this map on the integers. Now drop the integers from the domain.
– Steve D
Jul 29 at 17:30
2
2
I don't know who you are addressing your comment to, but @RocketMan 's comment gives you the solution, pure and simple.
– Arnaud Mortier
Jul 29 at 17:31
I don't know who you are addressing your comment to, but @RocketMan 's comment gives you the solution, pure and simple.
– Arnaud Mortier
Jul 29 at 17:31
6
6
Please don’t ask for help and then “Bro†prospective helpers. Thanks.
– Brian Tung
Jul 29 at 17:39
Please don’t ask for help and then “Bro†prospective helpers. Thanks.
– Brian Tung
Jul 29 at 17:39
See here for a more general result.
– Jyrki Lahtonen
Jul 29 at 18:10
See here for a more general result.
– Jyrki Lahtonen
Jul 29 at 18:10
add a comment |Â
5 Answers
5
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up vote
2
down vote
accepted
Something you can do is create extra gaps, and reduce the problem to a bijection between countable sets:
Take $phi$ to be the identity on $mathbb R - frac12 mathbb Z$. All you have to do then is find a bijection between $frac12 mathbb Z - mathbb Z$ and $frac12 mathbb Z - mathbb N$. On positive integers, take it to be the identity; on negative integers, construct a bijection between $-mathbb N$ and $-frac12mathbb N$.
answered Jul 29 at 17:45
barto
12.7k32478
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up vote
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The identity function almost works for $mathbb Rsetminusmathbb N to mathbb Rsetminus mathbb Z$, except that you need to avoid hitting the negative integers. Use Hilbert's Hotel to make room for each of them, e.g.
$$ f(x) = begincases x + c & textif x=-(n+1)+mctext for some n,minmathbb N \ x & textotherwise endcases $$
for some appropriate constant $c$.
Which numbers would work as $c$?
answered Jul 29 at 17:54
Henning Makholm
225k16290516
add a comment |Â
up vote
2
down vote
Let's start with the identity $f(x)=xquad forall xinmathbb R-mathbb Z$
$(mathbb R-mathbb Z)$ and $(mathbb R-mathbb N)$ differ by $mathbb
N_>0$.
So to have a bijection you must basically make every positive integer disappear!
But the black hole $mathbb N_>0mapstovarnothing$ is not a bijection.
Instead the idea is to hide them within the real numbers. We will send $1$ to a real which is not an integer, let say $phi(1)=frac 12$.
But location $frac 12$ is already occupied by $f(frac 12)=frac 12$, so we will push it deeper with $phi(frac 12)=frac 14$, and we have a conflict with $frac 14$ now, so we will push it also $phi(frac 14)=frac 18$ and so on...
Basically we shift the whole sequence $frac 12^kmapstofrac 12^k+1$ a step deeper, that is freeing a hole in $frac 12$ where we can put $1$.
Now where to put $2$ ?
Similarly we can shift the sequence of all $frac 13^k$, freeing a hole in $frac 13$ and place $2$ there $phi(2)=frac 13$.
On so on, we shift the sequences of powers of primes numbers one step deeper $phi(frac 1p_n^k)mapstofrac 1p_n^k+1$ this frees a hole in $frac 1p_n$ where we can put $n$, i.e $phi(n)=frac 1p_n$
This is only one example. For instance you can shift the sequences $n+frac 12^k$ instead and send $n$ to the center of interval $]n,n+1[$ instead of using prime numbers.
More sophisticated sequences are possible, like the one by J.C.Santos for instance, but the principle is the same.
answered Jul 29 at 17:35
zwim
11k627
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up vote
1
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You can define a bijection $psicolonmathbbRsetminusmathbbNlongrightarrowmathbbRsetminusmathbb Z$ defining $psi(x)=x$ if $xnotinleft,ninmathbb Nright$ and then:
- $psi(0)=frac12$;
- $psi(-1)=frac14$;
- $psi(-2)=frac16$
and so on. Besides
- $psileft(frac12right)=frac13$;
- $psileft(frac13right)=frac15$;
- $psileft(frac14right)=frac17$
and so on.
answered Jul 29 at 17:38
José Carlos Santos
112k1696173
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up vote
0
down vote
Rocket man commented that "Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way "
So $mathbb Rsetminus mathbb Z = cup_zin mathbb Z I_z$ where $I_z = (z-1,z)$
And $mathbb Rsetminus mathbb N = cup_nin mathbb N J_n$ where $J_n = (n-1, n)$ for $n>1$ and $J_1 = (-infty, 1)$.
You found a bijection $phi: mathbb N to mathbb Z$.
Now if you can find a sequence of bijections $psi_n: J_nto I_phi(n)$ we'd be done: We'd just need to define $psi:mathbb Rsetminus mathbb Nto mathbb Rsetminus mathbb Z$ as: $psi(x) = psi_n(x)$ when $x in J_n$.
....
to spell it out:
Let $psi: mathbb Rsetminus Nto mathbb Rsetminus Z$ be defined as follows:
If $x in mathbb Rsetminus N$ and $x > 1$ then there exists an $n$ so that $n-1 < x < n$. For this $x$ let $psi (x) = x - n + phi(n)$. Now $phi(n)-1< psi(x) < phi(n)$ so $psi(x) in Rsetminus Z$.
If $x in mathbb Rsetminus N$ and $x le 1$ then $x < 1$. So $x-1 < 0$ and $1-x > 0$ and $0 < frac 11-x < 1$ and $-1< frac 11-x-1< 0$. For this $x$ let $psi(x) = frac 11-x-1$. Now $-1 < psi(x) < 0$ so $psi(x) in Rsetminus Z$
We can verify this is a bijection:
It's one to one: If $xne y$. $x < 1$ and $y < 1$ the $psi(x) = frac 11-x$ and $psi (y) = frac 11-x$. $x= yiff 1-x = 1-y iff frac 11-x= frac 11-y$ so $psi(x) ne psi (y)$.
If one of $x$ or $y$ (wolog $x$) is less than $1$ and the other (wolog $y$) is not then there is an $n>1$ so that $n-1 < y < n$. And $phi(n) ne 0$. so $phi(n)-1 < psi(y) < phi(n)$ so $psi(y) not in (-1,0)$. Whereas $psi(x)in (-1,0)$ so $psi(x) ne psi(y)$.
If $n- 1 < x<n$ and $m-1< y < m$ and $m ne n$, then $phi(n) ne phi(m)$. And $phi(n)-1 < psi(x)< phi(n)$ while $phi(m)-1 < psi(y)< phi(m)$. So $psi(x)ne psi(y)$.
Finally if $n-1<x,y< n$ then $x = y iff x-n+phi(n) = y-n+phi(n) iff psi(x) = psi (y)$. So $psi(x)ne psi(y)$.
$psi$ is onto: If $xin mathbb Rsetminus Z$ then there is a unique $m$ so that $m-1< x < m$. And there is a unique $n$ so that $phi(n) = m$.
If $n ne 1$ then $y =x-m + n$ will mean $n-1< x < n$ and $psi(y) = x$.
If $n = 1$ then $m = 0$ and $-1 < x < 0$. And $0 < x + 1< 1$ and $frac 1x+1 > 1$ and let $y=1-frac 1x+1 < 0 < 1$. then $psi(y) = x$.
answered Jul 29 at 19:47
fleablood
60.2k22575
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Something you can do is create extra gaps, and reduce the problem to a bijection between countable sets:
Take $phi$ to be the identity on $mathbb R - frac12 mathbb Z$. All you have to do then is find a bijection between $frac12 mathbb Z - mathbb Z$ and $frac12 mathbb Z - mathbb N$. On positive integers, take it to be the identity; on negative integers, construct a bijection between $-mathbb N$ and $-frac12mathbb N$.
answered Jul 29 at 17:45
barto
12.7k32478
add a comment |Â
up vote
2
down vote
accepted
Something you can do is create extra gaps, and reduce the problem to a bijection between countable sets:
Take $phi$ to be the identity on $mathbb R - frac12 mathbb Z$. All you have to do then is find a bijection between $frac12 mathbb Z - mathbb Z$ and $frac12 mathbb Z - mathbb N$. On positive integers, take it to be the identity; on negative integers, construct a bijection between $-mathbb N$ and $-frac12mathbb N$.
answered Jul 29 at 17:45
barto
12.7k32478
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Something you can do is create extra gaps, and reduce the problem to a bijection between countable sets:
Take $phi$ to be the identity on $mathbb R - frac12 mathbb Z$. All you have to do then is find a bijection between $frac12 mathbb Z - mathbb Z$ and $frac12 mathbb Z - mathbb N$. On positive integers, take it to be the identity; on negative integers, construct a bijection between $-mathbb N$ and $-frac12mathbb N$.
answered Jul 29 at 17:45
barto
12.7k32478
Something you can do is create extra gaps, and reduce the problem to a bijection between countable sets:
Take $phi$ to be the identity on $mathbb R - frac12 mathbb Z$. All you have to do then is find a bijection between $frac12 mathbb Z - mathbb Z$ and $frac12 mathbb Z - mathbb N$. On positive integers, take it to be the identity; on negative integers, construct a bijection between $-mathbb N$ and $-frac12mathbb N$.
answered Jul 29 at 17:45
barto
12.7k32478
answered Jul 29 at 17:45
barto
12.7k32478
answered Jul 29 at 17:45
barto
12.7k32478
answered Jul 29 at 17:45
barto
12.7k32478
12.7k32478
add a comment |Â
add a comment |Â
up vote
3
down vote
The identity function almost works for $mathbb Rsetminusmathbb N to mathbb Rsetminus mathbb Z$, except that you need to avoid hitting the negative integers. Use Hilbert's Hotel to make room for each of them, e.g.
$$ f(x) = begincases x + c & textif x=-(n+1)+mctext for some n,minmathbb N \ x & textotherwise endcases $$
for some appropriate constant $c$.
Which numbers would work as $c$?
answered Jul 29 at 17:54
Henning Makholm
225k16290516
add a comment |Â
up vote
3
down vote
The identity function almost works for $mathbb Rsetminusmathbb N to mathbb Rsetminus mathbb Z$, except that you need to avoid hitting the negative integers. Use Hilbert's Hotel to make room for each of them, e.g.
$$ f(x) = begincases x + c & textif x=-(n+1)+mctext for some n,minmathbb N \ x & textotherwise endcases $$
for some appropriate constant $c$.
Which numbers would work as $c$?
answered Jul 29 at 17:54
Henning Makholm
225k16290516
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The identity function almost works for $mathbb Rsetminusmathbb N to mathbb Rsetminus mathbb Z$, except that you need to avoid hitting the negative integers. Use Hilbert's Hotel to make room for each of them, e.g.
$$ f(x) = begincases x + c & textif x=-(n+1)+mctext for some n,minmathbb N \ x & textotherwise endcases $$
for some appropriate constant $c$.
Which numbers would work as $c$?
answered Jul 29 at 17:54
Henning Makholm
225k16290516
The identity function almost works for $mathbb Rsetminusmathbb N to mathbb Rsetminus mathbb Z$, except that you need to avoid hitting the negative integers. Use Hilbert's Hotel to make room for each of them, e.g.
$$ f(x) = begincases x + c & textif x=-(n+1)+mctext for some n,minmathbb N \ x & textotherwise endcases $$
for some appropriate constant $c$.
Which numbers would work as $c$?
answered Jul 29 at 17:54
Henning Makholm
225k16290516
answered Jul 29 at 17:54
Henning Makholm
225k16290516
answered Jul 29 at 17:54
Henning Makholm
225k16290516
answered Jul 29 at 17:54
Henning Makholm
225k16290516
225k16290516
add a comment |Â
add a comment |Â
up vote
2
down vote
Let's start with the identity $f(x)=xquad forall xinmathbb R-mathbb Z$
$(mathbb R-mathbb Z)$ and $(mathbb R-mathbb N)$ differ by $mathbb
N_>0$.
So to have a bijection you must basically make every positive integer disappear!
But the black hole $mathbb N_>0mapstovarnothing$ is not a bijection.
Instead the idea is to hide them within the real numbers. We will send $1$ to a real which is not an integer, let say $phi(1)=frac 12$.
But location $frac 12$ is already occupied by $f(frac 12)=frac 12$, so we will push it deeper with $phi(frac 12)=frac 14$, and we have a conflict with $frac 14$ now, so we will push it also $phi(frac 14)=frac 18$ and so on...
Basically we shift the whole sequence $frac 12^kmapstofrac 12^k+1$ a step deeper, that is freeing a hole in $frac 12$ where we can put $1$.
Now where to put $2$ ?
Similarly we can shift the sequence of all $frac 13^k$, freeing a hole in $frac 13$ and place $2$ there $phi(2)=frac 13$.
On so on, we shift the sequences of powers of primes numbers one step deeper $phi(frac 1p_n^k)mapstofrac 1p_n^k+1$ this frees a hole in $frac 1p_n$ where we can put $n$, i.e $phi(n)=frac 1p_n$
This is only one example. For instance you can shift the sequences $n+frac 12^k$ instead and send $n$ to the center of interval $]n,n+1[$ instead of using prime numbers.
More sophisticated sequences are possible, like the one by J.C.Santos for instance, but the principle is the same.
answered Jul 29 at 17:35
zwim
11k627
add a comment |Â
up vote
2
down vote
Let's start with the identity $f(x)=xquad forall xinmathbb R-mathbb Z$
$(mathbb R-mathbb Z)$ and $(mathbb R-mathbb N)$ differ by $mathbb
N_>0$.
So to have a bijection you must basically make every positive integer disappear!
But the black hole $mathbb N_>0mapstovarnothing$ is not a bijection.
Instead the idea is to hide them within the real numbers. We will send $1$ to a real which is not an integer, let say $phi(1)=frac 12$.
But location $frac 12$ is already occupied by $f(frac 12)=frac 12$, so we will push it deeper with $phi(frac 12)=frac 14$, and we have a conflict with $frac 14$ now, so we will push it also $phi(frac 14)=frac 18$ and so on...
Basically we shift the whole sequence $frac 12^kmapstofrac 12^k+1$ a step deeper, that is freeing a hole in $frac 12$ where we can put $1$.
Now where to put $2$ ?
Similarly we can shift the sequence of all $frac 13^k$, freeing a hole in $frac 13$ and place $2$ there $phi(2)=frac 13$.
On so on, we shift the sequences of powers of primes numbers one step deeper $phi(frac 1p_n^k)mapstofrac 1p_n^k+1$ this frees a hole in $frac 1p_n$ where we can put $n$, i.e $phi(n)=frac 1p_n$
This is only one example. For instance you can shift the sequences $n+frac 12^k$ instead and send $n$ to the center of interval $]n,n+1[$ instead of using prime numbers.
More sophisticated sequences are possible, like the one by J.C.Santos for instance, but the principle is the same.
answered Jul 29 at 17:35
zwim
11k627
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let's start with the identity $f(x)=xquad forall xinmathbb R-mathbb Z$
$(mathbb R-mathbb Z)$ and $(mathbb R-mathbb N)$ differ by $mathbb
N_>0$.
So to have a bijection you must basically make every positive integer disappear!
But the black hole $mathbb N_>0mapstovarnothing$ is not a bijection.
Instead the idea is to hide them within the real numbers. We will send $1$ to a real which is not an integer, let say $phi(1)=frac 12$.
But location $frac 12$ is already occupied by $f(frac 12)=frac 12$, so we will push it deeper with $phi(frac 12)=frac 14$, and we have a conflict with $frac 14$ now, so we will push it also $phi(frac 14)=frac 18$ and so on...
Basically we shift the whole sequence $frac 12^kmapstofrac 12^k+1$ a step deeper, that is freeing a hole in $frac 12$ where we can put $1$.
Now where to put $2$ ?
Similarly we can shift the sequence of all $frac 13^k$, freeing a hole in $frac 13$ and place $2$ there $phi(2)=frac 13$.
On so on, we shift the sequences of powers of primes numbers one step deeper $phi(frac 1p_n^k)mapstofrac 1p_n^k+1$ this frees a hole in $frac 1p_n$ where we can put $n$, i.e $phi(n)=frac 1p_n$
This is only one example. For instance you can shift the sequences $n+frac 12^k$ instead and send $n$ to the center of interval $]n,n+1[$ instead of using prime numbers.
More sophisticated sequences are possible, like the one by J.C.Santos for instance, but the principle is the same.
answered Jul 29 at 17:35
zwim
11k627
Let's start with the identity $f(x)=xquad forall xinmathbb R-mathbb Z$
$(mathbb R-mathbb Z)$ and $(mathbb R-mathbb N)$ differ by $mathbb
N_>0$.
So to have a bijection you must basically make every positive integer disappear!
But the black hole $mathbb N_>0mapstovarnothing$ is not a bijection.
Instead the idea is to hide them within the real numbers. We will send $1$ to a real which is not an integer, let say $phi(1)=frac 12$.
But location $frac 12$ is already occupied by $f(frac 12)=frac 12$, so we will push it deeper with $phi(frac 12)=frac 14$, and we have a conflict with $frac 14$ now, so we will push it also $phi(frac 14)=frac 18$ and so on...
Basically we shift the whole sequence $frac 12^kmapstofrac 12^k+1$ a step deeper, that is freeing a hole in $frac 12$ where we can put $1$.
Now where to put $2$ ?
Similarly we can shift the sequence of all $frac 13^k$, freeing a hole in $frac 13$ and place $2$ there $phi(2)=frac 13$.
On so on, we shift the sequences of powers of primes numbers one step deeper $phi(frac 1p_n^k)mapstofrac 1p_n^k+1$ this frees a hole in $frac 1p_n$ where we can put $n$, i.e $phi(n)=frac 1p_n$
This is only one example. For instance you can shift the sequences $n+frac 12^k$ instead and send $n$ to the center of interval $]n,n+1[$ instead of using prime numbers.
More sophisticated sequences are possible, like the one by J.C.Santos for instance, but the principle is the same.
answered Jul 29 at 17:35
zwim
11k627
edited Jul 29 at 18:10
answered Jul 29 at 17:35
zwim
11k627
answered Jul 29 at 17:35
zwim
11k627
answered Jul 29 at 17:35
zwim
11k627
11k627
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You can define a bijection $psicolonmathbbRsetminusmathbbNlongrightarrowmathbbRsetminusmathbb Z$ defining $psi(x)=x$ if $xnotinleft,ninmathbb Nright$ and then:
- $psi(0)=frac12$;
- $psi(-1)=frac14$;
- $psi(-2)=frac16$
and so on. Besides
- $psileft(frac12right)=frac13$;
- $psileft(frac13right)=frac15$;
- $psileft(frac14right)=frac17$
and so on.
answered Jul 29 at 17:38
José Carlos Santos
112k1696173
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up vote
1
down vote
You can define a bijection $psicolonmathbbRsetminusmathbbNlongrightarrowmathbbRsetminusmathbb Z$ defining $psi(x)=x$ if $xnotinleft,ninmathbb Nright$ and then:
- $psi(0)=frac12$;
- $psi(-1)=frac14$;
- $psi(-2)=frac16$
and so on. Besides
- $psileft(frac12right)=frac13$;
- $psileft(frac13right)=frac15$;
- $psileft(frac14right)=frac17$
and so on.
answered Jul 29 at 17:38
José Carlos Santos
112k1696173
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can define a bijection $psicolonmathbbRsetminusmathbbNlongrightarrowmathbbRsetminusmathbb Z$ defining $psi(x)=x$ if $xnotinleft,ninmathbb Nright$ and then:
- $psi(0)=frac12$;
- $psi(-1)=frac14$;
- $psi(-2)=frac16$
and so on. Besides
- $psileft(frac12right)=frac13$;
- $psileft(frac13right)=frac15$;
- $psileft(frac14right)=frac17$
and so on.
answered Jul 29 at 17:38
José Carlos Santos
112k1696173
You can define a bijection $psicolonmathbbRsetminusmathbbNlongrightarrowmathbbRsetminusmathbb Z$ defining $psi(x)=x$ if $xnotinleft,ninmathbb Nright$ and then:
- $psi(0)=frac12$;
- $psi(-1)=frac14$;
- $psi(-2)=frac16$
and so on. Besides
- $psileft(frac12right)=frac13$;
- $psileft(frac13right)=frac15$;
- $psileft(frac14right)=frac17$
and so on.
answered Jul 29 at 17:38
José Carlos Santos
112k1696173
edited Jul 29 at 17:46
answered Jul 29 at 17:38
José Carlos Santos
112k1696173
answered Jul 29 at 17:38
José Carlos Santos
112k1696173
answered Jul 29 at 17:38
José Carlos Santos
112k1696173
112k1696173
add a comment |Â
add a comment |Â
up vote
0
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Rocket man commented that "Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way "
So $mathbb Rsetminus mathbb Z = cup_zin mathbb Z I_z$ where $I_z = (z-1,z)$
And $mathbb Rsetminus mathbb N = cup_nin mathbb N J_n$ where $J_n = (n-1, n)$ for $n>1$ and $J_1 = (-infty, 1)$.
You found a bijection $phi: mathbb N to mathbb Z$.
Now if you can find a sequence of bijections $psi_n: J_nto I_phi(n)$ we'd be done: We'd just need to define $psi:mathbb Rsetminus mathbb Nto mathbb Rsetminus mathbb Z$ as: $psi(x) = psi_n(x)$ when $x in J_n$.
....
to spell it out:
Let $psi: mathbb Rsetminus Nto mathbb Rsetminus Z$ be defined as follows:
If $x in mathbb Rsetminus N$ and $x > 1$ then there exists an $n$ so that $n-1 < x < n$. For this $x$ let $psi (x) = x - n + phi(n)$. Now $phi(n)-1< psi(x) < phi(n)$ so $psi(x) in Rsetminus Z$.
If $x in mathbb Rsetminus N$ and $x le 1$ then $x < 1$. So $x-1 < 0$ and $1-x > 0$ and $0 < frac 11-x < 1$ and $-1< frac 11-x-1< 0$. For this $x$ let $psi(x) = frac 11-x-1$. Now $-1 < psi(x) < 0$ so $psi(x) in Rsetminus Z$
We can verify this is a bijection:
It's one to one: If $xne y$. $x < 1$ and $y < 1$ the $psi(x) = frac 11-x$ and $psi (y) = frac 11-x$. $x= yiff 1-x = 1-y iff frac 11-x= frac 11-y$ so $psi(x) ne psi (y)$.
If one of $x$ or $y$ (wolog $x$) is less than $1$ and the other (wolog $y$) is not then there is an $n>1$ so that $n-1 < y < n$. And $phi(n) ne 0$. so $phi(n)-1 < psi(y) < phi(n)$ so $psi(y) not in (-1,0)$. Whereas $psi(x)in (-1,0)$ so $psi(x) ne psi(y)$.
If $n- 1 < x<n$ and $m-1< y < m$ and $m ne n$, then $phi(n) ne phi(m)$. And $phi(n)-1 < psi(x)< phi(n)$ while $phi(m)-1 < psi(y)< phi(m)$. So $psi(x)ne psi(y)$.
Finally if $n-1<x,y< n$ then $x = y iff x-n+phi(n) = y-n+phi(n) iff psi(x) = psi (y)$. So $psi(x)ne psi(y)$.
$psi$ is onto: If $xin mathbb Rsetminus Z$ then there is a unique $m$ so that $m-1< x < m$. And there is a unique $n$ so that $phi(n) = m$.
If $n ne 1$ then $y =x-m + n$ will mean $n-1< x < n$ and $psi(y) = x$.
If $n = 1$ then $m = 0$ and $-1 < x < 0$. And $0 < x + 1< 1$ and $frac 1x+1 > 1$ and let $y=1-frac 1x+1 < 0 < 1$. then $psi(y) = x$.
answered Jul 29 at 19:47
fleablood
60.2k22575
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Rocket man commented that "Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way "
So $mathbb Rsetminus mathbb Z = cup_zin mathbb Z I_z$ where $I_z = (z-1,z)$
And $mathbb Rsetminus mathbb N = cup_nin mathbb N J_n$ where $J_n = (n-1, n)$ for $n>1$ and $J_1 = (-infty, 1)$.
You found a bijection $phi: mathbb N to mathbb Z$.
Now if you can find a sequence of bijections $psi_n: J_nto I_phi(n)$ we'd be done: We'd just need to define $psi:mathbb Rsetminus mathbb Nto mathbb Rsetminus mathbb Z$ as: $psi(x) = psi_n(x)$ when $x in J_n$.
....
to spell it out:
Let $psi: mathbb Rsetminus Nto mathbb Rsetminus Z$ be defined as follows:
If $x in mathbb Rsetminus N$ and $x > 1$ then there exists an $n$ so that $n-1 < x < n$. For this $x$ let $psi (x) = x - n + phi(n)$. Now $phi(n)-1< psi(x) < phi(n)$ so $psi(x) in Rsetminus Z$.
If $x in mathbb Rsetminus N$ and $x le 1$ then $x < 1$. So $x-1 < 0$ and $1-x > 0$ and $0 < frac 11-x < 1$ and $-1< frac 11-x-1< 0$. For this $x$ let $psi(x) = frac 11-x-1$. Now $-1 < psi(x) < 0$ so $psi(x) in Rsetminus Z$
We can verify this is a bijection:
It's one to one: If $xne y$. $x < 1$ and $y < 1$ the $psi(x) = frac 11-x$ and $psi (y) = frac 11-x$. $x= yiff 1-x = 1-y iff frac 11-x= frac 11-y$ so $psi(x) ne psi (y)$.
If one of $x$ or $y$ (wolog $x$) is less than $1$ and the other (wolog $y$) is not then there is an $n>1$ so that $n-1 < y < n$. And $phi(n) ne 0$. so $phi(n)-1 < psi(y) < phi(n)$ so $psi(y) not in (-1,0)$. Whereas $psi(x)in (-1,0)$ so $psi(x) ne psi(y)$.
If $n- 1 < x<n$ and $m-1< y < m$ and $m ne n$, then $phi(n) ne phi(m)$. And $phi(n)-1 < psi(x)< phi(n)$ while $phi(m)-1 < psi(y)< phi(m)$. So $psi(x)ne psi(y)$.
Finally if $n-1<x,y< n$ then $x = y iff x-n+phi(n) = y-n+phi(n) iff psi(x) = psi (y)$. So $psi(x)ne psi(y)$.
$psi$ is onto: If $xin mathbb Rsetminus Z$ then there is a unique $m$ so that $m-1< x < m$. And there is a unique $n$ so that $phi(n) = m$.
If $n ne 1$ then $y =x-m + n$ will mean $n-1< x < n$ and $psi(y) = x$.
If $n = 1$ then $m = 0$ and $-1 < x < 0$. And $0 < x + 1< 1$ and $frac 1x+1 > 1$ and let $y=1-frac 1x+1 < 0 < 1$. then $psi(y) = x$.
answered Jul 29 at 19:47
fleablood
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Rocket man commented that "Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way "
So $mathbb Rsetminus mathbb Z = cup_zin mathbb Z I_z$ where $I_z = (z-1,z)$
And $mathbb Rsetminus mathbb N = cup_nin mathbb N J_n$ where $J_n = (n-1, n)$ for $n>1$ and $J_1 = (-infty, 1)$.
You found a bijection $phi: mathbb N to mathbb Z$.
Now if you can find a sequence of bijections $psi_n: J_nto I_phi(n)$ we'd be done: We'd just need to define $psi:mathbb Rsetminus mathbb Nto mathbb Rsetminus mathbb Z$ as: $psi(x) = psi_n(x)$ when $x in J_n$.
....
to spell it out:
Let $psi: mathbb Rsetminus Nto mathbb Rsetminus Z$ be defined as follows:
If $x in mathbb Rsetminus N$ and $x > 1$ then there exists an $n$ so that $n-1 < x < n$. For this $x$ let $psi (x) = x - n + phi(n)$. Now $phi(n)-1< psi(x) < phi(n)$ so $psi(x) in Rsetminus Z$.
If $x in mathbb Rsetminus N$ and $x le 1$ then $x < 1$. So $x-1 < 0$ and $1-x > 0$ and $0 < frac 11-x < 1$ and $-1< frac 11-x-1< 0$. For this $x$ let $psi(x) = frac 11-x-1$. Now $-1 < psi(x) < 0$ so $psi(x) in Rsetminus Z$
We can verify this is a bijection:
It's one to one: If $xne y$. $x < 1$ and $y < 1$ the $psi(x) = frac 11-x$ and $psi (y) = frac 11-x$. $x= yiff 1-x = 1-y iff frac 11-x= frac 11-y$ so $psi(x) ne psi (y)$.
If one of $x$ or $y$ (wolog $x$) is less than $1$ and the other (wolog $y$) is not then there is an $n>1$ so that $n-1 < y < n$. And $phi(n) ne 0$. so $phi(n)-1 < psi(y) < phi(n)$ so $psi(y) not in (-1,0)$. Whereas $psi(x)in (-1,0)$ so $psi(x) ne psi(y)$.
If $n- 1 < x<n$ and $m-1< y < m$ and $m ne n$, then $phi(n) ne phi(m)$. And $phi(n)-1 < psi(x)< phi(n)$ while $phi(m)-1 < psi(y)< phi(m)$. So $psi(x)ne psi(y)$.
Finally if $n-1<x,y< n$ then $x = y iff x-n+phi(n) = y-n+phi(n) iff psi(x) = psi (y)$. So $psi(x)ne psi(y)$.
$psi$ is onto: If $xin mathbb Rsetminus Z$ then there is a unique $m$ so that $m-1< x < m$. And there is a unique $n$ so that $phi(n) = m$.
If $n ne 1$ then $y =x-m + n$ will mean $n-1< x < n$ and $psi(y) = x$.
If $n = 1$ then $m = 0$ and $-1 < x < 0$. And $0 < x + 1< 1$ and $frac 1x+1 > 1$ and let $y=1-frac 1x+1 < 0 < 1$. then $psi(y) = x$.
answered Jul 29 at 19:47
fleablood
60.2k22575
Rocket man commented that "Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way "
So $mathbb Rsetminus mathbb Z = cup_zin mathbb Z I_z$ where $I_z = (z-1,z)$
And $mathbb Rsetminus mathbb N = cup_nin mathbb N J_n$ where $J_n = (n-1, n)$ for $n>1$ and $J_1 = (-infty, 1)$.
You found a bijection $phi: mathbb N to mathbb Z$.
Now if you can find a sequence of bijections $psi_n: J_nto I_phi(n)$ we'd be done: We'd just need to define $psi:mathbb Rsetminus mathbb Nto mathbb Rsetminus mathbb Z$ as: $psi(x) = psi_n(x)$ when $x in J_n$.
....
to spell it out:
Let $psi: mathbb Rsetminus Nto mathbb Rsetminus Z$ be defined as follows:
If $x in mathbb Rsetminus N$ and $x > 1$ then there exists an $n$ so that $n-1 < x < n$. For this $x$ let $psi (x) = x - n + phi(n)$. Now $phi(n)-1< psi(x) < phi(n)$ so $psi(x) in Rsetminus Z$.
If $x in mathbb Rsetminus N$ and $x le 1$ then $x < 1$. So $x-1 < 0$ and $1-x > 0$ and $0 < frac 11-x < 1$ and $-1< frac 11-x-1< 0$. For this $x$ let $psi(x) = frac 11-x-1$. Now $-1 < psi(x) < 0$ so $psi(x) in Rsetminus Z$
We can verify this is a bijection:
It's one to one: If $xne y$. $x < 1$ and $y < 1$ the $psi(x) = frac 11-x$ and $psi (y) = frac 11-x$. $x= yiff 1-x = 1-y iff frac 11-x= frac 11-y$ so $psi(x) ne psi (y)$.
If one of $x$ or $y$ (wolog $x$) is less than $1$ and the other (wolog $y$) is not then there is an $n>1$ so that $n-1 < y < n$. And $phi(n) ne 0$. so $phi(n)-1 < psi(y) < phi(n)$ so $psi(y) not in (-1,0)$. Whereas $psi(x)in (-1,0)$ so $psi(x) ne psi(y)$.
If $n- 1 < x<n$ and $m-1< y < m$ and $m ne n$, then $phi(n) ne phi(m)$. And $phi(n)-1 < psi(x)< phi(n)$ while $phi(m)-1 < psi(y)< phi(m)$. So $psi(x)ne psi(y)$.
Finally if $n-1<x,y< n$ then $x = y iff x-n+phi(n) = y-n+phi(n) iff psi(x) = psi (y)$. So $psi(x)ne psi(y)$.
$psi$ is onto: If $xin mathbb Rsetminus Z$ then there is a unique $m$ so that $m-1< x < m$. And there is a unique $n$ so that $phi(n) = m$.
If $n ne 1$ then $y =x-m + n$ will mean $n-1< x < n$ and $psi(y) = x$.
If $n = 1$ then $m = 0$ and $-1 < x < 0$. And $0 < x + 1< 1$ and $frac 1x+1 > 1$ and let $y=1-frac 1x+1 < 0 < 1$. then $psi(y) = x$.
answered Jul 29 at 19:47
fleablood
60.2k22575
answered Jul 29 at 19:47
fleablood
60.2k22575
answered Jul 29 at 19:47
fleablood
60.2k22575
answered Jul 29 at 19:47
fleablood
60.2k22575
60.2k22575
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5
Both of your sets are just a countable collection of disjoint intervals. Try thinking of your problem that way
– Rocket Man
Jul 29 at 17:27
1
Start with an identity map from reals to reals. Modify this map on the integers. Now drop the integers from the domain.
– Steve D
Jul 29 at 17:30
2
I don't know who you are addressing your comment to, but @RocketMan 's comment gives you the solution, pure and simple.
– Arnaud Mortier
Jul 29 at 17:31
6
Please don’t ask for help and then “Bro†prospective helpers. Thanks.
– Brian Tung
Jul 29 at 17:39
See here for a more general result.
– Jyrki Lahtonen
Jul 29 at 18:10