Is there a metric on $Bbb R$ based upon this measure of how close to odd and even a number is?
$begingroup$
Suppose a valuation $v(x)$ on $Bbb N_{>0}$ assigns $1$ if $x$ is even and $-1$ if $x$ is odd.
Let $v_2(x)$ be some natural or obvious extension of this valuation that continuously measures oddness vs evenness.
Is there a continuous extension of this valuation on $Bbb R$ that generates a metric space either on $(Bbb R,d)$, or $(Bbb R/Bbb Z,d)$?
Background, and my thoughts on the problem:
The motivation is Collatz-related.
My attempts:
A. I can come up with various continuous extensions of the valuation. Something like $v(x)=4(lfloor x/2rfloor-x/2)+1$, assigns a real number in the interval $[-1,1)$.
I was originally thinking of simply $d(x,y)=lvert v(x)-v(y)rvert$
But now I think $d(x,y)=lvert v(x-y)+1rvert$ yields a metric space on $Bbb R/2Bbb Z$
B. For a full metric space on $Bbb R$ my thoughts so far are: $v_2=frac{v_1+1}2$ assigns a valuation in $[0,1]$
But I'm not sure how to make a metric space. It would seem:
$d(x,0)=frac1{v_2}$ has a good chance but I'm struggling to consider how to treat numbers differing by an odd number, whose valuation would be infinite. A possible solution would seem to be to set their distance to zero $0$, but then the metric space would set all odd numbers equal.
C. Another option is a generalisation such as $v_2(x)=sin^2frac{picdot x}2$ which gives a continuous valuation for $xin Bbb [0,1]$ with the possible advantage of being differentiable throughout $Bbb R$, but again, I don't know how (or whether there's a way) to translate the valuation into a metric which would make $Bbb R$ a metric space.
metric-spaces
asked Dec 30 '18 at 21:14
user334732user334732
4,32211240
$endgroup$
|
show 1 more comment
$begingroup$
Suppose a valuation $v(x)$ on $Bbb N_{>0}$ assigns $1$ if $x$ is even and $-1$ if $x$ is odd.
Let $v_2(x)$ be some natural or obvious extension of this valuation that continuously measures oddness vs evenness.
Is there a continuous extension of this valuation on $Bbb R$ that generates a metric space either on $(Bbb R,d)$, or $(Bbb R/Bbb Z,d)$?
Background, and my thoughts on the problem:
The motivation is Collatz-related.
My attempts:
A. I can come up with various continuous extensions of the valuation. Something like $v(x)=4(lfloor x/2rfloor-x/2)+1$, assigns a real number in the interval $[-1,1)$.
I was originally thinking of simply $d(x,y)=lvert v(x)-v(y)rvert$
But now I think $d(x,y)=lvert v(x-y)+1rvert$ yields a metric space on $Bbb R/2Bbb Z$
B. For a full metric space on $Bbb R$ my thoughts so far are: $v_2=frac{v_1+1}2$ assigns a valuation in $[0,1]$
But I'm not sure how to make a metric space. It would seem:
$d(x,0)=frac1{v_2}$ has a good chance but I'm struggling to consider how to treat numbers differing by an odd number, whose valuation would be infinite. A possible solution would seem to be to set their distance to zero $0$, but then the metric space would set all odd numbers equal.
C. Another option is a generalisation such as $v_2(x)=sin^2frac{picdot x}2$ which gives a continuous valuation for $xin Bbb [0,1]$ with the possible advantage of being differentiable throughout $Bbb R$, but again, I don't know how (or whether there's a way) to translate the valuation into a metric which would make $Bbb R$ a metric space.
metric-spaces
asked Dec 30 '18 at 21:14
user334732user334732
4,32211240
$endgroup$
1
$begingroup$
What do you mean with the extension generating $mathbb R/mathbb Z,d$?
$endgroup$
– Wojowu
Dec 30 '18 at 21:16
$begingroup$
@Wojowu I mean the valuation naturally translates into a metric $d$ such that the given set is a metric space.
$endgroup$
– user334732
Dec 30 '18 at 21:21
$begingroup$
It seems that an extension must include the original space. So, if anything, it should be $mathbb{R}$ instead of $mathbb{R}setminusmathbb{Z}$.
$endgroup$
– D.B.
Dec 30 '18 at 21:24
$begingroup$
I don't have setminus, I have a quotient. I did this only because it was the best I can do because every odd integer has the same valuation but a metric space on all of $Bbb R$ would also be of interest.
$endgroup$
– user334732
Dec 30 '18 at 21:28
2
$begingroup$
The term "valuation" has a precise meaning in algebra and number theory. Or actually, unfortunately, two: "additive" vs. "multiplicative" valuations. One of the first beginners' tasks is to learn to switch between those two. -- Your $v$, however, violates at least one condition for either possible meaning, hence should not be called a valuation. It's just some function. Further, it is totally unclear how the metric is supposed to be related to that function $v$, hence what you are asking.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 20:52
|
show 1 more comment
$begingroup$
Suppose a valuation $v(x)$ on $Bbb N_{>0}$ assigns $1$ if $x$ is even and $-1$ if $x$ is odd.
Let $v_2(x)$ be some natural or obvious extension of this valuation that continuously measures oddness vs evenness.
Is there a continuous extension of this valuation on $Bbb R$ that generates a metric space either on $(Bbb R,d)$, or $(Bbb R/Bbb Z,d)$?
Background, and my thoughts on the problem:
The motivation is Collatz-related.
My attempts:
A. I can come up with various continuous extensions of the valuation. Something like $v(x)=4(lfloor x/2rfloor-x/2)+1$, assigns a real number in the interval $[-1,1)$.
I was originally thinking of simply $d(x,y)=lvert v(x)-v(y)rvert$
But now I think $d(x,y)=lvert v(x-y)+1rvert$ yields a metric space on $Bbb R/2Bbb Z$
B. For a full metric space on $Bbb R$ my thoughts so far are: $v_2=frac{v_1+1}2$ assigns a valuation in $[0,1]$
But I'm not sure how to make a metric space. It would seem:
$d(x,0)=frac1{v_2}$ has a good chance but I'm struggling to consider how to treat numbers differing by an odd number, whose valuation would be infinite. A possible solution would seem to be to set their distance to zero $0$, but then the metric space would set all odd numbers equal.
C. Another option is a generalisation such as $v_2(x)=sin^2frac{picdot x}2$ which gives a continuous valuation for $xin Bbb [0,1]$ with the possible advantage of being differentiable throughout $Bbb R$, but again, I don't know how (or whether there's a way) to translate the valuation into a metric which would make $Bbb R$ a metric space.
metric-spaces
asked Dec 30 '18 at 21:14
user334732user334732
4,32211240
$endgroup$
Suppose a valuation $v(x)$ on $Bbb N_{>0}$ assigns $1$ if $x$ is even and $-1$ if $x$ is odd.
Let $v_2(x)$ be some natural or obvious extension of this valuation that continuously measures oddness vs evenness.
Is there a continuous extension of this valuation on $Bbb R$ that generates a metric space either on $(Bbb R,d)$, or $(Bbb R/Bbb Z,d)$?
Background, and my thoughts on the problem:
The motivation is Collatz-related.
My attempts:
A. I can come up with various continuous extensions of the valuation. Something like $v(x)=4(lfloor x/2rfloor-x/2)+1$, assigns a real number in the interval $[-1,1)$.
I was originally thinking of simply $d(x,y)=lvert v(x)-v(y)rvert$
But now I think $d(x,y)=lvert v(x-y)+1rvert$ yields a metric space on $Bbb R/2Bbb Z$
B. For a full metric space on $Bbb R$ my thoughts so far are: $v_2=frac{v_1+1}2$ assigns a valuation in $[0,1]$
But I'm not sure how to make a metric space. It would seem:
$d(x,0)=frac1{v_2}$ has a good chance but I'm struggling to consider how to treat numbers differing by an odd number, whose valuation would be infinite. A possible solution would seem to be to set their distance to zero $0$, but then the metric space would set all odd numbers equal.
C. Another option is a generalisation such as $v_2(x)=sin^2frac{picdot x}2$ which gives a continuous valuation for $xin Bbb [0,1]$ with the possible advantage of being differentiable throughout $Bbb R$, but again, I don't know how (or whether there's a way) to translate the valuation into a metric which would make $Bbb R$ a metric space.
metric-spaces
metric-spaces
asked Dec 30 '18 at 21:14
user334732user334732
4,32211240
asked Dec 30 '18 at 21:14
user334732user334732
4,32211240
edited Dec 31 '18 at 10:58
user334732
asked Dec 30 '18 at 21:14
user334732user334732
4,32211240
asked Dec 30 '18 at 21:14
user334732user334732
4,32211240
asked Dec 30 '18 at 21:14
user334732user334732
4,32211240
4,32211240
1
$begingroup$
What do you mean with the extension generating $mathbb R/mathbb Z,d$?
$endgroup$
– Wojowu
Dec 30 '18 at 21:16
$begingroup$
@Wojowu I mean the valuation naturally translates into a metric $d$ such that the given set is a metric space.
$endgroup$
– user334732
Dec 30 '18 at 21:21
$begingroup$
It seems that an extension must include the original space. So, if anything, it should be $mathbb{R}$ instead of $mathbb{R}setminusmathbb{Z}$.
$endgroup$
– D.B.
Dec 30 '18 at 21:24
$begingroup$
I don't have setminus, I have a quotient. I did this only because it was the best I can do because every odd integer has the same valuation but a metric space on all of $Bbb R$ would also be of interest.
$endgroup$
– user334732
Dec 30 '18 at 21:28
2
$begingroup$
The term "valuation" has a precise meaning in algebra and number theory. Or actually, unfortunately, two: "additive" vs. "multiplicative" valuations. One of the first beginners' tasks is to learn to switch between those two. -- Your $v$, however, violates at least one condition for either possible meaning, hence should not be called a valuation. It's just some function. Further, it is totally unclear how the metric is supposed to be related to that function $v$, hence what you are asking.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 20:52
|
show 1 more comment
1
$begingroup$
What do you mean with the extension generating $mathbb R/mathbb Z,d$?
$endgroup$
– Wojowu
Dec 30 '18 at 21:16
$begingroup$
@Wojowu I mean the valuation naturally translates into a metric $d$ such that the given set is a metric space.
$endgroup$
– user334732
Dec 30 '18 at 21:21
$begingroup$
It seems that an extension must include the original space. So, if anything, it should be $mathbb{R}$ instead of $mathbb{R}setminusmathbb{Z}$.
$endgroup$
– D.B.
Dec 30 '18 at 21:24
$begingroup$
I don't have setminus, I have a quotient. I did this only because it was the best I can do because every odd integer has the same valuation but a metric space on all of $Bbb R$ would also be of interest.
$endgroup$
– user334732
Dec 30 '18 at 21:28
2
$begingroup$
The term "valuation" has a precise meaning in algebra and number theory. Or actually, unfortunately, two: "additive" vs. "multiplicative" valuations. One of the first beginners' tasks is to learn to switch between those two. -- Your $v$, however, violates at least one condition for either possible meaning, hence should not be called a valuation. It's just some function. Further, it is totally unclear how the metric is supposed to be related to that function $v$, hence what you are asking.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 20:52
1
1
$begingroup$
What do you mean with the extension generating $mathbb R/mathbb Z,d$?
$endgroup$
– Wojowu
Dec 30 '18 at 21:16
$begingroup$
What do you mean with the extension generating $mathbb R/mathbb Z,d$?
$endgroup$
– Wojowu
Dec 30 '18 at 21:16
$begingroup$
@Wojowu I mean the valuation naturally translates into a metric $d$ such that the given set is a metric space.
$endgroup$
– user334732
Dec 30 '18 at 21:21
$begingroup$
@Wojowu I mean the valuation naturally translates into a metric $d$ such that the given set is a metric space.
$endgroup$
– user334732
Dec 30 '18 at 21:21
$begingroup$
It seems that an extension must include the original space. So, if anything, it should be $mathbb{R}$ instead of $mathbb{R}setminusmathbb{Z}$.
$endgroup$
– D.B.
Dec 30 '18 at 21:24
$begingroup$
It seems that an extension must include the original space. So, if anything, it should be $mathbb{R}$ instead of $mathbb{R}setminusmathbb{Z}$.
$endgroup$
– D.B.
Dec 30 '18 at 21:24
$begingroup$
I don't have setminus, I have a quotient. I did this only because it was the best I can do because every odd integer has the same valuation but a metric space on all of $Bbb R$ would also be of interest.
$endgroup$
– user334732
Dec 30 '18 at 21:28
$begingroup$
I don't have setminus, I have a quotient. I did this only because it was the best I can do because every odd integer has the same valuation but a metric space on all of $Bbb R$ would also be of interest.
$endgroup$
– user334732
Dec 30 '18 at 21:28
2
2
$begingroup$
The term "valuation" has a precise meaning in algebra and number theory. Or actually, unfortunately, two: "additive" vs. "multiplicative" valuations. One of the first beginners' tasks is to learn to switch between those two. -- Your $v$, however, violates at least one condition for either possible meaning, hence should not be called a valuation. It's just some function. Further, it is totally unclear how the metric is supposed to be related to that function $v$, hence what you are asking.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 20:52
$begingroup$
The term "valuation" has a precise meaning in algebra and number theory. Or actually, unfortunately, two: "additive" vs. "multiplicative" valuations. One of the first beginners' tasks is to learn to switch between those two. -- Your $v$, however, violates at least one condition for either possible meaning, hence should not be called a valuation. It's just some function. Further, it is totally unclear how the metric is supposed to be related to that function $v$, hence what you are asking.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 20:52
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Given a function $v$, the two-variable function $d(x,y) = |v(x)-v(y)|$ can't possibly be a metric unless $v$ is injective (since otherwise there will be distinct points at "distance" $0$ from each other). Since you want $v$ to take the same value on all integers of a given parity, no extension of this function can yield a metric under your definition.
answered Dec 30 '18 at 21:24
Greg MartinGreg Martin
36.8k23565
$endgroup$
$begingroup$
You mean no metric on $Bbb R$? But it's possible on $Bbb R/2Bbb Z$, right?
$endgroup$
– user334732
Dec 30 '18 at 21:39
$begingroup$
@Dog_69 I was interested in all of $Bbb R$ but I couldn't make that work, then I thought a compromise would be $Bbb R/Bbb Z$, then what I worked out as a possible answer to my own question looks like it works for $Bbb R/2Bbb Z$
$endgroup$
– user334732
Dec 30 '18 at 22:04
$begingroup$
The 2-adic valuation doesn't inject, but that yields a metric space with no distinct points at distance $0$ from each other.
$endgroup$
– user334732
Dec 31 '18 at 10:29
$begingroup$
@user334732: Of course the $2$-adic metric is not induced by the $2$-adic valuation $v_2$ via $d(x,y) stackrel{?}= vert v_2(x)-v_2(y)vert$, but via $d(x,y)=2^{-v_2(x-y)}$.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 20:40
$begingroup$
@TorstenSchoeneberg yes, I already know that. I don't disagree with the first sentence in this answer. This question asks if a similar thing can be done with the valuation I give. Site policy requires me to give my own thoughts. I say in the question my own thoughts fail. This answer and your comment confirm my thoughts fail, but not that the question itself is impossible. My term 'extension' is vague and this answer interprets it more restrictively than I intended. My intended meaning would permit the kind of valuation to metric relationship you have described.
$endgroup$
– user334732
Jan 2 at 22:50
|
show 1 more comment
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given a function $v$, the two-variable function $d(x,y) = |v(x)-v(y)|$ can't possibly be a metric unless $v$ is injective (since otherwise there will be distinct points at "distance" $0$ from each other). Since you want $v$ to take the same value on all integers of a given parity, no extension of this function can yield a metric under your definition.
answered Dec 30 '18 at 21:24
Greg MartinGreg Martin
36.8k23565
$endgroup$
$begingroup$
You mean no metric on $Bbb R$? But it's possible on $Bbb R/2Bbb Z$, right?
$endgroup$
– user334732
Dec 30 '18 at 21:39
$begingroup$
@Dog_69 I was interested in all of $Bbb R$ but I couldn't make that work, then I thought a compromise would be $Bbb R/Bbb Z$, then what I worked out as a possible answer to my own question looks like it works for $Bbb R/2Bbb Z$
$endgroup$
– user334732
Dec 30 '18 at 22:04
$begingroup$
The 2-adic valuation doesn't inject, but that yields a metric space with no distinct points at distance $0$ from each other.
$endgroup$
– user334732
Dec 31 '18 at 10:29
$begingroup$
@user334732: Of course the $2$-adic metric is not induced by the $2$-adic valuation $v_2$ via $d(x,y) stackrel{?}= vert v_2(x)-v_2(y)vert$, but via $d(x,y)=2^{-v_2(x-y)}$.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 20:40
$begingroup$
@TorstenSchoeneberg yes, I already know that. I don't disagree with the first sentence in this answer. This question asks if a similar thing can be done with the valuation I give. Site policy requires me to give my own thoughts. I say in the question my own thoughts fail. This answer and your comment confirm my thoughts fail, but not that the question itself is impossible. My term 'extension' is vague and this answer interprets it more restrictively than I intended. My intended meaning would permit the kind of valuation to metric relationship you have described.
$endgroup$
– user334732
Jan 2 at 22:50
|
show 1 more comment
$begingroup$
Given a function $v$, the two-variable function $d(x,y) = |v(x)-v(y)|$ can't possibly be a metric unless $v$ is injective (since otherwise there will be distinct points at "distance" $0$ from each other). Since you want $v$ to take the same value on all integers of a given parity, no extension of this function can yield a metric under your definition.
answered Dec 30 '18 at 21:24
Greg MartinGreg Martin
36.8k23565
$endgroup$
$begingroup$
You mean no metric on $Bbb R$? But it's possible on $Bbb R/2Bbb Z$, right?
$endgroup$
– user334732
Dec 30 '18 at 21:39
$begingroup$
@Dog_69 I was interested in all of $Bbb R$ but I couldn't make that work, then I thought a compromise would be $Bbb R/Bbb Z$, then what I worked out as a possible answer to my own question looks like it works for $Bbb R/2Bbb Z$
$endgroup$
– user334732
Dec 30 '18 at 22:04
$begingroup$
The 2-adic valuation doesn't inject, but that yields a metric space with no distinct points at distance $0$ from each other.
$endgroup$
– user334732
Dec 31 '18 at 10:29
$begingroup$
@user334732: Of course the $2$-adic metric is not induced by the $2$-adic valuation $v_2$ via $d(x,y) stackrel{?}= vert v_2(x)-v_2(y)vert$, but via $d(x,y)=2^{-v_2(x-y)}$.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 20:40
$begingroup$
@TorstenSchoeneberg yes, I already know that. I don't disagree with the first sentence in this answer. This question asks if a similar thing can be done with the valuation I give. Site policy requires me to give my own thoughts. I say in the question my own thoughts fail. This answer and your comment confirm my thoughts fail, but not that the question itself is impossible. My term 'extension' is vague and this answer interprets it more restrictively than I intended. My intended meaning would permit the kind of valuation to metric relationship you have described.
$endgroup$
– user334732
Jan 2 at 22:50
|
show 1 more comment
$begingroup$
Given a function $v$, the two-variable function $d(x,y) = |v(x)-v(y)|$ can't possibly be a metric unless $v$ is injective (since otherwise there will be distinct points at "distance" $0$ from each other). Since you want $v$ to take the same value on all integers of a given parity, no extension of this function can yield a metric under your definition.
answered Dec 30 '18 at 21:24
Greg MartinGreg Martin
36.8k23565
$endgroup$
Given a function $v$, the two-variable function $d(x,y) = |v(x)-v(y)|$ can't possibly be a metric unless $v$ is injective (since otherwise there will be distinct points at "distance" $0$ from each other). Since you want $v$ to take the same value on all integers of a given parity, no extension of this function can yield a metric under your definition.
answered Dec 30 '18 at 21:24
Greg MartinGreg Martin
36.8k23565
answered Dec 30 '18 at 21:24
Greg MartinGreg Martin
36.8k23565
answered Dec 30 '18 at 21:24
Greg MartinGreg Martin
36.8k23565
answered Dec 30 '18 at 21:24
Greg MartinGreg Martin
36.8k23565
36.8k23565
$begingroup$
You mean no metric on $Bbb R$? But it's possible on $Bbb R/2Bbb Z$, right?
$endgroup$
– user334732
Dec 30 '18 at 21:39
$begingroup$
@Dog_69 I was interested in all of $Bbb R$ but I couldn't make that work, then I thought a compromise would be $Bbb R/Bbb Z$, then what I worked out as a possible answer to my own question looks like it works for $Bbb R/2Bbb Z$
$endgroup$
– user334732
Dec 30 '18 at 22:04
$begingroup$
The 2-adic valuation doesn't inject, but that yields a metric space with no distinct points at distance $0$ from each other.
$endgroup$
– user334732
Dec 31 '18 at 10:29
$begingroup$
@user334732: Of course the $2$-adic metric is not induced by the $2$-adic valuation $v_2$ via $d(x,y) stackrel{?}= vert v_2(x)-v_2(y)vert$, but via $d(x,y)=2^{-v_2(x-y)}$.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 20:40
$begingroup$
@TorstenSchoeneberg yes, I already know that. I don't disagree with the first sentence in this answer. This question asks if a similar thing can be done with the valuation I give. Site policy requires me to give my own thoughts. I say in the question my own thoughts fail. This answer and your comment confirm my thoughts fail, but not that the question itself is impossible. My term 'extension' is vague and this answer interprets it more restrictively than I intended. My intended meaning would permit the kind of valuation to metric relationship you have described.
$endgroup$
– user334732
Jan 2 at 22:50
|
show 1 more comment
$begingroup$
You mean no metric on $Bbb R$? But it's possible on $Bbb R/2Bbb Z$, right?
$endgroup$
– user334732
Dec 30 '18 at 21:39
$begingroup$
@Dog_69 I was interested in all of $Bbb R$ but I couldn't make that work, then I thought a compromise would be $Bbb R/Bbb Z$, then what I worked out as a possible answer to my own question looks like it works for $Bbb R/2Bbb Z$
$endgroup$
– user334732
Dec 30 '18 at 22:04
$begingroup$
The 2-adic valuation doesn't inject, but that yields a metric space with no distinct points at distance $0$ from each other.
$endgroup$
– user334732
Dec 31 '18 at 10:29
$begingroup$
@user334732: Of course the $2$-adic metric is not induced by the $2$-adic valuation $v_2$ via $d(x,y) stackrel{?}= vert v_2(x)-v_2(y)vert$, but via $d(x,y)=2^{-v_2(x-y)}$.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 20:40
$begingroup$
@TorstenSchoeneberg yes, I already know that. I don't disagree with the first sentence in this answer. This question asks if a similar thing can be done with the valuation I give. Site policy requires me to give my own thoughts. I say in the question my own thoughts fail. This answer and your comment confirm my thoughts fail, but not that the question itself is impossible. My term 'extension' is vague and this answer interprets it more restrictively than I intended. My intended meaning would permit the kind of valuation to metric relationship you have described.
$endgroup$
– user334732
Jan 2 at 22:50
$begingroup$
You mean no metric on $Bbb R$? But it's possible on $Bbb R/2Bbb Z$, right?
$endgroup$
– user334732
Dec 30 '18 at 21:39
$begingroup$
You mean no metric on $Bbb R$? But it's possible on $Bbb R/2Bbb Z$, right?
$endgroup$
– user334732
Dec 30 '18 at 21:39
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@Dog_69 I was interested in all of $Bbb R$ but I couldn't make that work, then I thought a compromise would be $Bbb R/Bbb Z$, then what I worked out as a possible answer to my own question looks like it works for $Bbb R/2Bbb Z$
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– user334732
Dec 30 '18 at 22:04
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@Dog_69 I was interested in all of $Bbb R$ but I couldn't make that work, then I thought a compromise would be $Bbb R/Bbb Z$, then what I worked out as a possible answer to my own question looks like it works for $Bbb R/2Bbb Z$
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– user334732
Dec 30 '18 at 22:04
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The 2-adic valuation doesn't inject, but that yields a metric space with no distinct points at distance $0$ from each other.
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– user334732
Dec 31 '18 at 10:29
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The 2-adic valuation doesn't inject, but that yields a metric space with no distinct points at distance $0$ from each other.
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– user334732
Dec 31 '18 at 10:29
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@user334732: Of course the $2$-adic metric is not induced by the $2$-adic valuation $v_2$ via $d(x,y) stackrel{?}= vert v_2(x)-v_2(y)vert$, but via $d(x,y)=2^{-v_2(x-y)}$.
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– Torsten Schoeneberg
Jan 2 at 20:40
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@user334732: Of course the $2$-adic metric is not induced by the $2$-adic valuation $v_2$ via $d(x,y) stackrel{?}= vert v_2(x)-v_2(y)vert$, but via $d(x,y)=2^{-v_2(x-y)}$.
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– Torsten Schoeneberg
Jan 2 at 20:40
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@TorstenSchoeneberg yes, I already know that. I don't disagree with the first sentence in this answer. This question asks if a similar thing can be done with the valuation I give. Site policy requires me to give my own thoughts. I say in the question my own thoughts fail. This answer and your comment confirm my thoughts fail, but not that the question itself is impossible. My term 'extension' is vague and this answer interprets it more restrictively than I intended. My intended meaning would permit the kind of valuation to metric relationship you have described.
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– user334732
Jan 2 at 22:50
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@TorstenSchoeneberg yes, I already know that. I don't disagree with the first sentence in this answer. This question asks if a similar thing can be done with the valuation I give. Site policy requires me to give my own thoughts. I say in the question my own thoughts fail. This answer and your comment confirm my thoughts fail, but not that the question itself is impossible. My term 'extension' is vague and this answer interprets it more restrictively than I intended. My intended meaning would permit the kind of valuation to metric relationship you have described.
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– user334732
Jan 2 at 22:50
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What do you mean with the extension generating $mathbb R/mathbb Z,d$?
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– Wojowu
Dec 30 '18 at 21:16
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@Wojowu I mean the valuation naturally translates into a metric $d$ such that the given set is a metric space.
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– user334732
Dec 30 '18 at 21:21
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It seems that an extension must include the original space. So, if anything, it should be $mathbb{R}$ instead of $mathbb{R}setminusmathbb{Z}$.
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– D.B.
Dec 30 '18 at 21:24
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I don't have setminus, I have a quotient. I did this only because it was the best I can do because every odd integer has the same valuation but a metric space on all of $Bbb R$ would also be of interest.
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– user334732
Dec 30 '18 at 21:28
2
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The term "valuation" has a precise meaning in algebra and number theory. Or actually, unfortunately, two: "additive" vs. "multiplicative" valuations. One of the first beginners' tasks is to learn to switch between those two. -- Your $v$, however, violates at least one condition for either possible meaning, hence should not be called a valuation. It's just some function. Further, it is totally unclear how the metric is supposed to be related to that function $v$, hence what you are asking.
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– Torsten Schoeneberg
Jan 2 at 20:52