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What's $mathbbR$ doing in the definition of a metric space?
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A metric space is often defined as a set $X$ along with a mapping $d: X^2rightarrowmathbb R$ obeying the following identities: beginalign*
&d(x,y)=0 Leftrightarrow x=y, \
&d(x,y)=d(y,x), \
&d(x,z) leq d(x,y)+d(y,z).
endalign*
This definition demands that $d$ map into $mathbb R$. I find this interesting because only a few properties of $mathbb R$ are necessary to make those three identities make sense $-$ presumably you could define something analogous to a metric space by replacing $mathbb R$ with another set $A$, so long as $A$ has some sort of order, some notion of addition, and an additive identity $0$.
What sort of things are lost in such a generalization? Surely something falls apart if $A$ does not have a total order. Additionally, I imagine the completeness of $mathbb R$ is important as well. But I don't have any specific examples. And I cannot really figure out why we need the field $mathbb R$ rather than just some group or ring.
So my question is: what attributes of metric spaces depend on $d$ mapping into $mathbb R$ rather than another, possibly quite similar, space? (e.g., ordered fields that are not complete; or complete totally ordered groups.) While there are some questions on the site discussing generalizations along the lines I've described (such as this one) they focus on the aspects of metric spaces that don't depend on $mathbb R$, whereas I'm interested in the ones that do.
general-topology metric-spaces
asked Jul 19 at 15:27
namsos
806
 |Â
show 2 more comments
up vote
7
down vote
favorite
A metric space is often defined as a set $X$ along with a mapping $d: X^2rightarrowmathbb R$ obeying the following identities: beginalign*
&d(x,y)=0 Leftrightarrow x=y, \
&d(x,y)=d(y,x), \
&d(x,z) leq d(x,y)+d(y,z).
endalign*
This definition demands that $d$ map into $mathbb R$. I find this interesting because only a few properties of $mathbb R$ are necessary to make those three identities make sense $-$ presumably you could define something analogous to a metric space by replacing $mathbb R$ with another set $A$, so long as $A$ has some sort of order, some notion of addition, and an additive identity $0$.
What sort of things are lost in such a generalization? Surely something falls apart if $A$ does not have a total order. Additionally, I imagine the completeness of $mathbb R$ is important as well. But I don't have any specific examples. And I cannot really figure out why we need the field $mathbb R$ rather than just some group or ring.
So my question is: what attributes of metric spaces depend on $d$ mapping into $mathbb R$ rather than another, possibly quite similar, space? (e.g., ordered fields that are not complete; or complete totally ordered groups.) While there are some questions on the site discussing generalizations along the lines I've described (such as this one) they focus on the aspects of metric spaces that don't depend on $mathbb R$, whereas I'm interested in the ones that do.
general-topology metric-spaces
asked Jul 19 at 15:27
namsos
806
6
You need an ordered Abelian group. You want to read about "valuations".
– Asaf Karagila♦
Jul 19 at 15:30
1
The image of the discrete metric is $mathbbZ/2mathbbZ$.
– gandalf61
Jul 19 at 15:42
3
Note that you don't have to go too far in the properties of an ordered group to 'force' $mathbbR$ - being abelian (a property you certainly need), complete, and Archimedean is enough to do it.
– Steven Stadnicki
Jul 19 at 15:45
1
@gandalf61 I'm not sure I agree with you about the discrete metric. It seems to me that its image is better understood as $0,1subsetmathbb R$, with $1+1=2$. If we use "addition" in the sense of $mathbb Z/2mathbb Z$ I do not think the triangle inequality holds: we have $d(a,b)+d(b,c)=1+1=0<d(a,c)$ for all distinct $a,b,c$. But I don't know much about this topic, so perhaps I'm missing something.
– namsos
Jul 19 at 18:29
1
You might like this.
– Mike Miller
Jul 19 at 21:45
 |Â
show 2 more comments
up vote
7
down vote
favorite
up vote
7
down vote
favorite
A metric space is often defined as a set $X$ along with a mapping $d: X^2rightarrowmathbb R$ obeying the following identities: beginalign*
&d(x,y)=0 Leftrightarrow x=y, \
&d(x,y)=d(y,x), \
&d(x,z) leq d(x,y)+d(y,z).
endalign*
This definition demands that $d$ map into $mathbb R$. I find this interesting because only a few properties of $mathbb R$ are necessary to make those three identities make sense $-$ presumably you could define something analogous to a metric space by replacing $mathbb R$ with another set $A$, so long as $A$ has some sort of order, some notion of addition, and an additive identity $0$.
What sort of things are lost in such a generalization? Surely something falls apart if $A$ does not have a total order. Additionally, I imagine the completeness of $mathbb R$ is important as well. But I don't have any specific examples. And I cannot really figure out why we need the field $mathbb R$ rather than just some group or ring.
So my question is: what attributes of metric spaces depend on $d$ mapping into $mathbb R$ rather than another, possibly quite similar, space? (e.g., ordered fields that are not complete; or complete totally ordered groups.) While there are some questions on the site discussing generalizations along the lines I've described (such as this one) they focus on the aspects of metric spaces that don't depend on $mathbb R$, whereas I'm interested in the ones that do.
general-topology metric-spaces
asked Jul 19 at 15:27
namsos
806
A metric space is often defined as a set $X$ along with a mapping $d: X^2rightarrowmathbb R$ obeying the following identities: beginalign*
&d(x,y)=0 Leftrightarrow x=y, \
&d(x,y)=d(y,x), \
&d(x,z) leq d(x,y)+d(y,z).
endalign*
This definition demands that $d$ map into $mathbb R$. I find this interesting because only a few properties of $mathbb R$ are necessary to make those three identities make sense $-$ presumably you could define something analogous to a metric space by replacing $mathbb R$ with another set $A$, so long as $A$ has some sort of order, some notion of addition, and an additive identity $0$.
What sort of things are lost in such a generalization? Surely something falls apart if $A$ does not have a total order. Additionally, I imagine the completeness of $mathbb R$ is important as well. But I don't have any specific examples. And I cannot really figure out why we need the field $mathbb R$ rather than just some group or ring.
So my question is: what attributes of metric spaces depend on $d$ mapping into $mathbb R$ rather than another, possibly quite similar, space? (e.g., ordered fields that are not complete; or complete totally ordered groups.) While there are some questions on the site discussing generalizations along the lines I've described (such as this one) they focus on the aspects of metric spaces that don't depend on $mathbb R$, whereas I'm interested in the ones that do.
general-topology metric-spaces
asked Jul 19 at 15:27
namsos
806
asked Jul 19 at 15:27
namsos
806
asked Jul 19 at 15:27
namsos
806
asked Jul 19 at 15:27
namsos
806
806
6
You need an ordered Abelian group. You want to read about "valuations".
– Asaf Karagila♦
Jul 19 at 15:30
1
The image of the discrete metric is $mathbbZ/2mathbbZ$.
– gandalf61
Jul 19 at 15:42
3
Note that you don't have to go too far in the properties of an ordered group to 'force' $mathbbR$ - being abelian (a property you certainly need), complete, and Archimedean is enough to do it.
– Steven Stadnicki
Jul 19 at 15:45
1
@gandalf61 I'm not sure I agree with you about the discrete metric. It seems to me that its image is better understood as $0,1subsetmathbb R$, with $1+1=2$. If we use "addition" in the sense of $mathbb Z/2mathbb Z$ I do not think the triangle inequality holds: we have $d(a,b)+d(b,c)=1+1=0<d(a,c)$ for all distinct $a,b,c$. But I don't know much about this topic, so perhaps I'm missing something.
– namsos
Jul 19 at 18:29
1
You might like this.
– Mike Miller
Jul 19 at 21:45
 |Â
show 2 more comments
6
You need an ordered Abelian group. You want to read about "valuations".
– Asaf Karagila♦
Jul 19 at 15:30
1
The image of the discrete metric is $mathbbZ/2mathbbZ$.
– gandalf61
Jul 19 at 15:42
3
Note that you don't have to go too far in the properties of an ordered group to 'force' $mathbbR$ - being abelian (a property you certainly need), complete, and Archimedean is enough to do it.
– Steven Stadnicki
Jul 19 at 15:45
1
@gandalf61 I'm not sure I agree with you about the discrete metric. It seems to me that its image is better understood as $0,1subsetmathbb R$, with $1+1=2$. If we use "addition" in the sense of $mathbb Z/2mathbb Z$ I do not think the triangle inequality holds: we have $d(a,b)+d(b,c)=1+1=0<d(a,c)$ for all distinct $a,b,c$. But I don't know much about this topic, so perhaps I'm missing something.
– namsos
Jul 19 at 18:29
1
You might like this.
– Mike Miller
Jul 19 at 21:45
6
6
You need an ordered Abelian group. You want to read about "valuations".
– Asaf Karagila♦
Jul 19 at 15:30
You need an ordered Abelian group. You want to read about "valuations".
– Asaf Karagila♦
Jul 19 at 15:30
1
1
The image of the discrete metric is $mathbbZ/2mathbbZ$.
– gandalf61
Jul 19 at 15:42
The image of the discrete metric is $mathbbZ/2mathbbZ$.
– gandalf61
Jul 19 at 15:42
3
3
Note that you don't have to go too far in the properties of an ordered group to 'force' $mathbbR$ - being abelian (a property you certainly need), complete, and Archimedean is enough to do it.
– Steven Stadnicki
Jul 19 at 15:45
Note that you don't have to go too far in the properties of an ordered group to 'force' $mathbbR$ - being abelian (a property you certainly need), complete, and Archimedean is enough to do it.
– Steven Stadnicki
Jul 19 at 15:45
1
1
@gandalf61 I'm not sure I agree with you about the discrete metric. It seems to me that its image is better understood as $0,1subsetmathbb R$, with $1+1=2$. If we use "addition" in the sense of $mathbb Z/2mathbb Z$ I do not think the triangle inequality holds: we have $d(a,b)+d(b,c)=1+1=0<d(a,c)$ for all distinct $a,b,c$. But I don't know much about this topic, so perhaps I'm missing something.
– namsos
Jul 19 at 18:29
@gandalf61 I'm not sure I agree with you about the discrete metric. It seems to me that its image is better understood as $0,1subsetmathbb R$, with $1+1=2$. If we use "addition" in the sense of $mathbb Z/2mathbb Z$ I do not think the triangle inequality holds: we have $d(a,b)+d(b,c)=1+1=0<d(a,c)$ for all distinct $a,b,c$. But I don't know much about this topic, so perhaps I'm missing something.
– namsos
Jul 19 at 18:29
1
1
You might like this.
– Mike Miller
Jul 19 at 21:45
You might like this.
– Mike Miller
Jul 19 at 21:45
 |Â
show 2 more comments
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6
You need an ordered Abelian group. You want to read about "valuations".
– Asaf Karagila♦
Jul 19 at 15:30
1
The image of the discrete metric is $mathbbZ/2mathbbZ$.
– gandalf61
Jul 19 at 15:42
3
Note that you don't have to go too far in the properties of an ordered group to 'force' $mathbbR$ - being abelian (a property you certainly need), complete, and Archimedean is enough to do it.
– Steven Stadnicki
Jul 19 at 15:45
1
@gandalf61 I'm not sure I agree with you about the discrete metric. It seems to me that its image is better understood as $0,1subsetmathbb R$, with $1+1=2$. If we use "addition" in the sense of $mathbb Z/2mathbb Z$ I do not think the triangle inequality holds: we have $d(a,b)+d(b,c)=1+1=0<d(a,c)$ for all distinct $a,b,c$. But I don't know much about this topic, so perhaps I'm missing something.
– namsos
Jul 19 at 18:29
1
You might like this.
– Mike Miller
Jul 19 at 21:45