What is the structure preserved by strong equivalence of metrics?












14












$begingroup$


Let $d_1$ and $d_2$ be two metrics on the same set $X$. Then $d_1$ and $d_2$ are (topologically) equivalent if the identity maps $i:(M,d_1)rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)rightarrow(M,d_1)$ are continuous. $d_1$ and $d_2$ are uniformly equivalent if $i$ and $i^{-1}$ are uniformly continuous. And $d_1$ and $d_2$ are strongly equivalent if there exist constants $alpha,beta>0$ such that $alpha d_1(x,y)leq d_2(x,y)leqbeta d_1(x,y)$ for all $x,yin X$.



All three of these are equivalence relations, so we can take equivalence classes under each one. If we take equivalence classes of metrics under equivalence, we can identify each equivalence class with a topology on $X$. If we take equivalence classes of metrics under uniform equivalence, we can identify each equivalence class with a uniformity on $X$. But my question is, if we take equivalence classes of metrics under strong equivalence, what kind of structure in $X$ can we identify each equivalence class with?



Note that I'm looking for a structure that makes no reference to metrics, just as you can define topological spaces and uniform spaces with no reference to metrics.






general-topology metric-spaces category-theory quotient-spaces uniform-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @SmileyCraft Thanks, I fixed it.
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 4:56






  • 1




    $begingroup$
    They have the same metric-bornology $mathcal{B}$, with $B in mathcal{B}$ iff $exists p in X, exists r>0: B subseteq B(p,r)$. You know a theorem that tells you that a bornology is induced by a metric. Maybe combine those facts?
    $endgroup$
    – Henno Brandsma
    Dec 23 '18 at 6:09






  • 1




    $begingroup$
    @HennoBrandsma But two metrics which induce the same uniformity and the same bornology need not be strongly equivalent; see here: math.stackexchange.com/a/3050424/71829 So that means that there must be some structure beyond the uniformity and the bornology that is preserved by strong equivalence.
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 17:22






  • 1




    $begingroup$
    that example is somewhat trivialised as$[0,1]$ has but one uniformity and all sunsets are bounded.
    $endgroup$
    – Henno Brandsma
    Dec 23 '18 at 17:25






  • 3




    $begingroup$
    I do not think this object has any name. You can call it a bilipschitz structure. One can identify such a structure with the isomorphism class of some Banach algebra (if you like functional analysis), see msp.org/pjm/1963/13-4/pjm-v13-n4-p31-p.pdf.
    $endgroup$
    – Moishe Kohan
    Dec 27 '18 at 4:28
















14












$begingroup$


Let $d_1$ and $d_2$ be two metrics on the same set $X$. Then $d_1$ and $d_2$ are (topologically) equivalent if the identity maps $i:(M,d_1)rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)rightarrow(M,d_1)$ are continuous. $d_1$ and $d_2$ are uniformly equivalent if $i$ and $i^{-1}$ are uniformly continuous. And $d_1$ and $d_2$ are strongly equivalent if there exist constants $alpha,beta>0$ such that $alpha d_1(x,y)leq d_2(x,y)leqbeta d_1(x,y)$ for all $x,yin X$.



All three of these are equivalence relations, so we can take equivalence classes under each one. If we take equivalence classes of metrics under equivalence, we can identify each equivalence class with a topology on $X$. If we take equivalence classes of metrics under uniform equivalence, we can identify each equivalence class with a uniformity on $X$. But my question is, if we take equivalence classes of metrics under strong equivalence, what kind of structure in $X$ can we identify each equivalence class with?



Note that I'm looking for a structure that makes no reference to metrics, just as you can define topological spaces and uniform spaces with no reference to metrics.






general-topology metric-spaces category-theory quotient-spaces uniform-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @SmileyCraft Thanks, I fixed it.
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 4:56






  • 1




    $begingroup$
    They have the same metric-bornology $mathcal{B}$, with $B in mathcal{B}$ iff $exists p in X, exists r>0: B subseteq B(p,r)$. You know a theorem that tells you that a bornology is induced by a metric. Maybe combine those facts?
    $endgroup$
    – Henno Brandsma
    Dec 23 '18 at 6:09






  • 1




    $begingroup$
    @HennoBrandsma But two metrics which induce the same uniformity and the same bornology need not be strongly equivalent; see here: math.stackexchange.com/a/3050424/71829 So that means that there must be some structure beyond the uniformity and the bornology that is preserved by strong equivalence.
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 17:22






  • 1




    $begingroup$
    that example is somewhat trivialised as$[0,1]$ has but one uniformity and all sunsets are bounded.
    $endgroup$
    – Henno Brandsma
    Dec 23 '18 at 17:25






  • 3




    $begingroup$
    I do not think this object has any name. You can call it a bilipschitz structure. One can identify such a structure with the isomorphism class of some Banach algebra (if you like functional analysis), see msp.org/pjm/1963/13-4/pjm-v13-n4-p31-p.pdf.
    $endgroup$
    – Moishe Kohan
    Dec 27 '18 at 4:28














14












14








14


3



$begingroup$


Let $d_1$ and $d_2$ be two metrics on the same set $X$. Then $d_1$ and $d_2$ are (topologically) equivalent if the identity maps $i:(M,d_1)rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)rightarrow(M,d_1)$ are continuous. $d_1$ and $d_2$ are uniformly equivalent if $i$ and $i^{-1}$ are uniformly continuous. And $d_1$ and $d_2$ are strongly equivalent if there exist constants $alpha,beta>0$ such that $alpha d_1(x,y)leq d_2(x,y)leqbeta d_1(x,y)$ for all $x,yin X$.



All three of these are equivalence relations, so we can take equivalence classes under each one. If we take equivalence classes of metrics under equivalence, we can identify each equivalence class with a topology on $X$. If we take equivalence classes of metrics under uniform equivalence, we can identify each equivalence class with a uniformity on $X$. But my question is, if we take equivalence classes of metrics under strong equivalence, what kind of structure in $X$ can we identify each equivalence class with?



Note that I'm looking for a structure that makes no reference to metrics, just as you can define topological spaces and uniform spaces with no reference to metrics.






general-topology metric-spaces category-theory quotient-spaces uniform-spaces







share|cite|improve this question











$endgroup$




Let $d_1$ and $d_2$ be two metrics on the same set $X$. Then $d_1$ and $d_2$ are (topologically) equivalent if the identity maps $i:(M,d_1)rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)rightarrow(M,d_1)$ are continuous. $d_1$ and $d_2$ are uniformly equivalent if $i$ and $i^{-1}$ are uniformly continuous. And $d_1$ and $d_2$ are strongly equivalent if there exist constants $alpha,beta>0$ such that $alpha d_1(x,y)leq d_2(x,y)leqbeta d_1(x,y)$ for all $x,yin X$.



All three of these are equivalence relations, so we can take equivalence classes under each one. If we take equivalence classes of metrics under equivalence, we can identify each equivalence class with a topology on $X$. If we take equivalence classes of metrics under uniform equivalence, we can identify each equivalence class with a uniformity on $X$. But my question is, if we take equivalence classes of metrics under strong equivalence, what kind of structure in $X$ can we identify each equivalence class with?



Note that I'm looking for a structure that makes no reference to metrics, just as you can define topological spaces and uniform spaces with no reference to metrics.






general-topology metric-spaces category-theory quotient-spaces uniform-spaces




general-topology metric-spaces category-theory quotient-spaces uniform-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 19:12







Keshav Srinivasan

















asked Dec 23 '18 at 4:42









Keshav SrinivasanKeshav Srinivasan

2,49921547




2,49921547








  • 1




    $begingroup$
    @SmileyCraft Thanks, I fixed it.
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 4:56






  • 1




    $begingroup$
    They have the same metric-bornology $mathcal{B}$, with $B in mathcal{B}$ iff $exists p in X, exists r>0: B subseteq B(p,r)$. You know a theorem that tells you that a bornology is induced by a metric. Maybe combine those facts?
    $endgroup$
    – Henno Brandsma
    Dec 23 '18 at 6:09






  • 1




    $begingroup$
    @HennoBrandsma But two metrics which induce the same uniformity and the same bornology need not be strongly equivalent; see here: math.stackexchange.com/a/3050424/71829 So that means that there must be some structure beyond the uniformity and the bornology that is preserved by strong equivalence.
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 17:22






  • 1




    $begingroup$
    that example is somewhat trivialised as$[0,1]$ has but one uniformity and all sunsets are bounded.
    $endgroup$
    – Henno Brandsma
    Dec 23 '18 at 17:25






  • 3




    $begingroup$
    I do not think this object has any name. You can call it a bilipschitz structure. One can identify such a structure with the isomorphism class of some Banach algebra (if you like functional analysis), see msp.org/pjm/1963/13-4/pjm-v13-n4-p31-p.pdf.
    $endgroup$
    – Moishe Kohan
    Dec 27 '18 at 4:28














  • 1




    $begingroup$
    @SmileyCraft Thanks, I fixed it.
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 4:56






  • 1




    $begingroup$
    They have the same metric-bornology $mathcal{B}$, with $B in mathcal{B}$ iff $exists p in X, exists r>0: B subseteq B(p,r)$. You know a theorem that tells you that a bornology is induced by a metric. Maybe combine those facts?
    $endgroup$
    – Henno Brandsma
    Dec 23 '18 at 6:09






  • 1




    $begingroup$
    @HennoBrandsma But two metrics which induce the same uniformity and the same bornology need not be strongly equivalent; see here: math.stackexchange.com/a/3050424/71829 So that means that there must be some structure beyond the uniformity and the bornology that is preserved by strong equivalence.
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 17:22






  • 1




    $begingroup$
    that example is somewhat trivialised as$[0,1]$ has but one uniformity and all sunsets are bounded.
    $endgroup$
    – Henno Brandsma
    Dec 23 '18 at 17:25






  • 3




    $begingroup$
    I do not think this object has any name. You can call it a bilipschitz structure. One can identify such a structure with the isomorphism class of some Banach algebra (if you like functional analysis), see msp.org/pjm/1963/13-4/pjm-v13-n4-p31-p.pdf.
    $endgroup$
    – Moishe Kohan
    Dec 27 '18 at 4:28








1




1




$begingroup$
@SmileyCraft Thanks, I fixed it.
$endgroup$
– Keshav Srinivasan
Dec 23 '18 at 4:56




$begingroup$
@SmileyCraft Thanks, I fixed it.
$endgroup$
– Keshav Srinivasan
Dec 23 '18 at 4:56




1




1




$begingroup$
They have the same metric-bornology $mathcal{B}$, with $B in mathcal{B}$ iff $exists p in X, exists r>0: B subseteq B(p,r)$. You know a theorem that tells you that a bornology is induced by a metric. Maybe combine those facts?
$endgroup$
– Henno Brandsma
Dec 23 '18 at 6:09




$begingroup$
They have the same metric-bornology $mathcal{B}$, with $B in mathcal{B}$ iff $exists p in X, exists r>0: B subseteq B(p,r)$. You know a theorem that tells you that a bornology is induced by a metric. Maybe combine those facts?
$endgroup$
– Henno Brandsma
Dec 23 '18 at 6:09




1




1




$begingroup$
@HennoBrandsma But two metrics which induce the same uniformity and the same bornology need not be strongly equivalent; see here: math.stackexchange.com/a/3050424/71829 So that means that there must be some structure beyond the uniformity and the bornology that is preserved by strong equivalence.
$endgroup$
– Keshav Srinivasan
Dec 23 '18 at 17:22




$begingroup$
@HennoBrandsma But two metrics which induce the same uniformity and the same bornology need not be strongly equivalent; see here: math.stackexchange.com/a/3050424/71829 So that means that there must be some structure beyond the uniformity and the bornology that is preserved by strong equivalence.
$endgroup$
– Keshav Srinivasan
Dec 23 '18 at 17:22




1




1




$begingroup$
that example is somewhat trivialised as$[0,1]$ has but one uniformity and all sunsets are bounded.
$endgroup$
– Henno Brandsma
Dec 23 '18 at 17:25




$begingroup$
that example is somewhat trivialised as$[0,1]$ has but one uniformity and all sunsets are bounded.
$endgroup$
– Henno Brandsma
Dec 23 '18 at 17:25




3




3




$begingroup$
I do not think this object has any name. You can call it a bilipschitz structure. One can identify such a structure with the isomorphism class of some Banach algebra (if you like functional analysis), see msp.org/pjm/1963/13-4/pjm-v13-n4-p31-p.pdf.
$endgroup$
– Moishe Kohan
Dec 27 '18 at 4:28




$begingroup$
I do not think this object has any name. You can call it a bilipschitz structure. One can identify such a structure with the isomorphism class of some Banach algebra (if you like functional analysis), see msp.org/pjm/1963/13-4/pjm-v13-n4-p31-p.pdf.
$endgroup$
– Moishe Kohan
Dec 27 '18 at 4:28










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