Is the notion of Cauchy sequences definable in a bornological topological space?
$begingroup$
Being a Cauchy sequence is not a topological property, i.e. two metrics can induce the same topology and yet a sequence which is Cauchy in one may not be Cauchy in the other. It is a uniform property though, i.e. if two metrics induce the same uniformity then they have the same set of Cauchy sequences. But I'm wondering if Cauchy sequences can be defined in weaker conditions than a uniform space.
Let $X$ be a topological space endowed with a bornology, i.e. a structure which defines a notion of bounded sets. My question is, is it possible to define the notion of Cauchy sequences in terms of this bornology? To put it another way, if two metrics induce both the same topology and the same bornology, then do they have the same set of Cauchy sequences?
general-topology metric-spaces cauchy-sequences uniform-spaces
asked Dec 19 '18 at 15:54
Keshav SrinivasanKeshav Srinivasan
2,31721445
$endgroup$
add a comment |
$begingroup$
Being a Cauchy sequence is not a topological property, i.e. two metrics can induce the same topology and yet a sequence which is Cauchy in one may not be Cauchy in the other. It is a uniform property though, i.e. if two metrics induce the same uniformity then they have the same set of Cauchy sequences. But I'm wondering if Cauchy sequences can be defined in weaker conditions than a uniform space.
Let $X$ be a topological space endowed with a bornology, i.e. a structure which defines a notion of bounded sets. My question is, is it possible to define the notion of Cauchy sequences in terms of this bornology? To put it another way, if two metrics induce both the same topology and the same bornology, then do they have the same set of Cauchy sequences?
general-topology metric-spaces cauchy-sequences uniform-spaces
asked Dec 19 '18 at 15:54
Keshav SrinivasanKeshav Srinivasan
2,31721445
$endgroup$
$begingroup$
A bornology seems to broad. All countable sets form a bornology e.g. For the question to make sense we need a uniformisable space $X$ (so Tychonoff) and a bornology somehow related to the topology.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 22:26
1
$begingroup$
@HennoBrandsma Well, for starters I’d just like to find out whether if two metric spaces for the same topology induce the same bornology, then they have the same set of Cauchy sequences. If there is a way to characterize that set of Cauchy sequences using just the topology and and the bornology, then I can see whether that characterization still make sense even if the topology is not metrizable or even uniformizable.
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 23:01
$begingroup$
Nice question. $;$
$endgroup$
– goblin
Dec 20 '18 at 9:20
$begingroup$
@HennoBrandsma By the way, the condition for a bornolovy to interact well with a topology seems to be that for any bounded set $S$, there exists a bounded set $T$ such that the closure of $S$ is a subset of the interior of $T$. I posted a question related to that: math.stackexchange.com/q/3047070/71829
$endgroup$
– Keshav Srinivasan
Dec 20 '18 at 15:36
add a comment |
$begingroup$
Being a Cauchy sequence is not a topological property, i.e. two metrics can induce the same topology and yet a sequence which is Cauchy in one may not be Cauchy in the other. It is a uniform property though, i.e. if two metrics induce the same uniformity then they have the same set of Cauchy sequences. But I'm wondering if Cauchy sequences can be defined in weaker conditions than a uniform space.
Let $X$ be a topological space endowed with a bornology, i.e. a structure which defines a notion of bounded sets. My question is, is it possible to define the notion of Cauchy sequences in terms of this bornology? To put it another way, if two metrics induce both the same topology and the same bornology, then do they have the same set of Cauchy sequences?
general-topology metric-spaces cauchy-sequences uniform-spaces
asked Dec 19 '18 at 15:54
Keshav SrinivasanKeshav Srinivasan
2,31721445
$endgroup$
Being a Cauchy sequence is not a topological property, i.e. two metrics can induce the same topology and yet a sequence which is Cauchy in one may not be Cauchy in the other. It is a uniform property though, i.e. if two metrics induce the same uniformity then they have the same set of Cauchy sequences. But I'm wondering if Cauchy sequences can be defined in weaker conditions than a uniform space.
Let $X$ be a topological space endowed with a bornology, i.e. a structure which defines a notion of bounded sets. My question is, is it possible to define the notion of Cauchy sequences in terms of this bornology? To put it another way, if two metrics induce both the same topology and the same bornology, then do they have the same set of Cauchy sequences?
general-topology metric-spaces cauchy-sequences uniform-spaces
general-topology metric-spaces cauchy-sequences uniform-spaces
asked Dec 19 '18 at 15:54
Keshav SrinivasanKeshav Srinivasan
2,31721445
asked Dec 19 '18 at 15:54
Keshav SrinivasanKeshav Srinivasan
2,31721445
asked Dec 19 '18 at 15:54
Keshav SrinivasanKeshav Srinivasan
2,31721445
asked Dec 19 '18 at 15:54
Keshav SrinivasanKeshav Srinivasan
2,31721445
asked Dec 19 '18 at 15:54
Keshav SrinivasanKeshav Srinivasan
2,31721445
2,31721445
$begingroup$
A bornology seems to broad. All countable sets form a bornology e.g. For the question to make sense we need a uniformisable space $X$ (so Tychonoff) and a bornology somehow related to the topology.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 22:26
1
$begingroup$
@HennoBrandsma Well, for starters I’d just like to find out whether if two metric spaces for the same topology induce the same bornology, then they have the same set of Cauchy sequences. If there is a way to characterize that set of Cauchy sequences using just the topology and and the bornology, then I can see whether that characterization still make sense even if the topology is not metrizable or even uniformizable.
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 23:01
$begingroup$
Nice question. $;$
$endgroup$
– goblin
Dec 20 '18 at 9:20
$begingroup$
@HennoBrandsma By the way, the condition for a bornolovy to interact well with a topology seems to be that for any bounded set $S$, there exists a bounded set $T$ such that the closure of $S$ is a subset of the interior of $T$. I posted a question related to that: math.stackexchange.com/q/3047070/71829
$endgroup$
– Keshav Srinivasan
Dec 20 '18 at 15:36
add a comment |
$begingroup$
A bornology seems to broad. All countable sets form a bornology e.g. For the question to make sense we need a uniformisable space $X$ (so Tychonoff) and a bornology somehow related to the topology.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 22:26
1
$begingroup$
@HennoBrandsma Well, for starters I’d just like to find out whether if two metric spaces for the same topology induce the same bornology, then they have the same set of Cauchy sequences. If there is a way to characterize that set of Cauchy sequences using just the topology and and the bornology, then I can see whether that characterization still make sense even if the topology is not metrizable or even uniformizable.
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 23:01
$begingroup$
Nice question. $;$
$endgroup$
– goblin
Dec 20 '18 at 9:20
$begingroup$
@HennoBrandsma By the way, the condition for a bornolovy to interact well with a topology seems to be that for any bounded set $S$, there exists a bounded set $T$ such that the closure of $S$ is a subset of the interior of $T$. I posted a question related to that: math.stackexchange.com/q/3047070/71829
$endgroup$
– Keshav Srinivasan
Dec 20 '18 at 15:36
$begingroup$
A bornology seems to broad. All countable sets form a bornology e.g. For the question to make sense we need a uniformisable space $X$ (so Tychonoff) and a bornology somehow related to the topology.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 22:26
$begingroup$
A bornology seems to broad. All countable sets form a bornology e.g. For the question to make sense we need a uniformisable space $X$ (so Tychonoff) and a bornology somehow related to the topology.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 22:26
1
1
$begingroup$
@HennoBrandsma Well, for starters I’d just like to find out whether if two metric spaces for the same topology induce the same bornology, then they have the same set of Cauchy sequences. If there is a way to characterize that set of Cauchy sequences using just the topology and and the bornology, then I can see whether that characterization still make sense even if the topology is not metrizable or even uniformizable.
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 23:01
$begingroup$
@HennoBrandsma Well, for starters I’d just like to find out whether if two metric spaces for the same topology induce the same bornology, then they have the same set of Cauchy sequences. If there is a way to characterize that set of Cauchy sequences using just the topology and and the bornology, then I can see whether that characterization still make sense even if the topology is not metrizable or even uniformizable.
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 23:01
$begingroup$
Nice question. $;$
$endgroup$
– goblin
Dec 20 '18 at 9:20
$begingroup$
Nice question. $;$
$endgroup$
– goblin
Dec 20 '18 at 9:20
$begingroup$
@HennoBrandsma By the way, the condition for a bornolovy to interact well with a topology seems to be that for any bounded set $S$, there exists a bounded set $T$ such that the closure of $S$ is a subset of the interior of $T$. I posted a question related to that: math.stackexchange.com/q/3047070/71829
$endgroup$
– Keshav Srinivasan
Dec 20 '18 at 15:36
$begingroup$
@HennoBrandsma By the way, the condition for a bornolovy to interact well with a topology seems to be that for any bounded set $S$, there exists a bounded set $T$ such that the closure of $S$ is a subset of the interior of $T$. I posted a question related to that: math.stackexchange.com/q/3047070/71829
$endgroup$
– Keshav Srinivasan
Dec 20 '18 at 15:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the function $f:(0,1)tomathbb{R}$ defined by $f(x)=sin(frac1x)$. Then $xmapsto(x,f(x))$ is a homeomorphism from $(0,1)$ to the graph of $f$. However, $a_n=frac2{npi}$ is a Cauchy sequence in $(0,1)$, while $f(a_n)$ is not a Cauchy sequence in the graph of $f$, since the $y$ values form the divergent sequence $0,1,0,-1,0,1,0,-1...$. Moreover, both $(0,1)$ and the graph of $f$ are bounded metric spaces, so they are not only homeomorphic, but also 'bornoleomorphic'.
answered Dec 22 '18 at 18:27
SmileyCraftSmileyCraft
3,611517
$endgroup$
$begingroup$
First of all, I was talking about two metrics on one and the same set. Second of all, is there an example where the two metrics are not bounded?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 18:49
$begingroup$
Your complaints are irrelevant. If you have a metric space homeomorphism $h:Vto W$ and you want two metrics on the same space, just use $d_1=d_V$ and $d_2(x,y)=d_W(h(x),h(y))$. If you want unbounded metric spaces, just take the disjoint union with some unbounded metric space. You probably think this is cheating, but Cauchy sequences have to be bounded anyways, so it makes no sense to want an unbounded metric space.
$endgroup$
– SmileyCraft
Dec 22 '18 at 20:37
$begingroup$
Hmm, good points. By the way, how do you define a metric on a disjoint union of metric spaces?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 20:42
$begingroup$
That is indeed not trivial. Let $V$ and $W$ be metric spaces. Choose any $vin V$ and $win W$. Then on the disjoint union of $V$ and $W$ we can define a metric as follows. If two elements are from the same space, use the old metric. Otherwise, for $xin V$ and $yin W$ define $d(x,y)=1+d_V(x,v)+d_W(y,w)$.
$endgroup$
– SmileyCraft
Dec 22 '18 at 21:09
$begingroup$
OK thanks for your help.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 21:13
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
Consider the function $f:(0,1)tomathbb{R}$ defined by $f(x)=sin(frac1x)$. Then $xmapsto(x,f(x))$ is a homeomorphism from $(0,1)$ to the graph of $f$. However, $a_n=frac2{npi}$ is a Cauchy sequence in $(0,1)$, while $f(a_n)$ is not a Cauchy sequence in the graph of $f$, since the $y$ values form the divergent sequence $0,1,0,-1,0,1,0,-1...$. Moreover, both $(0,1)$ and the graph of $f$ are bounded metric spaces, so they are not only homeomorphic, but also 'bornoleomorphic'.
answered Dec 22 '18 at 18:27
SmileyCraftSmileyCraft
3,611517
$endgroup$
$begingroup$
First of all, I was talking about two metrics on one and the same set. Second of all, is there an example where the two metrics are not bounded?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 18:49
$begingroup$
Your complaints are irrelevant. If you have a metric space homeomorphism $h:Vto W$ and you want two metrics on the same space, just use $d_1=d_V$ and $d_2(x,y)=d_W(h(x),h(y))$. If you want unbounded metric spaces, just take the disjoint union with some unbounded metric space. You probably think this is cheating, but Cauchy sequences have to be bounded anyways, so it makes no sense to want an unbounded metric space.
$endgroup$
– SmileyCraft
Dec 22 '18 at 20:37
$begingroup$
Hmm, good points. By the way, how do you define a metric on a disjoint union of metric spaces?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 20:42
$begingroup$
That is indeed not trivial. Let $V$ and $W$ be metric spaces. Choose any $vin V$ and $win W$. Then on the disjoint union of $V$ and $W$ we can define a metric as follows. If two elements are from the same space, use the old metric. Otherwise, for $xin V$ and $yin W$ define $d(x,y)=1+d_V(x,v)+d_W(y,w)$.
$endgroup$
– SmileyCraft
Dec 22 '18 at 21:09
$begingroup$
OK thanks for your help.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 21:13
|
show 1 more comment
$begingroup$
Consider the function $f:(0,1)tomathbb{R}$ defined by $f(x)=sin(frac1x)$. Then $xmapsto(x,f(x))$ is a homeomorphism from $(0,1)$ to the graph of $f$. However, $a_n=frac2{npi}$ is a Cauchy sequence in $(0,1)$, while $f(a_n)$ is not a Cauchy sequence in the graph of $f$, since the $y$ values form the divergent sequence $0,1,0,-1,0,1,0,-1...$. Moreover, both $(0,1)$ and the graph of $f$ are bounded metric spaces, so they are not only homeomorphic, but also 'bornoleomorphic'.
answered Dec 22 '18 at 18:27
SmileyCraftSmileyCraft
3,611517
$endgroup$
$begingroup$
First of all, I was talking about two metrics on one and the same set. Second of all, is there an example where the two metrics are not bounded?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 18:49
$begingroup$
Your complaints are irrelevant. If you have a metric space homeomorphism $h:Vto W$ and you want two metrics on the same space, just use $d_1=d_V$ and $d_2(x,y)=d_W(h(x),h(y))$. If you want unbounded metric spaces, just take the disjoint union with some unbounded metric space. You probably think this is cheating, but Cauchy sequences have to be bounded anyways, so it makes no sense to want an unbounded metric space.
$endgroup$
– SmileyCraft
Dec 22 '18 at 20:37
$begingroup$
Hmm, good points. By the way, how do you define a metric on a disjoint union of metric spaces?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 20:42
$begingroup$
That is indeed not trivial. Let $V$ and $W$ be metric spaces. Choose any $vin V$ and $win W$. Then on the disjoint union of $V$ and $W$ we can define a metric as follows. If two elements are from the same space, use the old metric. Otherwise, for $xin V$ and $yin W$ define $d(x,y)=1+d_V(x,v)+d_W(y,w)$.
$endgroup$
– SmileyCraft
Dec 22 '18 at 21:09
$begingroup$
OK thanks for your help.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 21:13
|
show 1 more comment
$begingroup$
Consider the function $f:(0,1)tomathbb{R}$ defined by $f(x)=sin(frac1x)$. Then $xmapsto(x,f(x))$ is a homeomorphism from $(0,1)$ to the graph of $f$. However, $a_n=frac2{npi}$ is a Cauchy sequence in $(0,1)$, while $f(a_n)$ is not a Cauchy sequence in the graph of $f$, since the $y$ values form the divergent sequence $0,1,0,-1,0,1,0,-1...$. Moreover, both $(0,1)$ and the graph of $f$ are bounded metric spaces, so they are not only homeomorphic, but also 'bornoleomorphic'.
answered Dec 22 '18 at 18:27
SmileyCraftSmileyCraft
3,611517
$endgroup$
Consider the function $f:(0,1)tomathbb{R}$ defined by $f(x)=sin(frac1x)$. Then $xmapsto(x,f(x))$ is a homeomorphism from $(0,1)$ to the graph of $f$. However, $a_n=frac2{npi}$ is a Cauchy sequence in $(0,1)$, while $f(a_n)$ is not a Cauchy sequence in the graph of $f$, since the $y$ values form the divergent sequence $0,1,0,-1,0,1,0,-1...$. Moreover, both $(0,1)$ and the graph of $f$ are bounded metric spaces, so they are not only homeomorphic, but also 'bornoleomorphic'.
answered Dec 22 '18 at 18:27
SmileyCraftSmileyCraft
3,611517
answered Dec 22 '18 at 18:27
SmileyCraftSmileyCraft
3,611517
answered Dec 22 '18 at 18:27
SmileyCraftSmileyCraft
3,611517
answered Dec 22 '18 at 18:27
SmileyCraftSmileyCraft
3,611517
3,611517
$begingroup$
First of all, I was talking about two metrics on one and the same set. Second of all, is there an example where the two metrics are not bounded?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 18:49
$begingroup$
Your complaints are irrelevant. If you have a metric space homeomorphism $h:Vto W$ and you want two metrics on the same space, just use $d_1=d_V$ and $d_2(x,y)=d_W(h(x),h(y))$. If you want unbounded metric spaces, just take the disjoint union with some unbounded metric space. You probably think this is cheating, but Cauchy sequences have to be bounded anyways, so it makes no sense to want an unbounded metric space.
$endgroup$
– SmileyCraft
Dec 22 '18 at 20:37
$begingroup$
Hmm, good points. By the way, how do you define a metric on a disjoint union of metric spaces?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 20:42
$begingroup$
That is indeed not trivial. Let $V$ and $W$ be metric spaces. Choose any $vin V$ and $win W$. Then on the disjoint union of $V$ and $W$ we can define a metric as follows. If two elements are from the same space, use the old metric. Otherwise, for $xin V$ and $yin W$ define $d(x,y)=1+d_V(x,v)+d_W(y,w)$.
$endgroup$
– SmileyCraft
Dec 22 '18 at 21:09
$begingroup$
OK thanks for your help.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 21:13
|
show 1 more comment
$begingroup$
First of all, I was talking about two metrics on one and the same set. Second of all, is there an example where the two metrics are not bounded?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 18:49
$begingroup$
Your complaints are irrelevant. If you have a metric space homeomorphism $h:Vto W$ and you want two metrics on the same space, just use $d_1=d_V$ and $d_2(x,y)=d_W(h(x),h(y))$. If you want unbounded metric spaces, just take the disjoint union with some unbounded metric space. You probably think this is cheating, but Cauchy sequences have to be bounded anyways, so it makes no sense to want an unbounded metric space.
$endgroup$
– SmileyCraft
Dec 22 '18 at 20:37
$begingroup$
Hmm, good points. By the way, how do you define a metric on a disjoint union of metric spaces?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 20:42
$begingroup$
That is indeed not trivial. Let $V$ and $W$ be metric spaces. Choose any $vin V$ and $win W$. Then on the disjoint union of $V$ and $W$ we can define a metric as follows. If two elements are from the same space, use the old metric. Otherwise, for $xin V$ and $yin W$ define $d(x,y)=1+d_V(x,v)+d_W(y,w)$.
$endgroup$
– SmileyCraft
Dec 22 '18 at 21:09
$begingroup$
OK thanks for your help.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 21:13
$begingroup$
First of all, I was talking about two metrics on one and the same set. Second of all, is there an example where the two metrics are not bounded?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 18:49
$begingroup$
First of all, I was talking about two metrics on one and the same set. Second of all, is there an example where the two metrics are not bounded?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 18:49
$begingroup$
Your complaints are irrelevant. If you have a metric space homeomorphism $h:Vto W$ and you want two metrics on the same space, just use $d_1=d_V$ and $d_2(x,y)=d_W(h(x),h(y))$. If you want unbounded metric spaces, just take the disjoint union with some unbounded metric space. You probably think this is cheating, but Cauchy sequences have to be bounded anyways, so it makes no sense to want an unbounded metric space.
$endgroup$
– SmileyCraft
Dec 22 '18 at 20:37
$begingroup$
Your complaints are irrelevant. If you have a metric space homeomorphism $h:Vto W$ and you want two metrics on the same space, just use $d_1=d_V$ and $d_2(x,y)=d_W(h(x),h(y))$. If you want unbounded metric spaces, just take the disjoint union with some unbounded metric space. You probably think this is cheating, but Cauchy sequences have to be bounded anyways, so it makes no sense to want an unbounded metric space.
$endgroup$
– SmileyCraft
Dec 22 '18 at 20:37
$begingroup$
Hmm, good points. By the way, how do you define a metric on a disjoint union of metric spaces?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 20:42
$begingroup$
Hmm, good points. By the way, how do you define a metric on a disjoint union of metric spaces?
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 20:42
$begingroup$
That is indeed not trivial. Let $V$ and $W$ be metric spaces. Choose any $vin V$ and $win W$. Then on the disjoint union of $V$ and $W$ we can define a metric as follows. If two elements are from the same space, use the old metric. Otherwise, for $xin V$ and $yin W$ define $d(x,y)=1+d_V(x,v)+d_W(y,w)$.
$endgroup$
– SmileyCraft
Dec 22 '18 at 21:09
$begingroup$
That is indeed not trivial. Let $V$ and $W$ be metric spaces. Choose any $vin V$ and $win W$. Then on the disjoint union of $V$ and $W$ we can define a metric as follows. If two elements are from the same space, use the old metric. Otherwise, for $xin V$ and $yin W$ define $d(x,y)=1+d_V(x,v)+d_W(y,w)$.
$endgroup$
– SmileyCraft
Dec 22 '18 at 21:09
$begingroup$
OK thanks for your help.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 21:13
$begingroup$
OK thanks for your help.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 21:13
|
show 1 more comment
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A bornology seems to broad. All countable sets form a bornology e.g. For the question to make sense we need a uniformisable space $X$ (so Tychonoff) and a bornology somehow related to the topology.
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– Henno Brandsma
Dec 19 '18 at 22:26
1
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@HennoBrandsma Well, for starters I’d just like to find out whether if two metric spaces for the same topology induce the same bornology, then they have the same set of Cauchy sequences. If there is a way to characterize that set of Cauchy sequences using just the topology and and the bornology, then I can see whether that characterization still make sense even if the topology is not metrizable or even uniformizable.
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– Keshav Srinivasan
Dec 19 '18 at 23:01
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Nice question. $;$
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– goblin
Dec 20 '18 at 9:20
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@HennoBrandsma By the way, the condition for a bornolovy to interact well with a topology seems to be that for any bounded set $S$, there exists a bounded set $T$ such that the closure of $S$ is a subset of the interior of $T$. I posted a question related to that: math.stackexchange.com/q/3047070/71829
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– Keshav Srinivasan
Dec 20 '18 at 15:36