Are there any sets that are not complete metric spaces under all possible metrics?
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I don't have any particular set in mind but this seemed interesthing since completeness depends on the metric.
general-topology metric-spaces complete-spaces
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1808
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add a comment |
$begingroup$
I don't have any particular set in mind but this seemed interesthing since completeness depends on the metric.
general-topology metric-spaces complete-spaces
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1808
$endgroup$
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
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– Alex Kruckman
Dec 19 '18 at 18:27
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@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31
add a comment |
$begingroup$
I don't have any particular set in mind but this seemed interesthing since completeness depends on the metric.
general-topology metric-spaces complete-spaces
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1808
$endgroup$
I don't have any particular set in mind but this seemed interesthing since completeness depends on the metric.
general-topology metric-spaces complete-spaces
general-topology metric-spaces complete-spaces
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1808
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1808
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1808
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1808
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1808
1808
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31
add a comment |
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since every set can be given the discrete metric $d(x,y)=left{begin{array}{ll} 1 & text{if $xneq y$}\0 &text{if $x = y$}end{array}right.$, and that for this metric every Cauchy sequence is stationnary and so convergent, you can't find a such example.
answered Dec 19 '18 at 14:45
BalloonBalloon
4,625822
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add a comment |
$begingroup$
Yes, the set of rationals $mathbb{Q}$ in its usual topology is not a Baire space, so whatever compatible metric you put on it, it will be non-complete.
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Since every set can be given the discrete metric $d(x,y)=left{begin{array}{ll} 1 & text{if $xneq y$}\0 &text{if $x = y$}end{array}right.$, and that for this metric every Cauchy sequence is stationnary and so convergent, you can't find a such example.
answered Dec 19 '18 at 14:45
BalloonBalloon
4,625822
$endgroup$
add a comment |
$begingroup$
Since every set can be given the discrete metric $d(x,y)=left{begin{array}{ll} 1 & text{if $xneq y$}\0 &text{if $x = y$}end{array}right.$, and that for this metric every Cauchy sequence is stationnary and so convergent, you can't find a such example.
answered Dec 19 '18 at 14:45
BalloonBalloon
4,625822
$endgroup$
add a comment |
$begingroup$
Since every set can be given the discrete metric $d(x,y)=left{begin{array}{ll} 1 & text{if $xneq y$}\0 &text{if $x = y$}end{array}right.$, and that for this metric every Cauchy sequence is stationnary and so convergent, you can't find a such example.
answered Dec 19 '18 at 14:45
BalloonBalloon
4,625822
$endgroup$
Since every set can be given the discrete metric $d(x,y)=left{begin{array}{ll} 1 & text{if $xneq y$}\0 &text{if $x = y$}end{array}right.$, and that for this metric every Cauchy sequence is stationnary and so convergent, you can't find a such example.
answered Dec 19 '18 at 14:45
BalloonBalloon
4,625822
answered Dec 19 '18 at 14:45
BalloonBalloon
4,625822
answered Dec 19 '18 at 14:45
BalloonBalloon
4,625822
answered Dec 19 '18 at 14:45
BalloonBalloon
4,625822
4,625822
add a comment |
add a comment |
$begingroup$
Yes, the set of rationals $mathbb{Q}$ in its usual topology is not a Baire space, so whatever compatible metric you put on it, it will be non-complete.
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
add a comment |
$begingroup$
Yes, the set of rationals $mathbb{Q}$ in its usual topology is not a Baire space, so whatever compatible metric you put on it, it will be non-complete.
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
add a comment |
$begingroup$
Yes, the set of rationals $mathbb{Q}$ in its usual topology is not a Baire space, so whatever compatible metric you put on it, it will be non-complete.
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
Yes, the set of rationals $mathbb{Q}$ in its usual topology is not a Baire space, so whatever compatible metric you put on it, it will be non-complete.
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
111k348118
111k348118
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
add a comment |
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
add a comment |
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$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31