Is there a metric for which the open unit interval is complete? The Next CEO of Stack OverflowWhich of the following metric spaces are complete?which of the following metric spaces are separable?Metric on half-open interval s.t. subset is open w.r.t. $d$ iff open w.r.t. Euclidean metricQuestion on complete metric spaces and whether the following is a complete metric space:Open interval $(0,1)$ with the usual topology admits a metric spaceComplete Metric Space examplechoose the complete metric space?Spaces of (complete) separable metric spacesLocally Complete Metric SpaceIs the family of all continuous functions from a compact metric space to a separable complete metric space separable?
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Is there a metric for which the open unit interval is complete?
The Next CEO of Stack OverflowWhich of the following metric spaces are complete?which of the following metric spaces are separable?Metric on half-open interval s.t. subset is open w.r.t. $d$ iff open w.r.t. Euclidean metricQuestion on complete metric spaces and whether the following is a complete metric space:Open interval $(0,1)$ with the usual topology admits a metric spaceComplete Metric Space examplechoose the complete metric space?Spaces of (complete) separable metric spacesLocally Complete Metric SpaceIs the family of all continuous functions from a compact metric space to a separable complete metric space separable?
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Let, $I= (0,1)$ It is well known that $I$ is not a complete with respect to the Euclidean metric $(x,y)mapsto |x-y|$.
However, $(I,|cdot|)$ is separable.
Question: Can we find a metric $d: Itimes I to(0,infty)$ for which, $(I,d)$ is separable and complete?
real-analysis analysis metric-spaces complete-spaces separable-spaces
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add a comment |
$begingroup$
Let, $I= (0,1)$ It is well known that $I$ is not a complete with respect to the Euclidean metric $(x,y)mapsto |x-y|$.
However, $(I,|cdot|)$ is separable.
Question: Can we find a metric $d: Itimes I to(0,infty)$ for which, $(I,d)$ is separable and complete?
real-analysis analysis metric-spaces complete-spaces separable-spaces
$endgroup$
2
$begingroup$
Pull back the metric from $mathbbR$ according to some homeomorphism between them.
$endgroup$
– user647486
Mar 19 at 20:52
add a comment |
$begingroup$
Let, $I= (0,1)$ It is well known that $I$ is not a complete with respect to the Euclidean metric $(x,y)mapsto |x-y|$.
However, $(I,|cdot|)$ is separable.
Question: Can we find a metric $d: Itimes I to(0,infty)$ for which, $(I,d)$ is separable and complete?
real-analysis analysis metric-spaces complete-spaces separable-spaces
$endgroup$
Let, $I= (0,1)$ It is well known that $I$ is not a complete with respect to the Euclidean metric $(x,y)mapsto |x-y|$.
However, $(I,|cdot|)$ is separable.
Question: Can we find a metric $d: Itimes I to(0,infty)$ for which, $(I,d)$ is separable and complete?
real-analysis analysis metric-spaces complete-spaces separable-spaces
real-analysis analysis metric-spaces complete-spaces separable-spaces
asked Mar 19 at 20:50
Guy FsoneGuy Fsone
17.3k43074
17.3k43074
2
$begingroup$
Pull back the metric from $mathbbR$ according to some homeomorphism between them.
$endgroup$
– user647486
Mar 19 at 20:52
add a comment |
2
$begingroup$
Pull back the metric from $mathbbR$ according to some homeomorphism between them.
$endgroup$
– user647486
Mar 19 at 20:52
2
2
$begingroup$
Pull back the metric from $mathbbR$ according to some homeomorphism between them.
$endgroup$
– user647486
Mar 19 at 20:52
$begingroup$
Pull back the metric from $mathbbR$ according to some homeomorphism between them.
$endgroup$
– user647486
Mar 19 at 20:52
add a comment |
2 Answers
2
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$begingroup$
Try $J = (-pi/2,pi/2)$ instead. How about $d(x,y) = |tan x - tan y|$?
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add a comment |
$begingroup$
Take a homeomorphism from the interval to $Bbb R$ and pull back the usual Euclidean metric.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Try $J = (-pi/2,pi/2)$ instead. How about $d(x,y) = |tan x - tan y|$?
$endgroup$
add a comment |
$begingroup$
Try $J = (-pi/2,pi/2)$ instead. How about $d(x,y) = |tan x - tan y|$?
$endgroup$
add a comment |
$begingroup$
Try $J = (-pi/2,pi/2)$ instead. How about $d(x,y) = |tan x - tan y|$?
$endgroup$
Try $J = (-pi/2,pi/2)$ instead. How about $d(x,y) = |tan x - tan y|$?
answered Mar 19 at 20:52
Umberto P.Umberto P.
40.2k13370
40.2k13370
add a comment |
add a comment |
$begingroup$
Take a homeomorphism from the interval to $Bbb R$ and pull back the usual Euclidean metric.
$endgroup$
add a comment |
$begingroup$
Take a homeomorphism from the interval to $Bbb R$ and pull back the usual Euclidean metric.
$endgroup$
add a comment |
$begingroup$
Take a homeomorphism from the interval to $Bbb R$ and pull back the usual Euclidean metric.
$endgroup$
Take a homeomorphism from the interval to $Bbb R$ and pull back the usual Euclidean metric.
answered Mar 19 at 20:52
Ted ShifrinTed Shifrin
64.7k44692
64.7k44692
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$begingroup$
Pull back the metric from $mathbbR$ according to some homeomorphism between them.
$endgroup$
– user647486
Mar 19 at 20:52