If two metric spaces are topologically equivalent (homeomorphic) imply that they are complete?Real numbers...
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If two metric spaces are topologically equivalent (homeomorphic) imply that they are complete?
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric spaceWhich of the following metric spaces are complete?Showing that a metric space is completeUnderstanding Complete Metric Spaces and Cauchy SequencesCompact homeomorphic non bilipschitz homeomorphic metric spacesA metric space is complete when every closed and bounded subset of it is compactTopologically equivalent metrics? Ceiling function of metric $d$Limit Points of closure of $A$ is subset of limit points of $A$If $d_1,d_2$ are not equivalent metrics, is it true $(X,d_1)$ is not homeomorphic to $(X,d_2)$?Under what metric spaces are pointwise and uniform convergence equivalent?How to *disprove* topological equivalence
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My definition of topologically equivalent is: every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit and vice versa.
I know that
Let $(X,d_A)$ and $(X, d_B)$ be two strongly equivalent metric spaces, if $(X, d_A)$ is complete then $(X, d_B)$ is complete.
I was thinking that if these spaces were just topologically equivalent yet would the statement be true?
My answer is no.
But I can't think of a counter example.
general-topology convergence metric-spaces
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add a comment |
$begingroup$
My definition of topologically equivalent is: every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit and vice versa.
I know that
Let $(X,d_A)$ and $(X, d_B)$ be two strongly equivalent metric spaces, if $(X, d_A)$ is complete then $(X, d_B)$ is complete.
I was thinking that if these spaces were just topologically equivalent yet would the statement be true?
My answer is no.
But I can't think of a counter example.
general-topology convergence metric-spaces
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2
$begingroup$
I think there is missing a part :definition is every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit - and vice versa
$endgroup$
– Maksim
16 hours ago
add a comment |
$begingroup$
My definition of topologically equivalent is: every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit and vice versa.
I know that
Let $(X,d_A)$ and $(X, d_B)$ be two strongly equivalent metric spaces, if $(X, d_A)$ is complete then $(X, d_B)$ is complete.
I was thinking that if these spaces were just topologically equivalent yet would the statement be true?
My answer is no.
But I can't think of a counter example.
general-topology convergence metric-spaces
$endgroup$
My definition of topologically equivalent is: every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit and vice versa.
I know that
Let $(X,d_A)$ and $(X, d_B)$ be two strongly equivalent metric spaces, if $(X, d_A)$ is complete then $(X, d_B)$ is complete.
I was thinking that if these spaces were just topologically equivalent yet would the statement be true?
My answer is no.
But I can't think of a counter example.
general-topology convergence metric-spaces
general-topology convergence metric-spaces
edited 9 hours ago
Juliana de Souza
asked 17 hours ago
Juliana de SouzaJuliana de Souza
877
877
2
$begingroup$
I think there is missing a part :definition is every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit - and vice versa
$endgroup$
– Maksim
16 hours ago
add a comment |
2
$begingroup$
I think there is missing a part :definition is every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit - and vice versa
$endgroup$
– Maksim
16 hours ago
2
2
$begingroup$
I think there is missing a part :definition is every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit - and vice versa
$endgroup$
– Maksim
16 hours ago
$begingroup$
I think there is missing a part :definition is every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit - and vice versa
$endgroup$
– Maksim
16 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider $(Bbb R,d)$ , where $d$ is the usual metric and $(Bbb R,rho)$ , where $rho$ is the metric defined by $$rho(x,y)=vert arctan x-arctan y vert$$ Then $(Bbb R,d)$ is topologically equivalent to $(Bbb R,rho)$(How ?), but one is complete while the other is not (How?)
Try to fill the gaps!
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1
$begingroup$
See math.stackexchange.com/questions/152243/… for the proof of non-completeness
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– Chinnapparaj R
16 hours ago
add a comment |
$begingroup$
$(-1,1)$ is topologically equivalent to $mathbb R$ via the homeomorphism $x to frac x {1-|x|}$. $(-1,1)$ is not complete but $mathbb R$ is complete. [Usual metric on both spaces].
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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$begingroup$
Consider $(Bbb R,d)$ , where $d$ is the usual metric and $(Bbb R,rho)$ , where $rho$ is the metric defined by $$rho(x,y)=vert arctan x-arctan y vert$$ Then $(Bbb R,d)$ is topologically equivalent to $(Bbb R,rho)$(How ?), but one is complete while the other is not (How?)
Try to fill the gaps!
$endgroup$
1
$begingroup$
See math.stackexchange.com/questions/152243/… for the proof of non-completeness
$endgroup$
– Chinnapparaj R
16 hours ago
add a comment |
$begingroup$
Consider $(Bbb R,d)$ , where $d$ is the usual metric and $(Bbb R,rho)$ , where $rho$ is the metric defined by $$rho(x,y)=vert arctan x-arctan y vert$$ Then $(Bbb R,d)$ is topologically equivalent to $(Bbb R,rho)$(How ?), but one is complete while the other is not (How?)
Try to fill the gaps!
$endgroup$
1
$begingroup$
See math.stackexchange.com/questions/152243/… for the proof of non-completeness
$endgroup$
– Chinnapparaj R
16 hours ago
add a comment |
$begingroup$
Consider $(Bbb R,d)$ , where $d$ is the usual metric and $(Bbb R,rho)$ , where $rho$ is the metric defined by $$rho(x,y)=vert arctan x-arctan y vert$$ Then $(Bbb R,d)$ is topologically equivalent to $(Bbb R,rho)$(How ?), but one is complete while the other is not (How?)
Try to fill the gaps!
$endgroup$
Consider $(Bbb R,d)$ , where $d$ is the usual metric and $(Bbb R,rho)$ , where $rho$ is the metric defined by $$rho(x,y)=vert arctan x-arctan y vert$$ Then $(Bbb R,d)$ is topologically equivalent to $(Bbb R,rho)$(How ?), but one is complete while the other is not (How?)
Try to fill the gaps!
answered 16 hours ago
Chinnapparaj RChinnapparaj R
5,9382928
5,9382928
1
$begingroup$
See math.stackexchange.com/questions/152243/… for the proof of non-completeness
$endgroup$
– Chinnapparaj R
16 hours ago
add a comment |
1
$begingroup$
See math.stackexchange.com/questions/152243/… for the proof of non-completeness
$endgroup$
– Chinnapparaj R
16 hours ago
1
1
$begingroup$
See math.stackexchange.com/questions/152243/… for the proof of non-completeness
$endgroup$
– Chinnapparaj R
16 hours ago
$begingroup$
See math.stackexchange.com/questions/152243/… for the proof of non-completeness
$endgroup$
– Chinnapparaj R
16 hours ago
add a comment |
$begingroup$
$(-1,1)$ is topologically equivalent to $mathbb R$ via the homeomorphism $x to frac x {1-|x|}$. $(-1,1)$ is not complete but $mathbb R$ is complete. [Usual metric on both spaces].
$endgroup$
add a comment |
$begingroup$
$(-1,1)$ is topologically equivalent to $mathbb R$ via the homeomorphism $x to frac x {1-|x|}$. $(-1,1)$ is not complete but $mathbb R$ is complete. [Usual metric on both spaces].
$endgroup$
add a comment |
$begingroup$
$(-1,1)$ is topologically equivalent to $mathbb R$ via the homeomorphism $x to frac x {1-|x|}$. $(-1,1)$ is not complete but $mathbb R$ is complete. [Usual metric on both spaces].
$endgroup$
$(-1,1)$ is topologically equivalent to $mathbb R$ via the homeomorphism $x to frac x {1-|x|}$. $(-1,1)$ is not complete but $mathbb R$ is complete. [Usual metric on both spaces].
answered 16 hours ago
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
72.9k53170
add a comment |
add a comment |
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$begingroup$
I think there is missing a part :definition is every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit - and vice versa
$endgroup$
– Maksim
16 hours ago