Complete metric spaces question
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If two metric spaces let's say $(X,d)$ and $(X,h)$ are complete, does it imply that $h$ and $d$ are topologically equivalent ?
general-topology
asked Jan 19 at 14:37
Pedro AlvarèsPedro Alvarès
636
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add a comment |
$begingroup$
If two metric spaces let's say $(X,d)$ and $(X,h)$ are complete, does it imply that $h$ and $d$ are topologically equivalent ?
general-topology
asked Jan 19 at 14:37
Pedro AlvarèsPedro Alvarès
636
$endgroup$
add a comment |
$begingroup$
If two metric spaces let's say $(X,d)$ and $(X,h)$ are complete, does it imply that $h$ and $d$ are topologically equivalent ?
general-topology
asked Jan 19 at 14:37
Pedro AlvarèsPedro Alvarès
636
$endgroup$
If two metric spaces let's say $(X,d)$ and $(X,h)$ are complete, does it imply that $h$ and $d$ are topologically equivalent ?
general-topology
general-topology
asked Jan 19 at 14:37
Pedro AlvarèsPedro Alvarès
636
asked Jan 19 at 14:37
Pedro AlvarèsPedro Alvarès
636
asked Jan 19 at 14:37
Pedro AlvarèsPedro Alvarès
636
asked Jan 19 at 14:37
Pedro AlvarèsPedro Alvarès
636
asked Jan 19 at 14:37
Pedro AlvarèsPedro Alvarès
636
636
add a comment |
add a comment |
1 Answer
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Of course not, let $X =mathbb{R}$ and $d(x,y)=|x-y|$ the usual metric and $h(x,y) = 0 (x=y), 1 (x neq y)$ the discrete metric. Both are complete, but they are very non-equivalent.
answered Jan 19 at 14:44
Henno BrandsmaHenno Brandsma
110k347116
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1 Answer
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1 Answer
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active
oldest
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active
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active
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$begingroup$
Of course not, let $X =mathbb{R}$ and $d(x,y)=|x-y|$ the usual metric and $h(x,y) = 0 (x=y), 1 (x neq y)$ the discrete metric. Both are complete, but they are very non-equivalent.
answered Jan 19 at 14:44
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Of course not, let $X =mathbb{R}$ and $d(x,y)=|x-y|$ the usual metric and $h(x,y) = 0 (x=y), 1 (x neq y)$ the discrete metric. Both are complete, but they are very non-equivalent.
answered Jan 19 at 14:44
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Of course not, let $X =mathbb{R}$ and $d(x,y)=|x-y|$ the usual metric and $h(x,y) = 0 (x=y), 1 (x neq y)$ the discrete metric. Both are complete, but they are very non-equivalent.
answered Jan 19 at 14:44
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
Of course not, let $X =mathbb{R}$ and $d(x,y)=|x-y|$ the usual metric and $h(x,y) = 0 (x=y), 1 (x neq y)$ the discrete metric. Both are complete, but they are very non-equivalent.
answered Jan 19 at 14:44
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 19 at 14:44
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 19 at 14:44
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 19 at 14:44
Henno BrandsmaHenno Brandsma
110k347116
110k347116
add a comment |
add a comment |
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