Metric on compact set with one point less
Let $K$ be a compact metric space. Now remove one point from it. Is there a metric on $K$ with one point removed that generates the same topology as the initial metric but is such that the space $K$ with one point removed is still complete?
calculus real-analysis general-topology analysis metric-spaces
asked Nov 29 at 0:27
Sascha
136318
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Let $K$ be a compact metric space. Now remove one point from it. Is there a metric on $K$ with one point removed that generates the same topology as the initial metric but is such that the space $K$ with one point removed is still complete?
calculus real-analysis general-topology analysis metric-spaces
asked Nov 29 at 0:27
Sascha
136318
add a comment |
Let $K$ be a compact metric space. Now remove one point from it. Is there a metric on $K$ with one point removed that generates the same topology as the initial metric but is such that the space $K$ with one point removed is still complete?
calculus real-analysis general-topology analysis metric-spaces
asked Nov 29 at 0:27
Sascha
136318
Let $K$ be a compact metric space. Now remove one point from it. Is there a metric on $K$ with one point removed that generates the same topology as the initial metric but is such that the space $K$ with one point removed is still complete?
calculus real-analysis general-topology analysis metric-spaces
calculus real-analysis general-topology analysis metric-spaces
asked Nov 29 at 0:27
Sascha
136318
asked Nov 29 at 0:27
Sascha
136318
asked Nov 29 at 0:27
Sascha
136318
asked Nov 29 at 0:27
Sascha
136318
asked Nov 29 at 0:27
Sascha
136318
136318
add a comment |
add a comment |
1 Answer
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If $(K,d)$ is any complete metric space and $x in K$ then $D(u,v)=d(u,v)+|frac 1 {d(u,x)} -frac 1 {d(v,x)}|$ makes $Ksetminus {x}$ a complete metric space and $d$ is equivalent to $D$ on $Ksetminus {x}$.
answered Nov 29 at 0:36
Kavi Rama Murthy
48.7k31854
1
This technique can be modified to work for all $G_delta$ subsets of a complete metric space.
– Henno Brandsma
Dec 2 at 10:11
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $(K,d)$ is any complete metric space and $x in K$ then $D(u,v)=d(u,v)+|frac 1 {d(u,x)} -frac 1 {d(v,x)}|$ makes $Ksetminus {x}$ a complete metric space and $d$ is equivalent to $D$ on $Ksetminus {x}$.
answered Nov 29 at 0:36
Kavi Rama Murthy
48.7k31854
1
This technique can be modified to work for all $G_delta$ subsets of a complete metric space.
– Henno Brandsma
Dec 2 at 10:11
add a comment |
If $(K,d)$ is any complete metric space and $x in K$ then $D(u,v)=d(u,v)+|frac 1 {d(u,x)} -frac 1 {d(v,x)}|$ makes $Ksetminus {x}$ a complete metric space and $d$ is equivalent to $D$ on $Ksetminus {x}$.
answered Nov 29 at 0:36
Kavi Rama Murthy
48.7k31854
1
This technique can be modified to work for all $G_delta$ subsets of a complete metric space.
– Henno Brandsma
Dec 2 at 10:11
add a comment |
If $(K,d)$ is any complete metric space and $x in K$ then $D(u,v)=d(u,v)+|frac 1 {d(u,x)} -frac 1 {d(v,x)}|$ makes $Ksetminus {x}$ a complete metric space and $d$ is equivalent to $D$ on $Ksetminus {x}$.
answered Nov 29 at 0:36
Kavi Rama Murthy
48.7k31854
If $(K,d)$ is any complete metric space and $x in K$ then $D(u,v)=d(u,v)+|frac 1 {d(u,x)} -frac 1 {d(v,x)}|$ makes $Ksetminus {x}$ a complete metric space and $d$ is equivalent to $D$ on $Ksetminus {x}$.
answered Nov 29 at 0:36
Kavi Rama Murthy
48.7k31854
answered Nov 29 at 0:36
Kavi Rama Murthy
48.7k31854
answered Nov 29 at 0:36
Kavi Rama Murthy
48.7k31854
answered Nov 29 at 0:36
Kavi Rama Murthy
48.7k31854
48.7k31854
1
This technique can be modified to work for all $G_delta$ subsets of a complete metric space.
– Henno Brandsma
Dec 2 at 10:11
add a comment |
1
This technique can be modified to work for all $G_delta$ subsets of a complete metric space.
– Henno Brandsma
Dec 2 at 10:11
1
1
This technique can be modified to work for all $G_delta$ subsets of a complete metric space.
– Henno Brandsma
Dec 2 at 10:11
This technique can be modified to work for all $G_delta$ subsets of a complete metric space.
– Henno Brandsma
Dec 2 at 10:11
add a comment |
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