Nearest point property and uniformly continuous image
0
$begingroup$
Show that a uniformly continuous image of a metric space that has the nearest point property need not have that property.
I have some trouble understanding the problem.
With the term ''uniformly continuous image'' does it mean that $f(X)=Y$ where $f$ is uniformly continuous ?
general-topology metric-spaces uniform-continuity
edited Dec 17 '18 at 18:33
Martin Sleziak
44.7k10119272
asked Dec 17 '18 at 12:56
argiriskarargiriskar
1409
$endgroup$
add a comment |
0
$begingroup$
Show that a uniformly continuous image of a metric space that has the nearest point property need not have that property.
I have some trouble understanding the problem.
With the term ''uniformly continuous image'' does it mean that $f(X)=Y$ where $f$ is uniformly continuous ?
general-topology metric-spaces uniform-continuity
edited Dec 17 '18 at 18:33
Martin Sleziak
44.7k10119272
asked Dec 17 '18 at 12:56
argiriskarargiriskar
1409
$endgroup$
$begingroup$
About the second question: Yes.
$endgroup$
– Yanko
Dec 17 '18 at 12:57
$begingroup$
So i need to show that $f$ does not need the property of nearest point to be uniformly continuous ?
$endgroup$
– argiriskar
Dec 17 '18 at 12:59
2
$begingroup$
As I understand this you need to find a metric space $X$ which satisfies the nearest point property and a uniformly continuous function $f$ such that $f(X)$ does not satisfy the nearest point property.
$endgroup$
– Yanko
Dec 17 '18 at 13:00
1
$begingroup$
What is the nearest point property?
$endgroup$
– Paul Frost
Dec 17 '18 at 13:57
add a comment |
0
0
0
$begingroup$
Show that a uniformly continuous image of a metric space that has the nearest point property need not have that property.
I have some trouble understanding the problem.
With the term ''uniformly continuous image'' does it mean that $f(X)=Y$ where $f$ is uniformly continuous ?
general-topology metric-spaces uniform-continuity
edited Dec 17 '18 at 18:33
Martin Sleziak
44.7k10119272
asked Dec 17 '18 at 12:56
argiriskarargiriskar
1409
$endgroup$
Show that a uniformly continuous image of a metric space that has the nearest point property need not have that property.
I have some trouble understanding the problem.
With the term ''uniformly continuous image'' does it mean that $f(X)=Y$ where $f$ is uniformly continuous ?
general-topology metric-spaces uniform-continuity
general-topology metric-spaces uniform-continuity
edited Dec 17 '18 at 18:33
Martin Sleziak
44.7k10119272
asked Dec 17 '18 at 12:56
argiriskarargiriskar
1409
edited Dec 17 '18 at 18:33
Martin Sleziak
44.7k10119272
asked Dec 17 '18 at 12:56
argiriskarargiriskar
1409
edited Dec 17 '18 at 18:33
Martin Sleziak
44.7k10119272
edited Dec 17 '18 at 18:33
Martin Sleziak
44.7k10119272
edited Dec 17 '18 at 18:33
Martin Sleziak
44.7k10119272
44.7k10119272
asked Dec 17 '18 at 12:56
argiriskarargiriskar
1409
asked Dec 17 '18 at 12:56
argiriskarargiriskar
1409
asked Dec 17 '18 at 12:56
argiriskarargiriskar
1409
1409
$begingroup$
About the second question: Yes.
$endgroup$
– Yanko
Dec 17 '18 at 12:57
$begingroup$
So i need to show that $f$ does not need the property of nearest point to be uniformly continuous ?
$endgroup$
– argiriskar
Dec 17 '18 at 12:59
2
$begingroup$
As I understand this you need to find a metric space $X$ which satisfies the nearest point property and a uniformly continuous function $f$ such that $f(X)$ does not satisfy the nearest point property.
$endgroup$
– Yanko
Dec 17 '18 at 13:00
1
$begingroup$
What is the nearest point property?
$endgroup$
– Paul Frost
Dec 17 '18 at 13:57
add a comment |
$begingroup$
About the second question: Yes.
$endgroup$
– Yanko
Dec 17 '18 at 12:57
$begingroup$
So i need to show that $f$ does not need the property of nearest point to be uniformly continuous ?
$endgroup$
– argiriskar
Dec 17 '18 at 12:59
2
$begingroup$
As I understand this you need to find a metric space $X$ which satisfies the nearest point property and a uniformly continuous function $f$ such that $f(X)$ does not satisfy the nearest point property.
$endgroup$
– Yanko
Dec 17 '18 at 13:00
1
$begingroup$
What is the nearest point property?
$endgroup$
– Paul Frost
Dec 17 '18 at 13:57
$begingroup$
About the second question: Yes.
$endgroup$
– Yanko
Dec 17 '18 at 12:57
$begingroup$
About the second question: Yes.
$endgroup$
– Yanko
Dec 17 '18 at 12:57
$begingroup$
So i need to show that $f$ does not need the property of nearest point to be uniformly continuous ?
$endgroup$
– argiriskar
Dec 17 '18 at 12:59
$begingroup$
So i need to show that $f$ does not need the property of nearest point to be uniformly continuous ?
$endgroup$
– argiriskar
Dec 17 '18 at 12:59
2
2
$begingroup$
As I understand this you need to find a metric space $X$ which satisfies the nearest point property and a uniformly continuous function $f$ such that $f(X)$ does not satisfy the nearest point property.
$endgroup$
– Yanko
Dec 17 '18 at 13:00
$begingroup$
As I understand this you need to find a metric space $X$ which satisfies the nearest point property and a uniformly continuous function $f$ such that $f(X)$ does not satisfy the nearest point property.
$endgroup$
– Yanko
Dec 17 '18 at 13:00
1
1
$begingroup$
What is the nearest point property?
$endgroup$
– Paul Frost
Dec 17 '18 at 13:57
$begingroup$
What is the nearest point property?
$endgroup$
– Paul Frost
Dec 17 '18 at 13:57
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
I did not know this nearest point property, but I found it in Google books and there the author gives the following equivalent formulations:
- Every infinite bounded subset of $X$ has an accumulation point in $X$.
- Every bounded sequence has a subsequence that converges in $X$.
$X$ is complete and every bounded subset is totally bounded.
Now consider $X = mathbb{N}$ in the usual metric, which has the nearest point property. And note that $f(n) = frac{1}{n}$ is uniformly continuous onto a subspace of $mathbb{R}$ that does not have that property.
answered Dec 17 '18 at 17:45
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
How about $(X,d_0)$ where $|X|=infty$? If we take $f:Xrightarrow X$ then $f$ is uniformly continuous but it does not have the nearest point property when $X$ is infinite
$endgroup$
– argiriskar
Dec 17 '18 at 18:09
$begingroup$
@argiriskar what is $d_0$? If it is the discrete metric then it does not obey the NPP
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:11
$begingroup$
@argiriskar nearest point property
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:21
$begingroup$
yes i undestood that after i posted the question. But that's the point of the problem. We don't want the function to obey the NPP
$endgroup$
– argiriskar
Dec 17 '18 at 18:22
$begingroup$
@argiriskar the domain has to obey NPP but the image not. Like in my example.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:24
|
show 3 more comments
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1
$begingroup$
I did not know this nearest point property, but I found it in Google books and there the author gives the following equivalent formulations:
- Every infinite bounded subset of $X$ has an accumulation point in $X$.
- Every bounded sequence has a subsequence that converges in $X$.
$X$ is complete and every bounded subset is totally bounded.
Now consider $X = mathbb{N}$ in the usual metric, which has the nearest point property. And note that $f(n) = frac{1}{n}$ is uniformly continuous onto a subspace of $mathbb{R}$ that does not have that property.
answered Dec 17 '18 at 17:45
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
How about $(X,d_0)$ where $|X|=infty$? If we take $f:Xrightarrow X$ then $f$ is uniformly continuous but it does not have the nearest point property when $X$ is infinite
$endgroup$
– argiriskar
Dec 17 '18 at 18:09
$begingroup$
@argiriskar what is $d_0$? If it is the discrete metric then it does not obey the NPP
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:11
$begingroup$
@argiriskar nearest point property
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:21
$begingroup$
yes i undestood that after i posted the question. But that's the point of the problem. We don't want the function to obey the NPP
$endgroup$
– argiriskar
Dec 17 '18 at 18:22
$begingroup$
@argiriskar the domain has to obey NPP but the image not. Like in my example.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:24
|
show 3 more comments
1
$begingroup$
I did not know this nearest point property, but I found it in Google books and there the author gives the following equivalent formulations:
- Every infinite bounded subset of $X$ has an accumulation point in $X$.
- Every bounded sequence has a subsequence that converges in $X$.
$X$ is complete and every bounded subset is totally bounded.
Now consider $X = mathbb{N}$ in the usual metric, which has the nearest point property. And note that $f(n) = frac{1}{n}$ is uniformly continuous onto a subspace of $mathbb{R}$ that does not have that property.
answered Dec 17 '18 at 17:45
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
How about $(X,d_0)$ where $|X|=infty$? If we take $f:Xrightarrow X$ then $f$ is uniformly continuous but it does not have the nearest point property when $X$ is infinite
$endgroup$
– argiriskar
Dec 17 '18 at 18:09
$begingroup$
@argiriskar what is $d_0$? If it is the discrete metric then it does not obey the NPP
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:11
$begingroup$
@argiriskar nearest point property
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:21
$begingroup$
yes i undestood that after i posted the question. But that's the point of the problem. We don't want the function to obey the NPP
$endgroup$
– argiriskar
Dec 17 '18 at 18:22
$begingroup$
@argiriskar the domain has to obey NPP but the image not. Like in my example.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:24
|
show 3 more comments
1
1
1
$begingroup$
I did not know this nearest point property, but I found it in Google books and there the author gives the following equivalent formulations:
- Every infinite bounded subset of $X$ has an accumulation point in $X$.
- Every bounded sequence has a subsequence that converges in $X$.
$X$ is complete and every bounded subset is totally bounded.
Now consider $X = mathbb{N}$ in the usual metric, which has the nearest point property. And note that $f(n) = frac{1}{n}$ is uniformly continuous onto a subspace of $mathbb{R}$ that does not have that property.
answered Dec 17 '18 at 17:45
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
I did not know this nearest point property, but I found it in Google books and there the author gives the following equivalent formulations:
- Every infinite bounded subset of $X$ has an accumulation point in $X$.
- Every bounded sequence has a subsequence that converges in $X$.
$X$ is complete and every bounded subset is totally bounded.
Now consider $X = mathbb{N}$ in the usual metric, which has the nearest point property. And note that $f(n) = frac{1}{n}$ is uniformly continuous onto a subspace of $mathbb{R}$ that does not have that property.
answered Dec 17 '18 at 17:45
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 17 '18 at 17:45
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 17 '18 at 17:45
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 17 '18 at 17:45
Henno BrandsmaHenno Brandsma
111k348118
111k348118
$begingroup$
How about $(X,d_0)$ where $|X|=infty$? If we take $f:Xrightarrow X$ then $f$ is uniformly continuous but it does not have the nearest point property when $X$ is infinite
$endgroup$
– argiriskar
Dec 17 '18 at 18:09
$begingroup$
@argiriskar what is $d_0$? If it is the discrete metric then it does not obey the NPP
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:11
$begingroup$
@argiriskar nearest point property
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:21
$begingroup$
yes i undestood that after i posted the question. But that's the point of the problem. We don't want the function to obey the NPP
$endgroup$
– argiriskar
Dec 17 '18 at 18:22
$begingroup$
@argiriskar the domain has to obey NPP but the image not. Like in my example.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:24
|
show 3 more comments
$begingroup$
How about $(X,d_0)$ where $|X|=infty$? If we take $f:Xrightarrow X$ then $f$ is uniformly continuous but it does not have the nearest point property when $X$ is infinite
$endgroup$
– argiriskar
Dec 17 '18 at 18:09
$begingroup$
@argiriskar what is $d_0$? If it is the discrete metric then it does not obey the NPP
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:11
$begingroup$
@argiriskar nearest point property
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:21
$begingroup$
yes i undestood that after i posted the question. But that's the point of the problem. We don't want the function to obey the NPP
$endgroup$
– argiriskar
Dec 17 '18 at 18:22
$begingroup$
@argiriskar the domain has to obey NPP but the image not. Like in my example.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:24
$begingroup$
How about $(X,d_0)$ where $|X|=infty$? If we take $f:Xrightarrow X$ then $f$ is uniformly continuous but it does not have the nearest point property when $X$ is infinite
$endgroup$
– argiriskar
Dec 17 '18 at 18:09
$begingroup$
How about $(X,d_0)$ where $|X|=infty$? If we take $f:Xrightarrow X$ then $f$ is uniformly continuous but it does not have the nearest point property when $X$ is infinite
$endgroup$
– argiriskar
Dec 17 '18 at 18:09
$begingroup$
@argiriskar what is $d_0$? If it is the discrete metric then it does not obey the NPP
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:11
$begingroup$
@argiriskar what is $d_0$? If it is the discrete metric then it does not obey the NPP
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:11
$begingroup$
@argiriskar nearest point property
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:21
$begingroup$
@argiriskar nearest point property
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:21
$begingroup$
yes i undestood that after i posted the question. But that's the point of the problem. We don't want the function to obey the NPP
$endgroup$
– argiriskar
Dec 17 '18 at 18:22
$begingroup$
yes i undestood that after i posted the question. But that's the point of the problem. We don't want the function to obey the NPP
$endgroup$
– argiriskar
Dec 17 '18 at 18:22
$begingroup$
@argiriskar the domain has to obey NPP but the image not. Like in my example.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:24
$begingroup$
@argiriskar the domain has to obey NPP but the image not. Like in my example.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 18:24
|
show 3 more comments
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$begingroup$
About the second question: Yes.
$endgroup$
– Yanko
Dec 17 '18 at 12:57
$begingroup$
So i need to show that $f$ does not need the property of nearest point to be uniformly continuous ?
$endgroup$
– argiriskar
Dec 17 '18 at 12:59
2
$begingroup$
As I understand this you need to find a metric space $X$ which satisfies the nearest point property and a uniformly continuous function $f$ such that $f(X)$ does not satisfy the nearest point property.
$endgroup$
– Yanko
Dec 17 '18 at 13:00
1
$begingroup$
What is the nearest point property?
$endgroup$
– Paul Frost
Dec 17 '18 at 13:57