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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 18:33









Martin Sleziak

44.9k10121274




44.9k10121274










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


















  • $begingroup$
    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










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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

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active

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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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 17:45









Henno BrandsmaHenno Brandsma

113k348122




113k348122












  • $begingroup$
    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










  • $begingroup$
    @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


















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