Proving Continuity of the Nearest Point function= Inf d(x,y) , y $in$ A , f:{$R^nto R$} (A is closed subset...












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  • Distance is (uniformly) continuous

    2 answers



  • Prove that $f_A (x) = d({{x}}, A)$, is continuous.

    2 answers




Let A be a non empty closed subset of $R^n$. $f:R^nto R$ by $f(x)= d(x,A)$ at each $xin R^n$ , where $d(x,A) = inf {d(x,y) , y in A}$.
Prove that $f$ is continuous on $R^n$.



It seems that it is true, but sets are not compact. I need some hints. Do you have any?










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marked as duplicate by José Carlos Santos, Aweygan, amWhy, Martin Sleziak, Davide Giraudo Nov 26 '18 at 21:55


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    Use the inequality $|d(x,A)-d(y,A)|leq d(x,y)$
    $endgroup$
    – UserS
    Nov 26 '18 at 17:49
















0












$begingroup$



This question already has an answer here:




  • Distance is (uniformly) continuous

    2 answers



  • Prove that $f_A (x) = d({{x}}, A)$, is continuous.

    2 answers




Let A be a non empty closed subset of $R^n$. $f:R^nto R$ by $f(x)= d(x,A)$ at each $xin R^n$ , where $d(x,A) = inf {d(x,y) , y in A}$.
Prove that $f$ is continuous on $R^n$.



It seems that it is true, but sets are not compact. I need some hints. Do you have any?










share|cite|improve this question











$endgroup$



marked as duplicate by José Carlos Santos, Aweygan, amWhy, Martin Sleziak, Davide Giraudo Nov 26 '18 at 21:55


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    Use the inequality $|d(x,A)-d(y,A)|leq d(x,y)$
    $endgroup$
    – UserS
    Nov 26 '18 at 17:49














0












0








0





$begingroup$



This question already has an answer here:




  • Distance is (uniformly) continuous

    2 answers



  • Prove that $f_A (x) = d({{x}}, A)$, is continuous.

    2 answers




Let A be a non empty closed subset of $R^n$. $f:R^nto R$ by $f(x)= d(x,A)$ at each $xin R^n$ , where $d(x,A) = inf {d(x,y) , y in A}$.
Prove that $f$ is continuous on $R^n$.



It seems that it is true, but sets are not compact. I need some hints. Do you have any?










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Distance is (uniformly) continuous

    2 answers



  • Prove that $f_A (x) = d({{x}}, A)$, is continuous.

    2 answers




Let A be a non empty closed subset of $R^n$. $f:R^nto R$ by $f(x)= d(x,A)$ at each $xin R^n$ , where $d(x,A) = inf {d(x,y) , y in A}$.
Prove that $f$ is continuous on $R^n$.



It seems that it is true, but sets are not compact. I need some hints. Do you have any?





This question already has an answer here:




  • Distance is (uniformly) continuous

    2 answers



  • Prove that $f_A (x) = d({{x}}, A)$, is continuous.

    2 answers








general-topology functions continuity metric-spaces






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share|cite|improve this question













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edited Nov 26 '18 at 19:52









Martin Sleziak

44.7k9117272




44.7k9117272










asked Nov 26 '18 at 17:47









EKarabulutEKarabulut

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marked as duplicate by José Carlos Santos, Aweygan, amWhy, Martin Sleziak, Davide Giraudo Nov 26 '18 at 21:55


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by José Carlos Santos, Aweygan, amWhy, Martin Sleziak, Davide Giraudo Nov 26 '18 at 21:55


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Use the inequality $|d(x,A)-d(y,A)|leq d(x,y)$
    $endgroup$
    – UserS
    Nov 26 '18 at 17:49


















  • $begingroup$
    Use the inequality $|d(x,A)-d(y,A)|leq d(x,y)$
    $endgroup$
    – UserS
    Nov 26 '18 at 17:49
















$begingroup$
Use the inequality $|d(x,A)-d(y,A)|leq d(x,y)$
$endgroup$
– UserS
Nov 26 '18 at 17:49




$begingroup$
Use the inequality $|d(x,A)-d(y,A)|leq d(x,y)$
$endgroup$
– UserS
Nov 26 '18 at 17:49










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