On a version of Urysohn lemma for complete metric spaces, involving uniform continuous functions












3












$begingroup$


Let $X$ be a complete metric space. Let us say that given a subset $Asubseteq X$, a point $ain X$ is a limit point of $A$ if for every $r>0, A cap B(a,r)setminus {a} ne phi$. Now let $A,B$ be disjoint closed subsets of a complete metric space $X$ such that $A,B$ has no limit point in $X$ and also assume that $A,B$ are infinite. Then, does there necessarily exist a uniformly continuous function $f: X to mathbb R$ such that $d(f(A),f(B)):=inf {|f(a)-f(b)| : ain A, bin B}>0$ ?






general-topology metric-spaces uniform-continuity complete-spaces







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  • $begingroup$
    You want $A$ and $B$ to only have isolated points?
    $endgroup$
    – mathworker21
    Dec 28 '18 at 2:16










  • $begingroup$
    @mathworker21: yes exactly right
    $endgroup$
    – user521337
    Dec 28 '18 at 2:28
















3












$begingroup$


Let $X$ be a complete metric space. Let us say that given a subset $Asubseteq X$, a point $ain X$ is a limit point of $A$ if for every $r>0, A cap B(a,r)setminus {a} ne phi$. Now let $A,B$ be disjoint closed subsets of a complete metric space $X$ such that $A,B$ has no limit point in $X$ and also assume that $A,B$ are infinite. Then, does there necessarily exist a uniformly continuous function $f: X to mathbb R$ such that $d(f(A),f(B)):=inf {|f(a)-f(b)| : ain A, bin B}>0$ ?






general-topology metric-spaces uniform-continuity complete-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    You want $A$ and $B$ to only have isolated points?
    $endgroup$
    – mathworker21
    Dec 28 '18 at 2:16










  • $begingroup$
    @mathworker21: yes exactly right
    $endgroup$
    – user521337
    Dec 28 '18 at 2:28














3












3








3





$begingroup$


Let $X$ be a complete metric space. Let us say that given a subset $Asubseteq X$, a point $ain X$ is a limit point of $A$ if for every $r>0, A cap B(a,r)setminus {a} ne phi$. Now let $A,B$ be disjoint closed subsets of a complete metric space $X$ such that $A,B$ has no limit point in $X$ and also assume that $A,B$ are infinite. Then, does there necessarily exist a uniformly continuous function $f: X to mathbb R$ such that $d(f(A),f(B)):=inf {|f(a)-f(b)| : ain A, bin B}>0$ ?






general-topology metric-spaces uniform-continuity complete-spaces







share|cite|improve this question









$endgroup$




Let $X$ be a complete metric space. Let us say that given a subset $Asubseteq X$, a point $ain X$ is a limit point of $A$ if for every $r>0, A cap B(a,r)setminus {a} ne phi$. Now let $A,B$ be disjoint closed subsets of a complete metric space $X$ such that $A,B$ has no limit point in $X$ and also assume that $A,B$ are infinite. Then, does there necessarily exist a uniformly continuous function $f: X to mathbb R$ such that $d(f(A),f(B)):=inf {|f(a)-f(b)| : ain A, bin B}>0$ ?






general-topology metric-spaces uniform-continuity complete-spaces




general-topology metric-spaces uniform-continuity complete-spaces






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Dec 28 '18 at 2:11









user521337user521337

1,1981417




1,1981417












  • $begingroup$
    You want $A$ and $B$ to only have isolated points?
    $endgroup$
    – mathworker21
    Dec 28 '18 at 2:16










  • $begingroup$
    @mathworker21: yes exactly right
    $endgroup$
    – user521337
    Dec 28 '18 at 2:28


















  • $begingroup$
    You want $A$ and $B$ to only have isolated points?
    $endgroup$
    – mathworker21
    Dec 28 '18 at 2:16










  • $begingroup$
    @mathworker21: yes exactly right
    $endgroup$
    – user521337
    Dec 28 '18 at 2:28
















$begingroup$
You want $A$ and $B$ to only have isolated points?
$endgroup$
– mathworker21
Dec 28 '18 at 2:16




$begingroup$
You want $A$ and $B$ to only have isolated points?
$endgroup$
– mathworker21
Dec 28 '18 at 2:16












$begingroup$
@mathworker21: yes exactly right
$endgroup$
– user521337
Dec 28 '18 at 2:28




$begingroup$
@mathworker21: yes exactly right
$endgroup$
– user521337
Dec 28 '18 at 2:28










1 Answer
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No. Take $X$ to be the real line, take $A$to be the set of positive integers, and take $B$ to be the set of numbers of the form $n+2^{-n}$ for positive integers $n$. If $f$ is uniformly continuous, then, for any $varepsilon>0$, we'll have $|f(n)-f(n+2^{-n})|<varepsilon$ for all sufficiently large $n$. So the infimum in your question will be zero.






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    1 Answer
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    $begingroup$

    No. Take $X$ to be the real line, take $A$to be the set of positive integers, and take $B$ to be the set of numbers of the form $n+2^{-n}$ for positive integers $n$. If $f$ is uniformly continuous, then, for any $varepsilon>0$, we'll have $|f(n)-f(n+2^{-n})|<varepsilon$ for all sufficiently large $n$. So the infimum in your question will be zero.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      No. Take $X$ to be the real line, take $A$to be the set of positive integers, and take $B$ to be the set of numbers of the form $n+2^{-n}$ for positive integers $n$. If $f$ is uniformly continuous, then, for any $varepsilon>0$, we'll have $|f(n)-f(n+2^{-n})|<varepsilon$ for all sufficiently large $n$. So the infimum in your question will be zero.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        No. Take $X$ to be the real line, take $A$to be the set of positive integers, and take $B$ to be the set of numbers of the form $n+2^{-n}$ for positive integers $n$. If $f$ is uniformly continuous, then, for any $varepsilon>0$, we'll have $|f(n)-f(n+2^{-n})|<varepsilon$ for all sufficiently large $n$. So the infimum in your question will be zero.






        share|cite|improve this answer









        $endgroup$



        No. Take $X$ to be the real line, take $A$to be the set of positive integers, and take $B$ to be the set of numbers of the form $n+2^{-n}$ for positive integers $n$. If $f$ is uniformly continuous, then, for any $varepsilon>0$, we'll have $|f(n)-f(n+2^{-n})|<varepsilon$ for all sufficiently large $n$. So the infimum in your question will be zero.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 3:44









        Andreas BlassAndreas Blass

        50.3k452109




        50.3k452109






























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