Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x in U$ and $B subseteq V$.












1












$begingroup$


Let $B$ be a compact subset of a metric space $X$. Let $x in X setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x in U$ and $B subseteq V$.



Is this an existence proof? If yes, do I get to define $U,V$?






real-analysis







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$endgroup$












  • $begingroup$
    you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
    $endgroup$
    – Martund
    Dec 20 '18 at 2:39








  • 1




    $begingroup$
    Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
    $endgroup$
    – Clayton
    Dec 20 '18 at 3:00
















1












$begingroup$


Let $B$ be a compact subset of a metric space $X$. Let $x in X setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x in U$ and $B subseteq V$.



Is this an existence proof? If yes, do I get to define $U,V$?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
    $endgroup$
    – Martund
    Dec 20 '18 at 2:39








  • 1




    $begingroup$
    Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
    $endgroup$
    – Clayton
    Dec 20 '18 at 3:00














1












1








1





$begingroup$


Let $B$ be a compact subset of a metric space $X$. Let $x in X setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x in U$ and $B subseteq V$.



Is this an existence proof? If yes, do I get to define $U,V$?






real-analysis







share|cite|improve this question











$endgroup$




Let $B$ be a compact subset of a metric space $X$. Let $x in X setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x in U$ and $B subseteq V$.



Is this an existence proof? If yes, do I get to define $U,V$?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 14:22







mathdaddy

















asked Dec 20 '18 at 2:25









mathdaddymathdaddy

133




133












  • $begingroup$
    you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
    $endgroup$
    – Martund
    Dec 20 '18 at 2:39








  • 1




    $begingroup$
    Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
    $endgroup$
    – Clayton
    Dec 20 '18 at 3:00


















  • $begingroup$
    you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
    $endgroup$
    – Martund
    Dec 20 '18 at 2:39








  • 1




    $begingroup$
    Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
    $endgroup$
    – Clayton
    Dec 20 '18 at 3:00
















$begingroup$
you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
$endgroup$
– Martund
Dec 20 '18 at 2:39






$begingroup$
you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
$endgroup$
– Martund
Dec 20 '18 at 2:39






1




1




$begingroup$
Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
$endgroup$
– Clayton
Dec 20 '18 at 3:00




$begingroup$
Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
$endgroup$
– Clayton
Dec 20 '18 at 3:00










2 Answers
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$B$ is compact so it is closed. Its complement is open so for any $xin B^c$ there exists $epsilongt 0$ such that the open ball of radius $epsilon$ around $x$ is a subset of $B^c$.



Let $U$ be the open ball of radius $frac{epsilon}{2}$ around $x$ and let $V={y:d(x,y)gtfrac{epsilon}{2}}$. Then $xin U$ and $Bsubset{y:d(x,y)geepsilon}subset{y:d(x,y)gtfrac{epsilon}{2}}=V$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)setminus{b}cap Bneqvarnothing;forall;r>0.$



    Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.



    Instead choose $delta=inf{d(x,y):::yin B}$. Consider $U=Bleft(x,dfrac{delta}{2}right)$ and $displaystyle V=bigcup_{bin B} Bleft(b,frac{delta}{2}right)$






    share|cite|improve this answer











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      2 Answers
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      active

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      2 Answers
      2






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      active

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      0












      $begingroup$

      $B$ is compact so it is closed. Its complement is open so for any $xin B^c$ there exists $epsilongt 0$ such that the open ball of radius $epsilon$ around $x$ is a subset of $B^c$.



      Let $U$ be the open ball of radius $frac{epsilon}{2}$ around $x$ and let $V={y:d(x,y)gtfrac{epsilon}{2}}$. Then $xin U$ and $Bsubset{y:d(x,y)geepsilon}subset{y:d(x,y)gtfrac{epsilon}{2}}=V$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $B$ is compact so it is closed. Its complement is open so for any $xin B^c$ there exists $epsilongt 0$ such that the open ball of radius $epsilon$ around $x$ is a subset of $B^c$.



        Let $U$ be the open ball of radius $frac{epsilon}{2}$ around $x$ and let $V={y:d(x,y)gtfrac{epsilon}{2}}$. Then $xin U$ and $Bsubset{y:d(x,y)geepsilon}subset{y:d(x,y)gtfrac{epsilon}{2}}=V$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $B$ is compact so it is closed. Its complement is open so for any $xin B^c$ there exists $epsilongt 0$ such that the open ball of radius $epsilon$ around $x$ is a subset of $B^c$.



          Let $U$ be the open ball of radius $frac{epsilon}{2}$ around $x$ and let $V={y:d(x,y)gtfrac{epsilon}{2}}$. Then $xin U$ and $Bsubset{y:d(x,y)geepsilon}subset{y:d(x,y)gtfrac{epsilon}{2}}=V$.






          share|cite|improve this answer









          $endgroup$



          $B$ is compact so it is closed. Its complement is open so for any $xin B^c$ there exists $epsilongt 0$ such that the open ball of radius $epsilon$ around $x$ is a subset of $B^c$.



          Let $U$ be the open ball of radius $frac{epsilon}{2}$ around $x$ and let $V={y:d(x,y)gtfrac{epsilon}{2}}$. Then $xin U$ and $Bsubset{y:d(x,y)geepsilon}subset{y:d(x,y)gtfrac{epsilon}{2}}=V$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 3:16









          John DoumaJohn Douma

          5,62711420




          5,62711420























              0












              $begingroup$

              You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)setminus{b}cap Bneqvarnothing;forall;r>0.$



              Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.



              Instead choose $delta=inf{d(x,y):::yin B}$. Consider $U=Bleft(x,dfrac{delta}{2}right)$ and $displaystyle V=bigcup_{bin B} Bleft(b,frac{delta}{2}right)$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)setminus{b}cap Bneqvarnothing;forall;r>0.$



                Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.



                Instead choose $delta=inf{d(x,y):::yin B}$. Consider $U=Bleft(x,dfrac{delta}{2}right)$ and $displaystyle V=bigcup_{bin B} Bleft(b,frac{delta}{2}right)$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)setminus{b}cap Bneqvarnothing;forall;r>0.$



                  Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.



                  Instead choose $delta=inf{d(x,y):::yin B}$. Consider $U=Bleft(x,dfrac{delta}{2}right)$ and $displaystyle V=bigcup_{bin B} Bleft(b,frac{delta}{2}right)$






                  share|cite|improve this answer











                  $endgroup$



                  You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)setminus{b}cap Bneqvarnothing;forall;r>0.$



                  Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.



                  Instead choose $delta=inf{d(x,y):::yin B}$. Consider $U=Bleft(x,dfrac{delta}{2}right)$ and $displaystyle V=bigcup_{bin B} Bleft(b,frac{delta}{2}right)$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 20 '18 at 3:00

























                  answered Dec 20 '18 at 2:31









                  Yadati KiranYadati Kiran

                  2,1121622




                  2,1121622






























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