If $C$ is a closed subset of a metric space, then $barB(C, r)$ is a closed subset also?

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Let $(X,d)$ be a metric space and $C$ be a closed subset of $X$. Let $barB(C, r) := exists c in C textits.t. d(x, c)≤r $. Then is $barB(C, r)$ a closed subset also?



Is the statement true if $X$ is a normed vector space over $mathbbR$ and $C$ is a closed, bounded convex subset of $X$?




real-analysis general-topology functional-analysis metric-spaces normed-spaces




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  • Let $C=y_1,y_2,...$ be such that the distance between any two of its elements if $1$. Let $x_1,x_2,...$ be such that $d(x_i,y_j)=2$, for all $i,j$, $d(x_i,x_j)=1$ for $ineq j$. Let $x_0$ be such that $d(x_0,x_i)=1+1/i$, $d(x_0,y_i)=2+1/i$.
    – user578878
    Jul 27 at 12:53










  • No - simple counterexample in the answer below. It's true if you assume $C$ is compact.
    – David C. Ullrich
    Jul 27 at 16:01















up vote
3
down vote

favorite












Let $(X,d)$ be a metric space and $C$ be a closed subset of $X$. Let $barB(C, r) := exists c in C textits.t. d(x, c)≤r $. Then is $barB(C, r)$ a closed subset also?



Is the statement true if $X$ is a normed vector space over $mathbbR$ and $C$ is a closed, bounded convex subset of $X$?




real-analysis general-topology functional-analysis metric-spaces normed-spaces




share|cite|improve this question



















  • Let $C=y_1,y_2,...$ be such that the distance between any two of its elements if $1$. Let $x_1,x_2,...$ be such that $d(x_i,y_j)=2$, for all $i,j$, $d(x_i,x_j)=1$ for $ineq j$. Let $x_0$ be such that $d(x_0,x_i)=1+1/i$, $d(x_0,y_i)=2+1/i$.
    – user578878
    Jul 27 at 12:53










  • No - simple counterexample in the answer below. It's true if you assume $C$ is compact.
    – David C. Ullrich
    Jul 27 at 16:01













up vote
3
down vote

favorite









up vote
3
down vote

favorite











Let $(X,d)$ be a metric space and $C$ be a closed subset of $X$. Let $barB(C, r) := exists c in C textits.t. d(x, c)≤r $. Then is $barB(C, r)$ a closed subset also?



Is the statement true if $X$ is a normed vector space over $mathbbR$ and $C$ is a closed, bounded convex subset of $X$?




real-analysis general-topology functional-analysis metric-spaces normed-spaces




share|cite|improve this question











Let $(X,d)$ be a metric space and $C$ be a closed subset of $X$. Let $barB(C, r) := exists c in C textits.t. d(x, c)≤r $. Then is $barB(C, r)$ a closed subset also?



Is the statement true if $X$ is a normed vector space over $mathbbR$ and $C$ is a closed, bounded convex subset of $X$?





real-analysis general-topology functional-analysis metric-spaces normed-spaces





share|cite|improve this question










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asked Jul 27 at 12:17









GouldBach

3368




3368











  • Let $C=y_1,y_2,...$ be such that the distance between any two of its elements if $1$. Let $x_1,x_2,...$ be such that $d(x_i,y_j)=2$, for all $i,j$, $d(x_i,x_j)=1$ for $ineq j$. Let $x_0$ be such that $d(x_0,x_i)=1+1/i$, $d(x_0,y_i)=2+1/i$.
    – user578878
    Jul 27 at 12:53










  • No - simple counterexample in the answer below. It's true if you assume $C$ is compact.
    – David C. Ullrich
    Jul 27 at 16:01

















  • Let $C=y_1,y_2,...$ be such that the distance between any two of its elements if $1$. Let $x_1,x_2,...$ be such that $d(x_i,y_j)=2$, for all $i,j$, $d(x_i,x_j)=1$ for $ineq j$. Let $x_0$ be such that $d(x_0,x_i)=1+1/i$, $d(x_0,y_i)=2+1/i$.
    – user578878
    Jul 27 at 12:53










  • No - simple counterexample in the answer below. It's true if you assume $C$ is compact.
    – David C. Ullrich
    Jul 27 at 16:01
















Let $C=y_1,y_2,...$ be such that the distance between any two of its elements if $1$. Let $x_1,x_2,...$ be such that $d(x_i,y_j)=2$, for all $i,j$, $d(x_i,x_j)=1$ for $ineq j$. Let $x_0$ be such that $d(x_0,x_i)=1+1/i$, $d(x_0,y_i)=2+1/i$.
– user578878
Jul 27 at 12:53




Let $C=y_1,y_2,...$ be such that the distance between any two of its elements if $1$. Let $x_1,x_2,...$ be such that $d(x_i,y_j)=2$, for all $i,j$, $d(x_i,x_j)=1$ for $ineq j$. Let $x_0$ be such that $d(x_0,x_i)=1+1/i$, $d(x_0,y_i)=2+1/i$.
– user578878
Jul 27 at 12:53












No - simple counterexample in the answer below. It's true if you assume $C$ is compact.
– David C. Ullrich
Jul 27 at 16:01





No - simple counterexample in the answer below. It's true if you assume $C$ is compact.
– David C. Ullrich
Jul 27 at 16:01











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Oberve the space $X=mathbb R-0$ equipped with subspace topology inherited from $mathbb R$.



Then $C:=(0,infty)$ is a closed subset.



But $overline B(C,1)=(-1,infty)setminus0$ is not closed.






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    Oberve the space $X=mathbb R-0$ equipped with subspace topology inherited from $mathbb R$.



    Then $C:=(0,infty)$ is a closed subset.



    But $overline B(C,1)=(-1,infty)setminus0$ is not closed.






    share|cite|improve this answer

























      up vote
      5
      down vote



      accepted










      Oberve the space $X=mathbb R-0$ equipped with subspace topology inherited from $mathbb R$.



      Then $C:=(0,infty)$ is a closed subset.



      But $overline B(C,1)=(-1,infty)setminus0$ is not closed.






      share|cite|improve this answer























        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        Oberve the space $X=mathbb R-0$ equipped with subspace topology inherited from $mathbb R$.



        Then $C:=(0,infty)$ is a closed subset.



        But $overline B(C,1)=(-1,infty)setminus0$ is not closed.






        share|cite|improve this answer













        Oberve the space $X=mathbb R-0$ equipped with subspace topology inherited from $mathbb R$.



        Then $C:=(0,infty)$ is a closed subset.



        But $overline B(C,1)=(-1,infty)setminus0$ is not closed.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 27 at 12:52









        drhab

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