Example of a metric space where diameter of a ball is not equal twice the radius

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Is there any metric space where the diameter of a ball is smaller than twice the radius?






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  • Is it correct to consider the discrete metric?
    – M.billy
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Is there any metric space where the diameter of a ball is smaller than twice the radius?






metric-spaces







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  • Is it correct to consider the discrete metric?
    – M.billy
    2 hours ago












up vote
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up vote
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down vote

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Is there any metric space where the diameter of a ball is smaller than twice the radius?






metric-spaces







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Is there any metric space where the diameter of a ball is smaller than twice the radius?






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  • Is it correct to consider the discrete metric?
    – M.billy
    2 hours ago
















  • Is it correct to consider the discrete metric?
    – M.billy
    2 hours ago















Is it correct to consider the discrete metric?
– M.billy
2 hours ago




Is it correct to consider the discrete metric?
– M.billy
2 hours ago










3 Answers
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Any ultrametric space would do, for example, the discrete metric on any set, or $mathbbZ$ with the $p$-adic metric.






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    How about the metric space $[0,infty)$ with the standard metric? The diameter of the ball $B(0,1)$ is $1$, not $2$ (which would be twice its radius)






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    • A problem with the question is that in general, in a metric space we can have $B(x,r)=B(x',r')$ with $xne x'$ or $rne r'$ or both, so an open ball may not have a unique radius or center. Your example is good, but if we ask about closed balls, then $overline B(0,1) =overline B(1/2,1/2)=[0,1]$.
      – DanielWainfleet
      5 mins ago


















    up vote
    1
    down vote













    Consider the discrete metric $d$ on a set $X$:



    $$d(x,y)=begincases
    0,&textif x=y\
    1,&textif xne y;.
    endcases$$



    Consider the ball of radius $r=1/2$ centered at $x$



    Then $B(x,r)=x$



    Now by definition, $operatornamediam A = sup d(a,b) : a, b in A $



    Applying it to our case where $A=B(x,r)$, we have diameter of $A$ equal to $0$






    share|cite|improve this answer




















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      3 Answers
      3






      active

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      3 Answers
      3






      active

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      active

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      active

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      up vote
      1
      down vote













      Any ultrametric space would do, for example, the discrete metric on any set, or $mathbbZ$ with the $p$-adic metric.






      share|cite|improve this answer
























        up vote
        1
        down vote













        Any ultrametric space would do, for example, the discrete metric on any set, or $mathbbZ$ with the $p$-adic metric.






        share|cite|improve this answer






















          up vote
          1
          down vote










          up vote
          1
          down vote









          Any ultrametric space would do, for example, the discrete metric on any set, or $mathbbZ$ with the $p$-adic metric.






          share|cite|improve this answer












          Any ultrametric space would do, for example, the discrete metric on any set, or $mathbbZ$ with the $p$-adic metric.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          user10354138

          3,405117




          3,405117




















              up vote
              1
              down vote













              How about the metric space $[0,infty)$ with the standard metric? The diameter of the ball $B(0,1)$ is $1$, not $2$ (which would be twice its radius)






              share|cite|improve this answer




















              • A problem with the question is that in general, in a metric space we can have $B(x,r)=B(x',r')$ with $xne x'$ or $rne r'$ or both, so an open ball may not have a unique radius or center. Your example is good, but if we ask about closed balls, then $overline B(0,1) =overline B(1/2,1/2)=[0,1]$.
                – DanielWainfleet
                5 mins ago















              up vote
              1
              down vote













              How about the metric space $[0,infty)$ with the standard metric? The diameter of the ball $B(0,1)$ is $1$, not $2$ (which would be twice its radius)






              share|cite|improve this answer




















              • A problem with the question is that in general, in a metric space we can have $B(x,r)=B(x',r')$ with $xne x'$ or $rne r'$ or both, so an open ball may not have a unique radius or center. Your example is good, but if we ask about closed balls, then $overline B(0,1) =overline B(1/2,1/2)=[0,1]$.
                – DanielWainfleet
                5 mins ago













              up vote
              1
              down vote










              up vote
              1
              down vote









              How about the metric space $[0,infty)$ with the standard metric? The diameter of the ball $B(0,1)$ is $1$, not $2$ (which would be twice its radius)






              share|cite|improve this answer












              How about the metric space $[0,infty)$ with the standard metric? The diameter of the ball $B(0,1)$ is $1$, not $2$ (which would be twice its radius)







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 2 hours ago









              5xum

              85.8k388154




              85.8k388154











              • A problem with the question is that in general, in a metric space we can have $B(x,r)=B(x',r')$ with $xne x'$ or $rne r'$ or both, so an open ball may not have a unique radius or center. Your example is good, but if we ask about closed balls, then $overline B(0,1) =overline B(1/2,1/2)=[0,1]$.
                – DanielWainfleet
                5 mins ago

















              • A problem with the question is that in general, in a metric space we can have $B(x,r)=B(x',r')$ with $xne x'$ or $rne r'$ or both, so an open ball may not have a unique radius or center. Your example is good, but if we ask about closed balls, then $overline B(0,1) =overline B(1/2,1/2)=[0,1]$.
                – DanielWainfleet
                5 mins ago
















              A problem with the question is that in general, in a metric space we can have $B(x,r)=B(x',r')$ with $xne x'$ or $rne r'$ or both, so an open ball may not have a unique radius or center. Your example is good, but if we ask about closed balls, then $overline B(0,1) =overline B(1/2,1/2)=[0,1]$.
              – DanielWainfleet
              5 mins ago





              A problem with the question is that in general, in a metric space we can have $B(x,r)=B(x',r')$ with $xne x'$ or $rne r'$ or both, so an open ball may not have a unique radius or center. Your example is good, but if we ask about closed balls, then $overline B(0,1) =overline B(1/2,1/2)=[0,1]$.
              – DanielWainfleet
              5 mins ago











              up vote
              1
              down vote













              Consider the discrete metric $d$ on a set $X$:



              $$d(x,y)=begincases
              0,&textif x=y\
              1,&textif xne y;.
              endcases$$



              Consider the ball of radius $r=1/2$ centered at $x$



              Then $B(x,r)=x$



              Now by definition, $operatornamediam A = sup d(a,b) : a, b in A $



              Applying it to our case where $A=B(x,r)$, we have diameter of $A$ equal to $0$






              share|cite|improve this answer
























                up vote
                1
                down vote













                Consider the discrete metric $d$ on a set $X$:



                $$d(x,y)=begincases
                0,&textif x=y\
                1,&textif xne y;.
                endcases$$



                Consider the ball of radius $r=1/2$ centered at $x$



                Then $B(x,r)=x$



                Now by definition, $operatornamediam A = sup d(a,b) : a, b in A $



                Applying it to our case where $A=B(x,r)$, we have diameter of $A$ equal to $0$






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Consider the discrete metric $d$ on a set $X$:



                  $$d(x,y)=begincases
                  0,&textif x=y\
                  1,&textif xne y;.
                  endcases$$



                  Consider the ball of radius $r=1/2$ centered at $x$



                  Then $B(x,r)=x$



                  Now by definition, $operatornamediam A = sup d(a,b) : a, b in A $



                  Applying it to our case where $A=B(x,r)$, we have diameter of $A$ equal to $0$






                  share|cite|improve this answer












                  Consider the discrete metric $d$ on a set $X$:



                  $$d(x,y)=begincases
                  0,&textif x=y\
                  1,&textif xne y;.
                  endcases$$



                  Consider the ball of radius $r=1/2$ centered at $x$



                  Then $B(x,r)=x$



                  Now by definition, $operatornamediam A = sup d(a,b) : a, b in A $



                  Applying it to our case where $A=B(x,r)$, we have diameter of $A$ equal to $0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  StammeringMathematician

                  1,738221




                  1,738221




















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