Compute $operatorname{diam}(A)$ of the set $A ={(0,0),(1,1),(1,2),(2,2)}$ in metric space$ (mathbb R^2,d),$...












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Compute the diameter of $A,;$ i.e. compute $operatorname{diam}(A)$ of the set $A ={(0,0),(1,1),(1,2),(2,2)}$ in metric space $(mathbb R^2, d),$ where $d$ is defined by the formula:



$$dbig((x_1,y_1),(x_2,y_2)big)=begin{cases} |y_1-y_2|, & x_1 = x_2\ \ |x_1-x_2|+|y_1|+|y_2|, & x_1neq x_2 end{cases}.$$



Can anyone explain this clearly?



Thanks .










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closed as off-topic by José Carlos Santos, amWhy, Davide Giraudo, Cesareo, Leucippus Jan 3 at 0:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, amWhy, Davide Giraudo, Cesareo, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Just substitute in the formula and calculate the diameter.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 17:46
















-1












$begingroup$


Compute the diameter of $A,;$ i.e. compute $operatorname{diam}(A)$ of the set $A ={(0,0),(1,1),(1,2),(2,2)}$ in metric space $(mathbb R^2, d),$ where $d$ is defined by the formula:



$$dbig((x_1,y_1),(x_2,y_2)big)=begin{cases} |y_1-y_2|, & x_1 = x_2\ \ |x_1-x_2|+|y_1|+|y_2|, & x_1neq x_2 end{cases}.$$



Can anyone explain this clearly?



Thanks .










share|cite|improve this question











$endgroup$



closed as off-topic by José Carlos Santos, amWhy, Davide Giraudo, Cesareo, Leucippus Jan 3 at 0:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, amWhy, Davide Giraudo, Cesareo, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Just substitute in the formula and calculate the diameter.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 17:46














-1












-1








-1





$begingroup$


Compute the diameter of $A,;$ i.e. compute $operatorname{diam}(A)$ of the set $A ={(0,0),(1,1),(1,2),(2,2)}$ in metric space $(mathbb R^2, d),$ where $d$ is defined by the formula:



$$dbig((x_1,y_1),(x_2,y_2)big)=begin{cases} |y_1-y_2|, & x_1 = x_2\ \ |x_1-x_2|+|y_1|+|y_2|, & x_1neq x_2 end{cases}.$$



Can anyone explain this clearly?



Thanks .










share|cite|improve this question











$endgroup$




Compute the diameter of $A,;$ i.e. compute $operatorname{diam}(A)$ of the set $A ={(0,0),(1,1),(1,2),(2,2)}$ in metric space $(mathbb R^2, d),$ where $d$ is defined by the formula:



$$dbig((x_1,y_1),(x_2,y_2)big)=begin{cases} |y_1-y_2|, & x_1 = x_2\ \ |x_1-x_2|+|y_1|+|y_2|, & x_1neq x_2 end{cases}.$$



Can anyone explain this clearly?



Thanks .







real-analysis metric-spaces






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edited Jan 2 at 21:38









Chris Custer

11.6k3824




11.6k3824










asked Jan 2 at 17:38









Arda Batuhan DemirArda Batuhan Demir

11




11




closed as off-topic by José Carlos Santos, amWhy, Davide Giraudo, Cesareo, Leucippus Jan 3 at 0:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, amWhy, Davide Giraudo, Cesareo, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by José Carlos Santos, amWhy, Davide Giraudo, Cesareo, Leucippus Jan 3 at 0:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, amWhy, Davide Giraudo, Cesareo, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Just substitute in the formula and calculate the diameter.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 17:46


















  • $begingroup$
    Just substitute in the formula and calculate the diameter.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 17:46
















$begingroup$
Just substitute in the formula and calculate the diameter.
$endgroup$
– Michael Rozenberg
Jan 2 at 17:46




$begingroup$
Just substitute in the formula and calculate the diameter.
$endgroup$
– Michael Rozenberg
Jan 2 at 17:46










2 Answers
2






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Compute all $binom{4}{2}=6$ distances between distinct points and see what the maximum is. I don't see the problem (but yes, it's work, do it!)



E.g. $d((0,0), (1,1)) = 2$ and $d((1,1), (1,2))=1$ etc.






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    0












    $begingroup$

    The diameter is $operatorname{max}{d((x_1,y_1),(x_2,y_2))mid (x_1,y_1),(x_2,y_2)in A}$.



    It looks like it's $5$. As pointed out by @Henno Brandsma, there are $6$ distances to check.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Compute all $binom{4}{2}=6$ distances between distinct points and see what the maximum is. I don't see the problem (but yes, it's work, do it!)



      E.g. $d((0,0), (1,1)) = 2$ and $d((1,1), (1,2))=1$ etc.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Compute all $binom{4}{2}=6$ distances between distinct points and see what the maximum is. I don't see the problem (but yes, it's work, do it!)



        E.g. $d((0,0), (1,1)) = 2$ and $d((1,1), (1,2))=1$ etc.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Compute all $binom{4}{2}=6$ distances between distinct points and see what the maximum is. I don't see the problem (but yes, it's work, do it!)



          E.g. $d((0,0), (1,1)) = 2$ and $d((1,1), (1,2))=1$ etc.






          share|cite|improve this answer









          $endgroup$



          Compute all $binom{4}{2}=6$ distances between distinct points and see what the maximum is. I don't see the problem (but yes, it's work, do it!)



          E.g. $d((0,0), (1,1)) = 2$ and $d((1,1), (1,2))=1$ etc.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 17:45









          Henno BrandsmaHenno Brandsma

          107k347114




          107k347114























              0












              $begingroup$

              The diameter is $operatorname{max}{d((x_1,y_1),(x_2,y_2))mid (x_1,y_1),(x_2,y_2)in A}$.



              It looks like it's $5$. As pointed out by @Henno Brandsma, there are $6$ distances to check.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The diameter is $operatorname{max}{d((x_1,y_1),(x_2,y_2))mid (x_1,y_1),(x_2,y_2)in A}$.



                It looks like it's $5$. As pointed out by @Henno Brandsma, there are $6$ distances to check.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The diameter is $operatorname{max}{d((x_1,y_1),(x_2,y_2))mid (x_1,y_1),(x_2,y_2)in A}$.



                  It looks like it's $5$. As pointed out by @Henno Brandsma, there are $6$ distances to check.






                  share|cite|improve this answer









                  $endgroup$



                  The diameter is $operatorname{max}{d((x_1,y_1),(x_2,y_2))mid (x_1,y_1),(x_2,y_2)in A}$.



                  It looks like it's $5$. As pointed out by @Henno Brandsma, there are $6$ distances to check.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 18:48









                  Chris CusterChris Custer

                  11.6k3824




                  11.6k3824















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