In the topology,Let $X={1,2,3}$ and let $d:X times X to[0,+ ∞)$ be a function defined by formulas












-1












$begingroup$


Let $X={1,2,3}$ and let $d:X times X to[0,+infty)$ be a function defined by formulas:
$$
d(1,1)=d(2,2)=d(3,3)=0,\
d(1,2)=d(2,1)=d(1,3)=d(3,1)=1,\
d(2,3)=d(3,2)=3.$$



Check if $d$ is a distance function on $X$.










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  • $begingroup$
    What have you tried so far? Do you remember the definition of distance functions?
    $endgroup$
    – Mindlack
    Dec 13 '18 at 18:11












  • $begingroup$
    Hello and welcome to Math.SE! In this community, when posting a question, you are encouraged to include in the body of the question some of your thoughts about the problem, some context about it. Check this link for some information about posting good questions.
    $endgroup$
    – Edmundo Martins
    Dec 13 '18 at 18:16










  • $begingroup$
    Yes I have 3 or 4 rules for this questions, but I can not be sure....Could you help me and explain the questions and answer ?
    $endgroup$
    – arbade
    Dec 13 '18 at 18:16
















-1












$begingroup$


Let $X={1,2,3}$ and let $d:X times X to[0,+infty)$ be a function defined by formulas:
$$
d(1,1)=d(2,2)=d(3,3)=0,\
d(1,2)=d(2,1)=d(1,3)=d(3,1)=1,\
d(2,3)=d(3,2)=3.$$



Check if $d$ is a distance function on $X$.










share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried so far? Do you remember the definition of distance functions?
    $endgroup$
    – Mindlack
    Dec 13 '18 at 18:11












  • $begingroup$
    Hello and welcome to Math.SE! In this community, when posting a question, you are encouraged to include in the body of the question some of your thoughts about the problem, some context about it. Check this link for some information about posting good questions.
    $endgroup$
    – Edmundo Martins
    Dec 13 '18 at 18:16










  • $begingroup$
    Yes I have 3 or 4 rules for this questions, but I can not be sure....Could you help me and explain the questions and answer ?
    $endgroup$
    – arbade
    Dec 13 '18 at 18:16














-1












-1








-1





$begingroup$


Let $X={1,2,3}$ and let $d:X times X to[0,+infty)$ be a function defined by formulas:
$$
d(1,1)=d(2,2)=d(3,3)=0,\
d(1,2)=d(2,1)=d(1,3)=d(3,1)=1,\
d(2,3)=d(3,2)=3.$$



Check if $d$ is a distance function on $X$.










share|cite|improve this question











$endgroup$




Let $X={1,2,3}$ and let $d:X times X to[0,+infty)$ be a function defined by formulas:
$$
d(1,1)=d(2,2)=d(3,3)=0,\
d(1,2)=d(2,1)=d(1,3)=d(3,1)=1,\
d(2,3)=d(3,2)=3.$$



Check if $d$ is a distance function on $X$.







general-topology






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share|cite|improve this question













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share|cite|improve this question








edited Dec 13 '18 at 22:33









Tianlalu

3,08421138




3,08421138










asked Dec 13 '18 at 18:08









arbadearbade

14




14












  • $begingroup$
    What have you tried so far? Do you remember the definition of distance functions?
    $endgroup$
    – Mindlack
    Dec 13 '18 at 18:11












  • $begingroup$
    Hello and welcome to Math.SE! In this community, when posting a question, you are encouraged to include in the body of the question some of your thoughts about the problem, some context about it. Check this link for some information about posting good questions.
    $endgroup$
    – Edmundo Martins
    Dec 13 '18 at 18:16










  • $begingroup$
    Yes I have 3 or 4 rules for this questions, but I can not be sure....Could you help me and explain the questions and answer ?
    $endgroup$
    – arbade
    Dec 13 '18 at 18:16


















  • $begingroup$
    What have you tried so far? Do you remember the definition of distance functions?
    $endgroup$
    – Mindlack
    Dec 13 '18 at 18:11












  • $begingroup$
    Hello and welcome to Math.SE! In this community, when posting a question, you are encouraged to include in the body of the question some of your thoughts about the problem, some context about it. Check this link for some information about posting good questions.
    $endgroup$
    – Edmundo Martins
    Dec 13 '18 at 18:16










  • $begingroup$
    Yes I have 3 or 4 rules for this questions, but I can not be sure....Could you help me and explain the questions and answer ?
    $endgroup$
    – arbade
    Dec 13 '18 at 18:16
















$begingroup$
What have you tried so far? Do you remember the definition of distance functions?
$endgroup$
– Mindlack
Dec 13 '18 at 18:11






$begingroup$
What have you tried so far? Do you remember the definition of distance functions?
$endgroup$
– Mindlack
Dec 13 '18 at 18:11














$begingroup$
Hello and welcome to Math.SE! In this community, when posting a question, you are encouraged to include in the body of the question some of your thoughts about the problem, some context about it. Check this link for some information about posting good questions.
$endgroup$
– Edmundo Martins
Dec 13 '18 at 18:16




$begingroup$
Hello and welcome to Math.SE! In this community, when posting a question, you are encouraged to include in the body of the question some of your thoughts about the problem, some context about it. Check this link for some information about posting good questions.
$endgroup$
– Edmundo Martins
Dec 13 '18 at 18:16












$begingroup$
Yes I have 3 or 4 rules for this questions, but I can not be sure....Could you help me and explain the questions and answer ?
$endgroup$
– arbade
Dec 13 '18 at 18:16




$begingroup$
Yes I have 3 or 4 rules for this questions, but I can not be sure....Could you help me and explain the questions and answer ?
$endgroup$
– arbade
Dec 13 '18 at 18:16










1 Answer
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$begingroup$

To check that $d$ is a distance on $X$ you need:

1) $d(x,y)geq 0$ $forall x,y in X$ and $d(x,y)=0$ iff $x=y$

2) $d(x,y)=d(y,x)$ $forall x,y in X$

3) $d(x,y) leq d(x,z)+d(z,y)$ $forall x,y,z in X$



The first line of formulas gives you 1), the second line gives you 2), but then
$3=d(2,3) leq d(2,1)+d(1,3)=1+1=2$ and that is a contradiction. Therefore $d$ is not a distance






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    1 Answer
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    $begingroup$

    To check that $d$ is a distance on $X$ you need:

    1) $d(x,y)geq 0$ $forall x,y in X$ and $d(x,y)=0$ iff $x=y$

    2) $d(x,y)=d(y,x)$ $forall x,y in X$

    3) $d(x,y) leq d(x,z)+d(z,y)$ $forall x,y,z in X$



    The first line of formulas gives you 1), the second line gives you 2), but then
    $3=d(2,3) leq d(2,1)+d(1,3)=1+1=2$ and that is a contradiction. Therefore $d$ is not a distance






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      To check that $d$ is a distance on $X$ you need:

      1) $d(x,y)geq 0$ $forall x,y in X$ and $d(x,y)=0$ iff $x=y$

      2) $d(x,y)=d(y,x)$ $forall x,y in X$

      3) $d(x,y) leq d(x,z)+d(z,y)$ $forall x,y,z in X$



      The first line of formulas gives you 1), the second line gives you 2), but then
      $3=d(2,3) leq d(2,1)+d(1,3)=1+1=2$ and that is a contradiction. Therefore $d$ is not a distance






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        To check that $d$ is a distance on $X$ you need:

        1) $d(x,y)geq 0$ $forall x,y in X$ and $d(x,y)=0$ iff $x=y$

        2) $d(x,y)=d(y,x)$ $forall x,y in X$

        3) $d(x,y) leq d(x,z)+d(z,y)$ $forall x,y,z in X$



        The first line of formulas gives you 1), the second line gives you 2), but then
        $3=d(2,3) leq d(2,1)+d(1,3)=1+1=2$ and that is a contradiction. Therefore $d$ is not a distance






        share|cite|improve this answer









        $endgroup$



        To check that $d$ is a distance on $X$ you need:

        1) $d(x,y)geq 0$ $forall x,y in X$ and $d(x,y)=0$ iff $x=y$

        2) $d(x,y)=d(y,x)$ $forall x,y in X$

        3) $d(x,y) leq d(x,z)+d(z,y)$ $forall x,y,z in X$



        The first line of formulas gives you 1), the second line gives you 2), but then
        $3=d(2,3) leq d(2,1)+d(1,3)=1+1=2$ and that is a contradiction. Therefore $d$ is not a distance







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 18:24









        user289143user289143

        1,002313




        1,002313






























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