Does every point in a metric space have another point a rational distance away?

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1
down vote

favorite












Is the following statement true or false?




Let $(X,d)$ be a metric space. Then for every $x in X$, there exists $y in X$ such that $d(x, y)$ is a non-zero rational number.




I'm not able to find a counter example.






general-topology metric-spaces







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  • 2




    What have you tried for finding counterexamples?
    – Dark Malthorp
    1 hour ago







  • 4




    Think about the discrete metric, but choose an irrational number instead of $1$.
    – Richard
    1 hour ago










  • @Richard good logics
    – jasmine
    1 hour ago










  • @NateEldredge okkss
    – jasmine
    1 hour ago














up vote
1
down vote

favorite












Is the following statement true or false?




Let $(X,d)$ be a metric space. Then for every $x in X$, there exists $y in X$ such that $d(x, y)$ is a non-zero rational number.




I'm not able to find a counter example.






general-topology metric-spaces







share|cite|improve this question



















  • 2




    What have you tried for finding counterexamples?
    – Dark Malthorp
    1 hour ago







  • 4




    Think about the discrete metric, but choose an irrational number instead of $1$.
    – Richard
    1 hour ago










  • @Richard good logics
    – jasmine
    1 hour ago










  • @NateEldredge okkss
    – jasmine
    1 hour ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Is the following statement true or false?




Let $(X,d)$ be a metric space. Then for every $x in X$, there exists $y in X$ such that $d(x, y)$ is a non-zero rational number.




I'm not able to find a counter example.






general-topology metric-spaces







share|cite|improve this question















Is the following statement true or false?




Let $(X,d)$ be a metric space. Then for every $x in X$, there exists $y in X$ such that $d(x, y)$ is a non-zero rational number.




I'm not able to find a counter example.






general-topology metric-spaces




general-topology metric-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 52 mins ago









Nate Eldredge

61.1k579165




61.1k579165










asked 1 hour ago









jasmine

1,258314




1,258314







  • 2




    What have you tried for finding counterexamples?
    – Dark Malthorp
    1 hour ago







  • 4




    Think about the discrete metric, but choose an irrational number instead of $1$.
    – Richard
    1 hour ago










  • @Richard good logics
    – jasmine
    1 hour ago










  • @NateEldredge okkss
    – jasmine
    1 hour ago












  • 2




    What have you tried for finding counterexamples?
    – Dark Malthorp
    1 hour ago







  • 4




    Think about the discrete metric, but choose an irrational number instead of $1$.
    – Richard
    1 hour ago










  • @Richard good logics
    – jasmine
    1 hour ago










  • @NateEldredge okkss
    – jasmine
    1 hour ago







2




2




What have you tried for finding counterexamples?
– Dark Malthorp
1 hour ago





What have you tried for finding counterexamples?
– Dark Malthorp
1 hour ago





4




4




Think about the discrete metric, but choose an irrational number instead of $1$.
– Richard
1 hour ago




Think about the discrete metric, but choose an irrational number instead of $1$.
– Richard
1 hour ago












@Richard good logics
– jasmine
1 hour ago




@Richard good logics
– jasmine
1 hour ago












@NateEldredge okkss
– jasmine
1 hour ago




@NateEldredge okkss
– jasmine
1 hour ago










3 Answers
3






active

oldest

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up vote
4
down vote



accepted










False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.






share|cite|improve this answer
















  • 2




    Alternatively, take $X = 0$ with trivial metric.
    – ConMan
    1 hour ago

















up vote
2
down vote













Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.






share|cite|improve this answer



























    up vote
    2
    down vote













    Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.



    • The two-point set $0, sqrt2$.


    • Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.


    • Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)






    share|cite|improve this answer






















    • Very clever example.
      – Matt Samuel
      36 mins ago










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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote



    accepted










    False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.






    share|cite|improve this answer
















    • 2




      Alternatively, take $X = 0$ with trivial metric.
      – ConMan
      1 hour ago














    up vote
    4
    down vote



    accepted










    False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.






    share|cite|improve this answer
















    • 2




      Alternatively, take $X = 0$ with trivial metric.
      – ConMan
      1 hour ago












    up vote
    4
    down vote



    accepted







    up vote
    4
    down vote



    accepted






    False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.






    share|cite|improve this answer












    False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    Noah Riggenbach

    63716




    63716







    • 2




      Alternatively, take $X = 0$ with trivial metric.
      – ConMan
      1 hour ago












    • 2




      Alternatively, take $X = 0$ with trivial metric.
      – ConMan
      1 hour ago







    2




    2




    Alternatively, take $X = 0$ with trivial metric.
    – ConMan
    1 hour ago




    Alternatively, take $X = 0$ with trivial metric.
    – ConMan
    1 hour ago










    up vote
    2
    down vote













    Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.






    share|cite|improve this answer
























      up vote
      2
      down vote













      Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.






        share|cite|improve this answer












        Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Mjiig

        34727




        34727




















            up vote
            2
            down vote













            Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.



            • The two-point set $0, sqrt2$.


            • Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.


            • Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)






            share|cite|improve this answer






















            • Very clever example.
              – Matt Samuel
              36 mins ago














            up vote
            2
            down vote













            Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.



            • The two-point set $0, sqrt2$.


            • Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.


            • Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)






            share|cite|improve this answer






















            • Very clever example.
              – Matt Samuel
              36 mins ago












            up vote
            2
            down vote










            up vote
            2
            down vote









            Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.



            • The two-point set $0, sqrt2$.


            • Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.


            • Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)






            share|cite|improve this answer














            Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.



            • The two-point set $0, sqrt2$.


            • Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.


            • Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 24 mins ago

























            answered 1 hour ago









            Nate Eldredge

            61.1k579165




            61.1k579165











            • Very clever example.
              – Matt Samuel
              36 mins ago
















            • Very clever example.
              – Matt Samuel
              36 mins ago















            Very clever example.
            – Matt Samuel
            36 mins ago




            Very clever example.
            – Matt Samuel
            36 mins ago

















             

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