In discrete topology, every neighborhood of $Delta_X$ is an entourage?












0












$begingroup$


Let $X={x_n}_{nin
mathbb{N}}$ where $x_n=sum_{i=1}^{n}(frac{1}{i})$ given the
metric $d$ inherited from $mathbb{R}$. Also let $mathcal{U}(d)$ be uniformity generated by metric $d$. We know that every entourage $Dinmathcal{U}(d)$ is a neighborhood of $Delta_X$.



In a paper, Authors claimed that since $X$ has discrete topology,
there is a neighborhood $U$ of $Delta_X$ such that $U_x={x }$ for all
$xin X$.
In the following, we give a proof:
Since $x_{n-1}<x_n<x_{n+1}$, hence we can choice $delta_n>0$ such that $B(x_n, delta_n)={x_n}$. Take
$U= bigcup_{n=1}^{infty}B(x_n, delta_n)times B(x_n, delta_n)$. Then $U$ is an open set and $Delta_Xsubseteq U$.



I do not know if my proof is true and I do not know if there is an entourage $Dinmathcal{U}(d)$ with $D=U$.



Please help me to know it.






general-topology uniform-spaces







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $X={x_n}_{nin
    mathbb{N}}$     where $x_n=sum_{i=1}^{n}(frac{1}{i})$ given the
    metric $d$ inherited from $mathbb{R}$. Also let $mathcal{U}(d)$ be uniformity generated by metric $d$. We know that every entourage $Dinmathcal{U}(d)$ is a neighborhood of $Delta_X$.



    In a paper, Authors claimed that since $X$ has discrete topology,
    there is a neighborhood $U$ of $Delta_X$ such that $U_x={x }$ for all
    $xin X$.
    In the following, we give a proof:
    Since $x_{n-1}<x_n<x_{n+1}$, hence we can choice $delta_n>0$ such that $B(x_n, delta_n)={x_n}$. Take
    $U= bigcup_{n=1}^{infty}B(x_n, delta_n)times B(x_n, delta_n)$. Then $U$ is an open set and $Delta_Xsubseteq U$.



    I do not know if my proof is true and I do not know if there is an entourage $Dinmathcal{U}(d)$ with $D=U$.



    Please help me to know it.






    general-topology uniform-spaces







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $X={x_n}_{nin
      mathbb{N}}$       where $x_n=sum_{i=1}^{n}(frac{1}{i})$ given the
      metric $d$ inherited from $mathbb{R}$. Also let $mathcal{U}(d)$ be uniformity generated by metric $d$. We know that every entourage $Dinmathcal{U}(d)$ is a neighborhood of $Delta_X$.



      In a paper, Authors claimed that since $X$ has discrete topology,
      there is a neighborhood $U$ of $Delta_X$ such that $U_x={x }$ for all
      $xin X$.
      In the following, we give a proof:
      Since $x_{n-1}<x_n<x_{n+1}$, hence we can choice $delta_n>0$ such that $B(x_n, delta_n)={x_n}$. Take
      $U= bigcup_{n=1}^{infty}B(x_n, delta_n)times B(x_n, delta_n)$. Then $U$ is an open set and $Delta_Xsubseteq U$.



      I do not know if my proof is true and I do not know if there is an entourage $Dinmathcal{U}(d)$ with $D=U$.



      Please help me to know it.






      general-topology uniform-spaces







      share|cite|improve this question











      $endgroup$




      Let $X={x_n}_{nin
      mathbb{N}}$       where $x_n=sum_{i=1}^{n}(frac{1}{i})$ given the
      metric $d$ inherited from $mathbb{R}$. Also let $mathcal{U}(d)$ be uniformity generated by metric $d$. We know that every entourage $Dinmathcal{U}(d)$ is a neighborhood of $Delta_X$.



      In a paper, Authors claimed that since $X$ has discrete topology,
      there is a neighborhood $U$ of $Delta_X$ such that $U_x={x }$ for all
      $xin X$.
      In the following, we give a proof:
      Since $x_{n-1}<x_n<x_{n+1}$, hence we can choice $delta_n>0$ such that $B(x_n, delta_n)={x_n}$. Take
      $U= bigcup_{n=1}^{infty}B(x_n, delta_n)times B(x_n, delta_n)$. Then $U$ is an open set and $Delta_Xsubseteq U$.



      I do not know if my proof is true and I do not know if there is an entourage $Dinmathcal{U}(d)$ with $D=U$.



      Please help me to know it.






      general-topology uniform-spaces




      general-topology uniform-spaces






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      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 3 at 18:16







      Ali Barzanouni

















      asked Jan 3 at 16:03









      Ali BarzanouniAli Barzanouni

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          Indeed $X$ as a topological space is discrete: it's an infinite sequence that increases to $infty$ and all such sequences (considered as sets) are discrete as a subspace of $mathbb{R}$. Hence so is its square. This means that $D subseteq X times X$ is open and so its own open neighbourhood.



          But any basic entourage under $d$, so any set $U(varepsilon)={(x,x') in X times X: d(x,x') < varepsilon}$ can never just contain $D$, as there are always terms $x_n$ in $X$ closer together than $varepsilon$, for any $varepsilon>0$. So maximal uniformity for $X$ (the discrete one, of all subsets of $X^2$ containing $D$) is different from the induced by $d$, although the topologies are the same, which is probably the point of this exercise.






          share|cite|improve this answer









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            $begingroup$

            Indeed $X$ as a topological space is discrete: it's an infinite sequence that increases to $infty$ and all such sequences (considered as sets) are discrete as a subspace of $mathbb{R}$. Hence so is its square. This means that $D subseteq X times X$ is open and so its own open neighbourhood.



            But any basic entourage under $d$, so any set $U(varepsilon)={(x,x') in X times X: d(x,x') < varepsilon}$ can never just contain $D$, as there are always terms $x_n$ in $X$ closer together than $varepsilon$, for any $varepsilon>0$. So maximal uniformity for $X$ (the discrete one, of all subsets of $X^2$ containing $D$) is different from the induced by $d$, although the topologies are the same, which is probably the point of this exercise.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Indeed $X$ as a topological space is discrete: it's an infinite sequence that increases to $infty$ and all such sequences (considered as sets) are discrete as a subspace of $mathbb{R}$. Hence so is its square. This means that $D subseteq X times X$ is open and so its own open neighbourhood.



              But any basic entourage under $d$, so any set $U(varepsilon)={(x,x') in X times X: d(x,x') < varepsilon}$ can never just contain $D$, as there are always terms $x_n$ in $X$ closer together than $varepsilon$, for any $varepsilon>0$. So maximal uniformity for $X$ (the discrete one, of all subsets of $X^2$ containing $D$) is different from the induced by $d$, although the topologies are the same, which is probably the point of this exercise.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Indeed $X$ as a topological space is discrete: it's an infinite sequence that increases to $infty$ and all such sequences (considered as sets) are discrete as a subspace of $mathbb{R}$. Hence so is its square. This means that $D subseteq X times X$ is open and so its own open neighbourhood.



                But any basic entourage under $d$, so any set $U(varepsilon)={(x,x') in X times X: d(x,x') < varepsilon}$ can never just contain $D$, as there are always terms $x_n$ in $X$ closer together than $varepsilon$, for any $varepsilon>0$. So maximal uniformity for $X$ (the discrete one, of all subsets of $X^2$ containing $D$) is different from the induced by $d$, although the topologies are the same, which is probably the point of this exercise.






                share|cite|improve this answer









                $endgroup$



                Indeed $X$ as a topological space is discrete: it's an infinite sequence that increases to $infty$ and all such sequences (considered as sets) are discrete as a subspace of $mathbb{R}$. Hence so is its square. This means that $D subseteq X times X$ is open and so its own open neighbourhood.



                But any basic entourage under $d$, so any set $U(varepsilon)={(x,x') in X times X: d(x,x') < varepsilon}$ can never just contain $D$, as there are always terms $x_n$ in $X$ closer together than $varepsilon$, for any $varepsilon>0$. So maximal uniformity for $X$ (the discrete one, of all subsets of $X^2$ containing $D$) is different from the induced by $d$, although the topologies are the same, which is probably the point of this exercise.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 3 at 22:35









                Henno BrandsmaHenno Brandsma

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