In discrete topology, every neighborhood of $Delta_X$ is an entourage?
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
asked 2 days ago
Ali Barzanouni
62
add a comment |
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
asked 2 days ago
Ali Barzanouni
62
add a comment |
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
asked 2 days ago
Ali Barzanouni
62
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
asked 2 days ago
Ali Barzanouni
62
asked 2 days ago
Ali Barzanouni
62
edited 2 days ago
asked 2 days ago
Ali Barzanouni
62
asked 2 days ago
Ali Barzanouni
62
asked 2 days ago
Ali Barzanouni
62
62
add a comment |
add a comment |
1 Answer
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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.
answered yesterday
Henno Brandsma
105k347114
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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.
answered yesterday
Henno Brandsma
105k347114
add a comment |
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.
answered yesterday
Henno Brandsma
105k347114
add a comment |
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.
answered yesterday
Henno Brandsma
105k347114
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.
answered yesterday
Henno Brandsma
105k347114
answered yesterday
Henno Brandsma
105k347114
answered yesterday
Henno Brandsma
105k347114
answered yesterday
Henno Brandsma
105k347114
105k347114
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