Mixing
In discrete topology, every neighborhood of $Delta_X$ is an entourage?
$begingroup$
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
asked Jan 3 at 16:03
Ali BarzanouniAli Barzanouni
62
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 16:03
Ali BarzanouniAli Barzanouni
62
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 16:03
Ali BarzanouniAli Barzanouni
62
$endgroup$
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
asked Jan 3 at 16:03
Ali BarzanouniAli Barzanouni
62
asked Jan 3 at 16:03
Ali BarzanouniAli Barzanouni
62
edited Jan 3 at 18:16
Ali Barzanouni
asked Jan 3 at 16:03
Ali BarzanouniAli Barzanouni
62
asked Jan 3 at 16:03
Ali BarzanouniAli Barzanouni
62
asked Jan 3 at 16:03
Ali BarzanouniAli Barzanouni
62
62
add a comment |
add a comment |
1 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.
answered Jan 3 at 22:35
Henno BrandsmaHenno Brandsma
106k347114
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1 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.
answered Jan 3 at 22:35
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 3 at 22:35
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 3 at 22:35
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
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.
answered Jan 3 at 22:35
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 3 at 22:35
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 3 at 22:35
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 3 at 22:35
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
add a comment |
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