separated and dense in locally compact is discrete? [on hold]
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It is true that if $Asubset mathbb{R}$ $delta$-separated and $epsilon$-dense in $mathbb{R}$ then $A$ is discrete? (This because of $mathbb{R}$ is locally compact?
I ask this since $mathbb{Z}$ is 1-separated and 1-dense in $mathbb{R}$ and is discrete
real-analysis analysis compactness
asked yesterday
eraldcoil
17218
put on hold as unclear what you're asking by Mirko, Masacroso, Lord Shark the Unknown, José Carlos Santos, amWhy yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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0
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favorite
It is true that if $Asubset mathbb{R}$ $delta$-separated and $epsilon$-dense in $mathbb{R}$ then $A$ is discrete? (This because of $mathbb{R}$ is locally compact?
I ask this since $mathbb{Z}$ is 1-separated and 1-dense in $mathbb{R}$ and is discrete
real-analysis analysis compactness
asked yesterday
eraldcoil
17218
put on hold as unclear what you're asking by Mirko, Masacroso, Lord Shark the Unknown, José Carlos Santos, amWhy yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
What is the definition of $delta$-separated? If my guess is correct, then it implies discrete.
– Mirko
yesterday
$A$ $delta$-separated if $d(a,a')geq delta forall aneq a'in A$
– eraldcoil
yesterday
1
So then $A$ is closed and discrete since for every $xinmathbb R$ the $delta/2$ ball centered at $x$ meets at most one point of $A$.
– Mirko
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
It is true that if $Asubset mathbb{R}$ $delta$-separated and $epsilon$-dense in $mathbb{R}$ then $A$ is discrete? (This because of $mathbb{R}$ is locally compact?
I ask this since $mathbb{Z}$ is 1-separated and 1-dense in $mathbb{R}$ and is discrete
real-analysis analysis compactness
asked yesterday
eraldcoil
17218
It is true that if $Asubset mathbb{R}$ $delta$-separated and $epsilon$-dense in $mathbb{R}$ then $A$ is discrete? (This because of $mathbb{R}$ is locally compact?
I ask this since $mathbb{Z}$ is 1-separated and 1-dense in $mathbb{R}$ and is discrete
real-analysis analysis compactness
real-analysis analysis compactness
asked yesterday
eraldcoil
17218
asked yesterday
eraldcoil
17218
asked yesterday
eraldcoil
17218
asked yesterday
eraldcoil
17218
asked yesterday
eraldcoil
17218
17218
put on hold as unclear what you're asking by Mirko, Masacroso, Lord Shark the Unknown, José Carlos Santos, amWhy yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Mirko, Masacroso, Lord Shark the Unknown, José Carlos Santos, amWhy yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
What is the definition of $delta$-separated? If my guess is correct, then it implies discrete.
– Mirko
yesterday
$A$ $delta$-separated if $d(a,a')geq delta forall aneq a'in A$
– eraldcoil
yesterday
1
So then $A$ is closed and discrete since for every $xinmathbb R$ the $delta/2$ ball centered at $x$ meets at most one point of $A$.
– Mirko
yesterday
add a comment |
1
What is the definition of $delta$-separated? If my guess is correct, then it implies discrete.
– Mirko
yesterday
$A$ $delta$-separated if $d(a,a')geq delta forall aneq a'in A$
– eraldcoil
yesterday
1
So then $A$ is closed and discrete since for every $xinmathbb R$ the $delta/2$ ball centered at $x$ meets at most one point of $A$.
– Mirko
yesterday
1
1
What is the definition of $delta$-separated? If my guess is correct, then it implies discrete.
– Mirko
yesterday
What is the definition of $delta$-separated? If my guess is correct, then it implies discrete.
– Mirko
yesterday
$A$ $delta$-separated if $d(a,a')geq delta forall aneq a'in A$
– eraldcoil
yesterday
$A$ $delta$-separated if $d(a,a')geq delta forall aneq a'in A$
– eraldcoil
yesterday
1
1
So then $A$ is closed and discrete since for every $xinmathbb R$ the $delta/2$ ball centered at $x$ meets at most one point of $A$.
– Mirko
yesterday
So then $A$ is closed and discrete since for every $xinmathbb R$ the $delta/2$ ball centered at $x$ meets at most one point of $A$.
– Mirko
yesterday
add a comment |
1 Answer
1
active
oldest
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up vote
2
down vote
If $delta >0$ the any $delta$ separated set is discrete because it has no limit points.
answered yesterday
Kavi Rama Murthy
39k31748
It is also discrete as a sub$space$ because in the sub-space topology any non-empty open ball of radius $delta/2$ contains exactly one point...............+1
– DanielWainfleet
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If $delta >0$ the any $delta$ separated set is discrete because it has no limit points.
answered yesterday
Kavi Rama Murthy
39k31748
It is also discrete as a sub$space$ because in the sub-space topology any non-empty open ball of radius $delta/2$ contains exactly one point...............+1
– DanielWainfleet
yesterday
add a comment |
up vote
2
down vote
If $delta >0$ the any $delta$ separated set is discrete because it has no limit points.
answered yesterday
Kavi Rama Murthy
39k31748
It is also discrete as a sub$space$ because in the sub-space topology any non-empty open ball of radius $delta/2$ contains exactly one point...............+1
– DanielWainfleet
yesterday
add a comment |
up vote
2
down vote
up vote
2
down vote
If $delta >0$ the any $delta$ separated set is discrete because it has no limit points.
answered yesterday
Kavi Rama Murthy
39k31748
If $delta >0$ the any $delta$ separated set is discrete because it has no limit points.
answered yesterday
Kavi Rama Murthy
39k31748
answered yesterday
Kavi Rama Murthy
39k31748
answered yesterday
Kavi Rama Murthy
39k31748
answered yesterday
Kavi Rama Murthy
39k31748
39k31748
It is also discrete as a sub$space$ because in the sub-space topology any non-empty open ball of radius $delta/2$ contains exactly one point...............+1
– DanielWainfleet
yesterday
add a comment |
It is also discrete as a sub$space$ because in the sub-space topology any non-empty open ball of radius $delta/2$ contains exactly one point...............+1
– DanielWainfleet
yesterday
It is also discrete as a sub$space$ because in the sub-space topology any non-empty open ball of radius $delta/2$ contains exactly one point...............+1
– DanielWainfleet
yesterday
It is also discrete as a sub$space$ because in the sub-space topology any non-empty open ball of radius $delta/2$ contains exactly one point...............+1
– DanielWainfleet
yesterday
add a comment |
1
What is the definition of $delta$-separated? If my guess is correct, then it implies discrete.
– Mirko
yesterday
$A$ $delta$-separated if $d(a,a')geq delta forall aneq a'in A$
– eraldcoil
yesterday
1
So then $A$ is closed and discrete since for every $xinmathbb R$ the $delta/2$ ball centered at $x$ meets at most one point of $A$.
– Mirko
yesterday