Find the closure , Frontier and boundary of the following set
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Find the closure , Frontier and boundary of the following set in $ mathbb{R}^2$ given by
$$ S={(0,1) cup (2,0) } cup {(x,y): |x|+|y|< 1 }$$
Answer:
The closure is given by
$Cl (A) ={(0,1) cup (2,0) } cup {(x,y): |x|+|y| leq 1 }$
But I thing the closure can be
$ {(x,y):|x|+|y| leq 1 } cup {(0,2) }$ , because $ (0,1) in {(x,y):|x|+|y| leq 1 }$.
Am I right?
Also help me with the frontier and boundary of the set.
real-analysis general-topology
asked Nov 20 at 0:02
M. A. SARKAR
2,1201619
|
show 1 more comment
up vote
0
down vote
favorite
Find the closure , Frontier and boundary of the following set in $ mathbb{R}^2$ given by
$$ S={(0,1) cup (2,0) } cup {(x,y): |x|+|y|< 1 }$$
Answer:
The closure is given by
$Cl (A) ={(0,1) cup (2,0) } cup {(x,y): |x|+|y| leq 1 }$
But I thing the closure can be
$ {(x,y):|x|+|y| leq 1 } cup {(0,2) }$ , because $ (0,1) in {(x,y):|x|+|y| leq 1 }$.
Am I right?
Also help me with the frontier and boundary of the set.
real-analysis general-topology
asked Nov 20 at 0:02
M. A. SARKAR
2,1201619
2
You're definitely right when it comes to the closure. As far as the bondary/frontier goes, how do you distinguish between them? I'm under the impression that they're the same thing.
– AlkaKadri
Nov 20 at 0:06
@AlkaKadri, No there is a small difference between frontier of a set and boundary set as we can see from the example- "The frontier set of a closed unit ball is the empty set while the boundary of the unit closed ball is the unit circle". That is, frontier set does not include the boundary
– M. A. SARKAR
Nov 20 at 0:10
1
Okay, thanks for clarifying. Note that most people use 'frontier' and 'boundary' interchangeably, but I assume you're using frontier to mean the set of boundary points not included in the set itself? i.e., $partial S setminus S$?
– AlkaKadri
Nov 20 at 0:13
2
It's important to look at your definitions. Write down the definition of the exterior of a set. According to en.wikipedia.org/wiki/Interior_(topology), it's just the complement of the closure (which you've already obtained), but depending on what definition you're using there may be some more steps you have to show.
– AlkaKadri
Nov 20 at 0:18
1
Your notation for $S$ is weird: you cannot take a union of two points. You mean ${(0,1)} cup {(2,0)} = {(0,1),(2,0)}$ probably.
– Henno Brandsma
Nov 20 at 16:41
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find the closure , Frontier and boundary of the following set in $ mathbb{R}^2$ given by
$$ S={(0,1) cup (2,0) } cup {(x,y): |x|+|y|< 1 }$$
Answer:
The closure is given by
$Cl (A) ={(0,1) cup (2,0) } cup {(x,y): |x|+|y| leq 1 }$
But I thing the closure can be
$ {(x,y):|x|+|y| leq 1 } cup {(0,2) }$ , because $ (0,1) in {(x,y):|x|+|y| leq 1 }$.
Am I right?
Also help me with the frontier and boundary of the set.
real-analysis general-topology
asked Nov 20 at 0:02
M. A. SARKAR
2,1201619
Find the closure , Frontier and boundary of the following set in $ mathbb{R}^2$ given by
$$ S={(0,1) cup (2,0) } cup {(x,y): |x|+|y|< 1 }$$
Answer:
The closure is given by
$Cl (A) ={(0,1) cup (2,0) } cup {(x,y): |x|+|y| leq 1 }$
But I thing the closure can be
$ {(x,y):|x|+|y| leq 1 } cup {(0,2) }$ , because $ (0,1) in {(x,y):|x|+|y| leq 1 }$.
Am I right?
Also help me with the frontier and boundary of the set.
real-analysis general-topology
real-analysis general-topology
asked Nov 20 at 0:02
M. A. SARKAR
2,1201619
asked Nov 20 at 0:02
M. A. SARKAR
2,1201619
asked Nov 20 at 0:02
M. A. SARKAR
2,1201619
asked Nov 20 at 0:02
M. A. SARKAR
2,1201619
asked Nov 20 at 0:02
M. A. SARKAR
2,1201619
2,1201619
2
You're definitely right when it comes to the closure. As far as the bondary/frontier goes, how do you distinguish between them? I'm under the impression that they're the same thing.
– AlkaKadri
Nov 20 at 0:06
@AlkaKadri, No there is a small difference between frontier of a set and boundary set as we can see from the example- "The frontier set of a closed unit ball is the empty set while the boundary of the unit closed ball is the unit circle". That is, frontier set does not include the boundary
– M. A. SARKAR
Nov 20 at 0:10
1
Okay, thanks for clarifying. Note that most people use 'frontier' and 'boundary' interchangeably, but I assume you're using frontier to mean the set of boundary points not included in the set itself? i.e., $partial S setminus S$?
– AlkaKadri
Nov 20 at 0:13
2
It's important to look at your definitions. Write down the definition of the exterior of a set. According to en.wikipedia.org/wiki/Interior_(topology), it's just the complement of the closure (which you've already obtained), but depending on what definition you're using there may be some more steps you have to show.
– AlkaKadri
Nov 20 at 0:18
1
Your notation for $S$ is weird: you cannot take a union of two points. You mean ${(0,1)} cup {(2,0)} = {(0,1),(2,0)}$ probably.
– Henno Brandsma
Nov 20 at 16:41
|
show 1 more comment
2
You're definitely right when it comes to the closure. As far as the bondary/frontier goes, how do you distinguish between them? I'm under the impression that they're the same thing.
– AlkaKadri
Nov 20 at 0:06
@AlkaKadri, No there is a small difference between frontier of a set and boundary set as we can see from the example- "The frontier set of a closed unit ball is the empty set while the boundary of the unit closed ball is the unit circle". That is, frontier set does not include the boundary
– M. A. SARKAR
Nov 20 at 0:10
1
Okay, thanks for clarifying. Note that most people use 'frontier' and 'boundary' interchangeably, but I assume you're using frontier to mean the set of boundary points not included in the set itself? i.e., $partial S setminus S$?
– AlkaKadri
Nov 20 at 0:13
2
It's important to look at your definitions. Write down the definition of the exterior of a set. According to en.wikipedia.org/wiki/Interior_(topology), it's just the complement of the closure (which you've already obtained), but depending on what definition you're using there may be some more steps you have to show.
– AlkaKadri
Nov 20 at 0:18
1
Your notation for $S$ is weird: you cannot take a union of two points. You mean ${(0,1)} cup {(2,0)} = {(0,1),(2,0)}$ probably.
– Henno Brandsma
Nov 20 at 16:41
2
2
You're definitely right when it comes to the closure. As far as the bondary/frontier goes, how do you distinguish between them? I'm under the impression that they're the same thing.
– AlkaKadri
Nov 20 at 0:06
You're definitely right when it comes to the closure. As far as the bondary/frontier goes, how do you distinguish between them? I'm under the impression that they're the same thing.
– AlkaKadri
Nov 20 at 0:06
@AlkaKadri, No there is a small difference between frontier of a set and boundary set as we can see from the example- "The frontier set of a closed unit ball is the empty set while the boundary of the unit closed ball is the unit circle". That is, frontier set does not include the boundary
– M. A. SARKAR
Nov 20 at 0:10
@AlkaKadri, No there is a small difference between frontier of a set and boundary set as we can see from the example- "The frontier set of a closed unit ball is the empty set while the boundary of the unit closed ball is the unit circle". That is, frontier set does not include the boundary
– M. A. SARKAR
Nov 20 at 0:10
1
1
Okay, thanks for clarifying. Note that most people use 'frontier' and 'boundary' interchangeably, but I assume you're using frontier to mean the set of boundary points not included in the set itself? i.e., $partial S setminus S$?
– AlkaKadri
Nov 20 at 0:13
Okay, thanks for clarifying. Note that most people use 'frontier' and 'boundary' interchangeably, but I assume you're using frontier to mean the set of boundary points not included in the set itself? i.e., $partial S setminus S$?
– AlkaKadri
Nov 20 at 0:13
2
2
It's important to look at your definitions. Write down the definition of the exterior of a set. According to en.wikipedia.org/wiki/Interior_(topology), it's just the complement of the closure (which you've already obtained), but depending on what definition you're using there may be some more steps you have to show.
– AlkaKadri
Nov 20 at 0:18
It's important to look at your definitions. Write down the definition of the exterior of a set. According to en.wikipedia.org/wiki/Interior_(topology), it's just the complement of the closure (which you've already obtained), but depending on what definition you're using there may be some more steps you have to show.
– AlkaKadri
Nov 20 at 0:18
1
1
Your notation for $S$ is weird: you cannot take a union of two points. You mean ${(0,1)} cup {(2,0)} = {(0,1),(2,0)}$ probably.
– Henno Brandsma
Nov 20 at 16:41
Your notation for $S$ is weird: you cannot take a union of two points. You mean ${(0,1)} cup {(2,0)} = {(0,1),(2,0)}$ probably.
– Henno Brandsma
Nov 20 at 16:41
|
show 1 more comment
1 Answer
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oldest
votes
up vote
2
down vote
The frontier of $S$ is (probably, your example fits it, and it is a usage that occurs) $operatorname{Cl}{S}setminus S$ (all points in the closure that were not in the original set) so in this case $$operatorname{Fr}(S) = {(x,y): |x| + |y| =1}setminus {(0,1)}$$
clearly:
$$operatorname{Int}(S) = {(x,y): |x| + |y| < 1}$$
So $$operatorname{Bd}(S) =operatorname{Cl}(S) setminus operatorname{Int}(S) = {(2,0)} cup {(x,y): |x| + |y| = 1 }$$
So all these sets can be computed/determined one we know the closure and the interior.
answered Nov 20 at 17:30
Henno Brandsma
103k345112
add a comment |
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The frontier of $S$ is (probably, your example fits it, and it is a usage that occurs) $operatorname{Cl}{S}setminus S$ (all points in the closure that were not in the original set) so in this case $$operatorname{Fr}(S) = {(x,y): |x| + |y| =1}setminus {(0,1)}$$
clearly:
$$operatorname{Int}(S) = {(x,y): |x| + |y| < 1}$$
So $$operatorname{Bd}(S) =operatorname{Cl}(S) setminus operatorname{Int}(S) = {(2,0)} cup {(x,y): |x| + |y| = 1 }$$
So all these sets can be computed/determined one we know the closure and the interior.
answered Nov 20 at 17:30
Henno Brandsma
103k345112
add a comment |
up vote
2
down vote
The frontier of $S$ is (probably, your example fits it, and it is a usage that occurs) $operatorname{Cl}{S}setminus S$ (all points in the closure that were not in the original set) so in this case $$operatorname{Fr}(S) = {(x,y): |x| + |y| =1}setminus {(0,1)}$$
clearly:
$$operatorname{Int}(S) = {(x,y): |x| + |y| < 1}$$
So $$operatorname{Bd}(S) =operatorname{Cl}(S) setminus operatorname{Int}(S) = {(2,0)} cup {(x,y): |x| + |y| = 1 }$$
So all these sets can be computed/determined one we know the closure and the interior.
answered Nov 20 at 17:30
Henno Brandsma
103k345112
add a comment |
up vote
2
down vote
up vote
2
down vote
The frontier of $S$ is (probably, your example fits it, and it is a usage that occurs) $operatorname{Cl}{S}setminus S$ (all points in the closure that were not in the original set) so in this case $$operatorname{Fr}(S) = {(x,y): |x| + |y| =1}setminus {(0,1)}$$
clearly:
$$operatorname{Int}(S) = {(x,y): |x| + |y| < 1}$$
So $$operatorname{Bd}(S) =operatorname{Cl}(S) setminus operatorname{Int}(S) = {(2,0)} cup {(x,y): |x| + |y| = 1 }$$
So all these sets can be computed/determined one we know the closure and the interior.
answered Nov 20 at 17:30
Henno Brandsma
103k345112
The frontier of $S$ is (probably, your example fits it, and it is a usage that occurs) $operatorname{Cl}{S}setminus S$ (all points in the closure that were not in the original set) so in this case $$operatorname{Fr}(S) = {(x,y): |x| + |y| =1}setminus {(0,1)}$$
clearly:
$$operatorname{Int}(S) = {(x,y): |x| + |y| < 1}$$
So $$operatorname{Bd}(S) =operatorname{Cl}(S) setminus operatorname{Int}(S) = {(2,0)} cup {(x,y): |x| + |y| = 1 }$$
So all these sets can be computed/determined one we know the closure and the interior.
answered Nov 20 at 17:30
Henno Brandsma
103k345112
answered Nov 20 at 17:30
Henno Brandsma
103k345112
answered Nov 20 at 17:30
Henno Brandsma
103k345112
answered Nov 20 at 17:30
Henno Brandsma
103k345112
103k345112
add a comment |
add a comment |
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2
You're definitely right when it comes to the closure. As far as the bondary/frontier goes, how do you distinguish between them? I'm under the impression that they're the same thing.
– AlkaKadri
Nov 20 at 0:06
@AlkaKadri, No there is a small difference between frontier of a set and boundary set as we can see from the example- "The frontier set of a closed unit ball is the empty set while the boundary of the unit closed ball is the unit circle". That is, frontier set does not include the boundary
– M. A. SARKAR
Nov 20 at 0:10
1
Okay, thanks for clarifying. Note that most people use 'frontier' and 'boundary' interchangeably, but I assume you're using frontier to mean the set of boundary points not included in the set itself? i.e., $partial S setminus S$?
– AlkaKadri
Nov 20 at 0:13
2
It's important to look at your definitions. Write down the definition of the exterior of a set. According to en.wikipedia.org/wiki/Interior_(topology), it's just the complement of the closure (which you've already obtained), but depending on what definition you're using there may be some more steps you have to show.
– AlkaKadri
Nov 20 at 0:18
1
Your notation for $S$ is weird: you cannot take a union of two points. You mean ${(0,1)} cup {(2,0)} = {(0,1),(2,0)}$ probably.
– Henno Brandsma
Nov 20 at 16:41