Does a neighbourhood have to extend from the point $a$ to both directions equally?
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The $epsilon$-neighbourhood of a point "$a$" in $mathbb R$ is defined as the set of points $x$ about a point "$a$" such that $|x-a|<epsilon$, $epsilon>0$.
But this means that the point "$a$" will be in the middle of the neighbourhood?
real-analysis definition
edited Sep 4 at 11:19
Chinnapparaj R
2,123420
asked Sep 4 at 10:50
Shahbaz Shaik
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The $epsilon$-neighbourhood of a point "$a$" in $mathbb R$ is defined as the set of points $x$ about a point "$a$" such that $|x-a|<epsilon$, $epsilon>0$.
But this means that the point "$a$" will be in the middle of the neighbourhood?
real-analysis definition
edited Sep 4 at 11:19
Chinnapparaj R
2,123420
asked Sep 4 at 10:50
Shahbaz Shaik
1
1
The definition by wiki: >If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V$ of $X$ that includes an open set $U$ containing $p$ . So no.
– Holo
Sep 4 at 10:59
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The $epsilon$-neighbourhood of a point "$a$" in $mathbb R$ is defined as the set of points $x$ about a point "$a$" such that $|x-a|<epsilon$, $epsilon>0$.
But this means that the point "$a$" will be in the middle of the neighbourhood?
real-analysis definition
edited Sep 4 at 11:19
Chinnapparaj R
2,123420
asked Sep 4 at 10:50
Shahbaz Shaik
1
The $epsilon$-neighbourhood of a point "$a$" in $mathbb R$ is defined as the set of points $x$ about a point "$a$" such that $|x-a|<epsilon$, $epsilon>0$.
But this means that the point "$a$" will be in the middle of the neighbourhood?
real-analysis definition
real-analysis definition
edited Sep 4 at 11:19
Chinnapparaj R
2,123420
asked Sep 4 at 10:50
Shahbaz Shaik
1
edited Sep 4 at 11:19
Chinnapparaj R
2,123420
asked Sep 4 at 10:50
Shahbaz Shaik
1
edited Sep 4 at 11:19
Chinnapparaj R
2,123420
edited Sep 4 at 11:19
Chinnapparaj R
2,123420
edited Sep 4 at 11:19
Chinnapparaj R
2,123420
2,123420
asked Sep 4 at 10:50
Shahbaz Shaik
1
asked Sep 4 at 10:50
Shahbaz Shaik
1
asked Sep 4 at 10:50
Shahbaz Shaik
1
1
1
The definition by wiki: >If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V$ of $X$ that includes an open set $U$ containing $p$ . So no.
– Holo
Sep 4 at 10:59
add a comment |Â
1
The definition by wiki: >If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V$ of $X$ that includes an open set $U$ containing $p$ . So no.
– Holo
Sep 4 at 10:59
1
1
The definition by wiki: >If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V$ of $X$ that includes an open set $U$ containing $p$ . So no.
– Holo
Sep 4 at 10:59
The definition by wiki: >If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V$ of $X$ that includes an open set $U$ containing $p$ . So no.
– Holo
Sep 4 at 10:59
add a comment |Â
2 Answers
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No, not at all. A neighborhood $N$ of $a$ must contain some interval $(a-r,a+r)$, with $r>0$. But $N$ doesn't have to have a middle point and, even if it has one, the middle point doesn't have to be $a$. For instance, $(1,+infty)$ is a neighborhood of $2$.
answered Sep 4 at 10:53
José Carlos Santos
122k16101186
Thanks, that does clear things up quite a bit.
– Shahbaz Shaik
Sep 4 at 10:58
add a comment |Â
up vote
1
down vote
Your instinct to investigate this idea is a good one. As the subject develops, there are distinctions to be made, and the idea of a neighbourhood of a point becomes any open set containing it. A collection of open sets with appropriate properties is called a topology, and can be defined without any reference to distance.
What you have defined is best called an open ball around $x$ - because it is an open set containing $x$ it is a neighbourhood of $x$ and remains a neighbourhood as the definition and context change - it is the classic example of a neighbourhood in some ways. The definition of a ball does put $x$ at the centre.
answered Sep 4 at 11:04
Mark Bennet
77.4k774173
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
No, not at all. A neighborhood $N$ of $a$ must contain some interval $(a-r,a+r)$, with $r>0$. But $N$ doesn't have to have a middle point and, even if it has one, the middle point doesn't have to be $a$. For instance, $(1,+infty)$ is a neighborhood of $2$.
answered Sep 4 at 10:53
José Carlos Santos
122k16101186
Thanks, that does clear things up quite a bit.
– Shahbaz Shaik
Sep 4 at 10:58
add a comment |Â
up vote
1
down vote
No, not at all. A neighborhood $N$ of $a$ must contain some interval $(a-r,a+r)$, with $r>0$. But $N$ doesn't have to have a middle point and, even if it has one, the middle point doesn't have to be $a$. For instance, $(1,+infty)$ is a neighborhood of $2$.
answered Sep 4 at 10:53
José Carlos Santos
122k16101186
Thanks, that does clear things up quite a bit.
– Shahbaz Shaik
Sep 4 at 10:58
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No, not at all. A neighborhood $N$ of $a$ must contain some interval $(a-r,a+r)$, with $r>0$. But $N$ doesn't have to have a middle point and, even if it has one, the middle point doesn't have to be $a$. For instance, $(1,+infty)$ is a neighborhood of $2$.
answered Sep 4 at 10:53
José Carlos Santos
122k16101186
No, not at all. A neighborhood $N$ of $a$ must contain some interval $(a-r,a+r)$, with $r>0$. But $N$ doesn't have to have a middle point and, even if it has one, the middle point doesn't have to be $a$. For instance, $(1,+infty)$ is a neighborhood of $2$.
answered Sep 4 at 10:53
José Carlos Santos
122k16101186
answered Sep 4 at 10:53
José Carlos Santos
122k16101186
answered Sep 4 at 10:53
José Carlos Santos
122k16101186
answered Sep 4 at 10:53
José Carlos Santos
122k16101186
122k16101186
Thanks, that does clear things up quite a bit.
– Shahbaz Shaik
Sep 4 at 10:58
add a comment |Â
Thanks, that does clear things up quite a bit.
– Shahbaz Shaik
Sep 4 at 10:58
Thanks, that does clear things up quite a bit.
– Shahbaz Shaik
Sep 4 at 10:58
Thanks, that does clear things up quite a bit.
– Shahbaz Shaik
Sep 4 at 10:58
add a comment |Â
up vote
1
down vote
Your instinct to investigate this idea is a good one. As the subject develops, there are distinctions to be made, and the idea of a neighbourhood of a point becomes any open set containing it. A collection of open sets with appropriate properties is called a topology, and can be defined without any reference to distance.
What you have defined is best called an open ball around $x$ - because it is an open set containing $x$ it is a neighbourhood of $x$ and remains a neighbourhood as the definition and context change - it is the classic example of a neighbourhood in some ways. The definition of a ball does put $x$ at the centre.
answered Sep 4 at 11:04
Mark Bennet
77.4k774173
add a comment |Â
up vote
1
down vote
Your instinct to investigate this idea is a good one. As the subject develops, there are distinctions to be made, and the idea of a neighbourhood of a point becomes any open set containing it. A collection of open sets with appropriate properties is called a topology, and can be defined without any reference to distance.
What you have defined is best called an open ball around $x$ - because it is an open set containing $x$ it is a neighbourhood of $x$ and remains a neighbourhood as the definition and context change - it is the classic example of a neighbourhood in some ways. The definition of a ball does put $x$ at the centre.
answered Sep 4 at 11:04
Mark Bennet
77.4k774173
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your instinct to investigate this idea is a good one. As the subject develops, there are distinctions to be made, and the idea of a neighbourhood of a point becomes any open set containing it. A collection of open sets with appropriate properties is called a topology, and can be defined without any reference to distance.
What you have defined is best called an open ball around $x$ - because it is an open set containing $x$ it is a neighbourhood of $x$ and remains a neighbourhood as the definition and context change - it is the classic example of a neighbourhood in some ways. The definition of a ball does put $x$ at the centre.
answered Sep 4 at 11:04
Mark Bennet
77.4k774173
Your instinct to investigate this idea is a good one. As the subject develops, there are distinctions to be made, and the idea of a neighbourhood of a point becomes any open set containing it. A collection of open sets with appropriate properties is called a topology, and can be defined without any reference to distance.
What you have defined is best called an open ball around $x$ - because it is an open set containing $x$ it is a neighbourhood of $x$ and remains a neighbourhood as the definition and context change - it is the classic example of a neighbourhood in some ways. The definition of a ball does put $x$ at the centre.
answered Sep 4 at 11:04
Mark Bennet
77.4k774173
answered Sep 4 at 11:04
Mark Bennet
77.4k774173
answered Sep 4 at 11:04
Mark Bennet
77.4k774173
answered Sep 4 at 11:04
Mark Bennet
77.4k774173
77.4k774173
add a comment |Â
add a comment |Â
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1
The definition by wiki: >If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V$ of $X$ that includes an open set $U$ containing $p$ . So no.
– Holo
Sep 4 at 10:59