Is infinity part of the boundary of R^n?
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I am having a set $S= xin(-1,1), yin(0,infty) $. Does $bar S$S include the set of points for which $ytoinfty$; figuratively speaking, does it include the "boundary" at infinity?
Thanks for explanations!
general-topology infinity
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I am having a set $S= xin(-1,1), yin(0,infty) $. Does $bar S$S include the set of points for which $ytoinfty$; figuratively speaking, does it include the "boundary" at infinity?
Thanks for explanations!
general-topology infinity
2
Talking about a closure $overline S$ presupposes an ambient topological space in which the closure is taken. If this space is $Bbb R^2$, then the answer is "no".
â Lord Shark the Unknown
Sep 2 at 10:59
As an easier example you may consider $(0,infty)subseteqmathbbR$ the closure of this set does not contain infinity simply because $inftynotinmathbbR$.
â Yanko
Sep 2 at 11:12
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up vote
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down vote
favorite
I am having a set $S= xin(-1,1), yin(0,infty) $. Does $bar S$S include the set of points for which $ytoinfty$; figuratively speaking, does it include the "boundary" at infinity?
Thanks for explanations!
general-topology infinity
I am having a set $S= xin(-1,1), yin(0,infty) $. Does $bar S$S include the set of points for which $ytoinfty$; figuratively speaking, does it include the "boundary" at infinity?
Thanks for explanations!
general-topology infinity
general-topology infinity
asked Sep 2 at 10:57
Britzel
31
31
2
Talking about a closure $overline S$ presupposes an ambient topological space in which the closure is taken. If this space is $Bbb R^2$, then the answer is "no".
â Lord Shark the Unknown
Sep 2 at 10:59
As an easier example you may consider $(0,infty)subseteqmathbbR$ the closure of this set does not contain infinity simply because $inftynotinmathbbR$.
â Yanko
Sep 2 at 11:12
add a comment |Â
2
Talking about a closure $overline S$ presupposes an ambient topological space in which the closure is taken. If this space is $Bbb R^2$, then the answer is "no".
â Lord Shark the Unknown
Sep 2 at 10:59
As an easier example you may consider $(0,infty)subseteqmathbbR$ the closure of this set does not contain infinity simply because $inftynotinmathbbR$.
â Yanko
Sep 2 at 11:12
2
2
Talking about a closure $overline S$ presupposes an ambient topological space in which the closure is taken. If this space is $Bbb R^2$, then the answer is "no".
â Lord Shark the Unknown
Sep 2 at 10:59
Talking about a closure $overline S$ presupposes an ambient topological space in which the closure is taken. If this space is $Bbb R^2$, then the answer is "no".
â Lord Shark the Unknown
Sep 2 at 10:59
As an easier example you may consider $(0,infty)subseteqmathbbR$ the closure of this set does not contain infinity simply because $inftynotinmathbbR$.
â Yanko
Sep 2 at 11:12
As an easier example you may consider $(0,infty)subseteqmathbbR$ the closure of this set does not contain infinity simply because $inftynotinmathbbR$.
â Yanko
Sep 2 at 11:12
add a comment |Â
1 Answer
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Normally it doesn't. The overall space you are considering does not contain infinity either as a point itself. Normally for you to have infinity you would do a one point compactification of your space, very similar to the extended complex plane.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Normally it doesn't. The overall space you are considering does not contain infinity either as a point itself. Normally for you to have infinity you would do a one point compactification of your space, very similar to the extended complex plane.
add a comment |Â
up vote
0
down vote
accepted
Normally it doesn't. The overall space you are considering does not contain infinity either as a point itself. Normally for you to have infinity you would do a one point compactification of your space, very similar to the extended complex plane.
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Normally it doesn't. The overall space you are considering does not contain infinity either as a point itself. Normally for you to have infinity you would do a one point compactification of your space, very similar to the extended complex plane.
Normally it doesn't. The overall space you are considering does not contain infinity either as a point itself. Normally for you to have infinity you would do a one point compactification of your space, very similar to the extended complex plane.
answered Sep 2 at 11:01
Fundamentals
835
835
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2
Talking about a closure $overline S$ presupposes an ambient topological space in which the closure is taken. If this space is $Bbb R^2$, then the answer is "no".
â Lord Shark the Unknown
Sep 2 at 10:59
As an easier example you may consider $(0,infty)subseteqmathbbR$ the closure of this set does not contain infinity simply because $inftynotinmathbbR$.
â Yanko
Sep 2 at 11:12