If $f$ is a loop in $mathbb{S}^{n}$, then is $f^{-1}({x})$ a compact set in $[0, 1]$? [closed]
If $f$ is a loop in $mathbb{S}^{n}$, then is $f^{-1}({x})$, $x in mathbb{S}^{n}$, a compact set in $[0, 1]$?
general-topology
asked Dec 5 '14 at 16:47
BehroozBehrooz
400318
closed as off-topic by Brahadeesh, Lord_Farin, Lord Shark the Unknown, Shailesh, KReiser Dec 4 '18 at 1:20
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If $f$ is a loop in $mathbb{S}^{n}$, then is $f^{-1}({x})$, $x in mathbb{S}^{n}$, a compact set in $[0, 1]$?
general-topology
asked Dec 5 '14 at 16:47
BehroozBehrooz
400318
closed as off-topic by Brahadeesh, Lord_Farin, Lord Shark the Unknown, Shailesh, KReiser Dec 4 '18 at 1:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Lord_Farin, Lord Shark the Unknown, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
4
It's a closed subset of a compact space.
– Daniel Fischer♦
Dec 5 '14 at 16:50
@ Daniel Fischer Why is $f^{-1}({x})$ closed in $[0, 1]$?
– Behrooz
Dec 5 '14 at 16:56
1
$f$ is continuous, and $mathbb{S}^n$ is Hausdorff.
– Daniel Fischer♦
Dec 5 '14 at 16:57
@ Daniel Fischer Thank you!
– Behrooz
Dec 5 '14 at 16:58
add a comment |
If $f$ is a loop in $mathbb{S}^{n}$, then is $f^{-1}({x})$, $x in mathbb{S}^{n}$, a compact set in $[0, 1]$?
general-topology
asked Dec 5 '14 at 16:47
BehroozBehrooz
400318
If $f$ is a loop in $mathbb{S}^{n}$, then is $f^{-1}({x})$, $x in mathbb{S}^{n}$, a compact set in $[0, 1]$?
general-topology
general-topology
asked Dec 5 '14 at 16:47
BehroozBehrooz
400318
asked Dec 5 '14 at 16:47
BehroozBehrooz
400318
asked Dec 5 '14 at 16:47
BehroozBehrooz
400318
asked Dec 5 '14 at 16:47
BehroozBehrooz
400318
asked Dec 5 '14 at 16:47
BehroozBehrooz
400318
400318
closed as off-topic by Brahadeesh, Lord_Farin, Lord Shark the Unknown, Shailesh, KReiser Dec 4 '18 at 1:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Lord_Farin, Lord Shark the Unknown, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brahadeesh, Lord_Farin, Lord Shark the Unknown, Shailesh, KReiser Dec 4 '18 at 1:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Lord_Farin, Lord Shark the Unknown, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
4
It's a closed subset of a compact space.
– Daniel Fischer♦
Dec 5 '14 at 16:50
@ Daniel Fischer Why is $f^{-1}({x})$ closed in $[0, 1]$?
– Behrooz
Dec 5 '14 at 16:56
1
$f$ is continuous, and $mathbb{S}^n$ is Hausdorff.
– Daniel Fischer♦
Dec 5 '14 at 16:57
@ Daniel Fischer Thank you!
– Behrooz
Dec 5 '14 at 16:58
add a comment |
4
It's a closed subset of a compact space.
– Daniel Fischer♦
Dec 5 '14 at 16:50
@ Daniel Fischer Why is $f^{-1}({x})$ closed in $[0, 1]$?
– Behrooz
Dec 5 '14 at 16:56
1
$f$ is continuous, and $mathbb{S}^n$ is Hausdorff.
– Daniel Fischer♦
Dec 5 '14 at 16:57
@ Daniel Fischer Thank you!
– Behrooz
Dec 5 '14 at 16:58
4
4
It's a closed subset of a compact space.
– Daniel Fischer♦
Dec 5 '14 at 16:50
It's a closed subset of a compact space.
– Daniel Fischer♦
Dec 5 '14 at 16:50
@ Daniel Fischer Why is $f^{-1}({x})$ closed in $[0, 1]$?
– Behrooz
Dec 5 '14 at 16:56
@ Daniel Fischer Why is $f^{-1}({x})$ closed in $[0, 1]$?
– Behrooz
Dec 5 '14 at 16:56
1
1
$f$ is continuous, and $mathbb{S}^n$ is Hausdorff.
– Daniel Fischer♦
Dec 5 '14 at 16:57
$f$ is continuous, and $mathbb{S}^n$ is Hausdorff.
– Daniel Fischer♦
Dec 5 '14 at 16:57
@ Daniel Fischer Thank you!
– Behrooz
Dec 5 '14 at 16:58
@ Daniel Fischer Thank you!
– Behrooz
Dec 5 '14 at 16:58
add a comment |
1 Answer
1
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oldest
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This community wiki solution is intended to clear the question from the unanswered queue.
Since $mathbb S^n$ is Hausdorff, singletons are closed.
The loop $f : [0, 1] to mathbb S^n$ is a continuous map, so the inverse image of closed sets is closed, in particular $f^{-1}({x})$ is closed.
But $[0, 1]$ is compact and closed subsets of compact sets are compact.
Conclude that $f^{-1}({x})$ is compact.
Clearing the queue of unanswered questions deserves great respect! +1
– Paul Frost
Dec 3 '18 at 18:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This community wiki solution is intended to clear the question from the unanswered queue.
Since $mathbb S^n$ is Hausdorff, singletons are closed.
The loop $f : [0, 1] to mathbb S^n$ is a continuous map, so the inverse image of closed sets is closed, in particular $f^{-1}({x})$ is closed.
But $[0, 1]$ is compact and closed subsets of compact sets are compact.
Conclude that $f^{-1}({x})$ is compact.
Clearing the queue of unanswered questions deserves great respect! +1
– Paul Frost
Dec 3 '18 at 18:34
add a comment |
This community wiki solution is intended to clear the question from the unanswered queue.
Since $mathbb S^n$ is Hausdorff, singletons are closed.
The loop $f : [0, 1] to mathbb S^n$ is a continuous map, so the inverse image of closed sets is closed, in particular $f^{-1}({x})$ is closed.
But $[0, 1]$ is compact and closed subsets of compact sets are compact.
Conclude that $f^{-1}({x})$ is compact.
Clearing the queue of unanswered questions deserves great respect! +1
– Paul Frost
Dec 3 '18 at 18:34
add a comment |
This community wiki solution is intended to clear the question from the unanswered queue.
Since $mathbb S^n$ is Hausdorff, singletons are closed.
The loop $f : [0, 1] to mathbb S^n$ is a continuous map, so the inverse image of closed sets is closed, in particular $f^{-1}({x})$ is closed.
But $[0, 1]$ is compact and closed subsets of compact sets are compact.
Conclude that $f^{-1}({x})$ is compact.
This community wiki solution is intended to clear the question from the unanswered queue.
Since $mathbb S^n$ is Hausdorff, singletons are closed.
The loop $f : [0, 1] to mathbb S^n$ is a continuous map, so the inverse image of closed sets is closed, in particular $f^{-1}({x})$ is closed.
But $[0, 1]$ is compact and closed subsets of compact sets are compact.
Conclude that $f^{-1}({x})$ is compact.
answered Dec 3 '18 at 15:03
community wiki
Robert Cardona
Clearing the queue of unanswered questions deserves great respect! +1
– Paul Frost
Dec 3 '18 at 18:34
add a comment |
Clearing the queue of unanswered questions deserves great respect! +1
– Paul Frost
Dec 3 '18 at 18:34
Clearing the queue of unanswered questions deserves great respect! +1
– Paul Frost
Dec 3 '18 at 18:34
Clearing the queue of unanswered questions deserves great respect! +1
– Paul Frost
Dec 3 '18 at 18:34
add a comment |
4
It's a closed subset of a compact space.
– Daniel Fischer♦
Dec 5 '14 at 16:50
@ Daniel Fischer Why is $f^{-1}({x})$ closed in $[0, 1]$?
– Behrooz
Dec 5 '14 at 16:56
1
$f$ is continuous, and $mathbb{S}^n$ is Hausdorff.
– Daniel Fischer♦
Dec 5 '14 at 16:57
@ Daniel Fischer Thank you!
– Behrooz
Dec 5 '14 at 16:58