why the set ${(x,y,0) : x^2 +y^2 < 1}$ is neither open nor closed? [closed]
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why the set ${(x,y,0) : x^2 +y^2 < 1}$ is neither open nor closed ?
My attempt : it will not closed because it is locally compact
i don't know how to proved that it will not open ?
Any hints/solution will be appreciated
thanks u
general-topology
asked Dec 5 '18 at 3:32
jasminejasmine
1,689416
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closed as off-topic by user21820, Saad, Jyrki Lahtonen, Brahadeesh, DRF Dec 8 '18 at 14:24
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Saad, Jyrki Lahtonen, Brahadeesh, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
why the set ${(x,y,0) : x^2 +y^2 < 1}$ is neither open nor closed ?
My attempt : it will not closed because it is locally compact
i don't know how to proved that it will not open ?
Any hints/solution will be appreciated
thanks u
general-topology
asked Dec 5 '18 at 3:32
jasminejasmine
1,689416
$endgroup$
closed as off-topic by user21820, Saad, Jyrki Lahtonen, Brahadeesh, DRF Dec 8 '18 at 14:24
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Saad, Jyrki Lahtonen, Brahadeesh, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Who says a locally compact set can't be closed? Every compact set is both locally compact and closed.
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– bof
Dec 5 '18 at 3:36
3
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The set $S={(x,y,0):x^2+y^2lt1}$ is not closed because $$left(frac12,0,0right),left(frac23,0,0right),left(frac34,0,0right),dots,left(frac n{n+1},0,0right),dots$$ is a sequence of points in $S$ converging to the point $(1,0,0)$ which is not in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:41
1
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And it’s not open because every neighborhood of the origin contains points not in the set.
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– Lubin
Dec 5 '18 at 3:45
2
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The set $S={(x,y,0):x^2+y^2lt1}$ is not open because $$(0,0,1),(0,0,1/2),(0,0,1/3),dots,(0,0,1/n),dots$$ is a sequence of points outside of $S$ congerging to the point $(0,0,0)$ in $S$.
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– bof
Dec 5 '18 at 3:45
$begingroup$
Is using the definition of open/closed not an option at all? I am frankly baffled why anyone would try and use the concept of locally compact here, when all you need to do is to check the definition?
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 9:46
add a comment |
$begingroup$
why the set ${(x,y,0) : x^2 +y^2 < 1}$ is neither open nor closed ?
My attempt : it will not closed because it is locally compact
i don't know how to proved that it will not open ?
Any hints/solution will be appreciated
thanks u
general-topology
asked Dec 5 '18 at 3:32
jasminejasmine
1,689416
$endgroup$
why the set ${(x,y,0) : x^2 +y^2 < 1}$ is neither open nor closed ?
My attempt : it will not closed because it is locally compact
i don't know how to proved that it will not open ?
Any hints/solution will be appreciated
thanks u
general-topology
general-topology
asked Dec 5 '18 at 3:32
jasminejasmine
1,689416
asked Dec 5 '18 at 3:32
jasminejasmine
1,689416
asked Dec 5 '18 at 3:32
jasminejasmine
1,689416
asked Dec 5 '18 at 3:32
jasminejasmine
1,689416
asked Dec 5 '18 at 3:32
jasminejasmine
1,689416
1,689416
closed as off-topic by user21820, Saad, Jyrki Lahtonen, Brahadeesh, DRF Dec 8 '18 at 14:24
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Saad, Jyrki Lahtonen, Brahadeesh, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, Saad, Jyrki Lahtonen, Brahadeesh, DRF Dec 8 '18 at 14:24
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Saad, Jyrki Lahtonen, Brahadeesh, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Who says a locally compact set can't be closed? Every compact set is both locally compact and closed.
$endgroup$
– bof
Dec 5 '18 at 3:36
3
$begingroup$
The set $S={(x,y,0):x^2+y^2lt1}$ is not closed because $$left(frac12,0,0right),left(frac23,0,0right),left(frac34,0,0right),dots,left(frac n{n+1},0,0right),dots$$ is a sequence of points in $S$ converging to the point $(1,0,0)$ which is not in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:41
1
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And it’s not open because every neighborhood of the origin contains points not in the set.
$endgroup$
– Lubin
Dec 5 '18 at 3:45
2
$begingroup$
The set $S={(x,y,0):x^2+y^2lt1}$ is not open because $$(0,0,1),(0,0,1/2),(0,0,1/3),dots,(0,0,1/n),dots$$ is a sequence of points outside of $S$ congerging to the point $(0,0,0)$ in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:45
$begingroup$
Is using the definition of open/closed not an option at all? I am frankly baffled why anyone would try and use the concept of locally compact here, when all you need to do is to check the definition?
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 9:46
add a comment |
1
$begingroup$
Who says a locally compact set can't be closed? Every compact set is both locally compact and closed.
$endgroup$
– bof
Dec 5 '18 at 3:36
3
$begingroup$
The set $S={(x,y,0):x^2+y^2lt1}$ is not closed because $$left(frac12,0,0right),left(frac23,0,0right),left(frac34,0,0right),dots,left(frac n{n+1},0,0right),dots$$ is a sequence of points in $S$ converging to the point $(1,0,0)$ which is not in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:41
1
$begingroup$
And it’s not open because every neighborhood of the origin contains points not in the set.
$endgroup$
– Lubin
Dec 5 '18 at 3:45
2
$begingroup$
The set $S={(x,y,0):x^2+y^2lt1}$ is not open because $$(0,0,1),(0,0,1/2),(0,0,1/3),dots,(0,0,1/n),dots$$ is a sequence of points outside of $S$ congerging to the point $(0,0,0)$ in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:45
$begingroup$
Is using the definition of open/closed not an option at all? I am frankly baffled why anyone would try and use the concept of locally compact here, when all you need to do is to check the definition?
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 9:46
1
1
$begingroup$
Who says a locally compact set can't be closed? Every compact set is both locally compact and closed.
$endgroup$
– bof
Dec 5 '18 at 3:36
$begingroup$
Who says a locally compact set can't be closed? Every compact set is both locally compact and closed.
$endgroup$
– bof
Dec 5 '18 at 3:36
3
3
$begingroup$
The set $S={(x,y,0):x^2+y^2lt1}$ is not closed because $$left(frac12,0,0right),left(frac23,0,0right),left(frac34,0,0right),dots,left(frac n{n+1},0,0right),dots$$ is a sequence of points in $S$ converging to the point $(1,0,0)$ which is not in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:41
$begingroup$
The set $S={(x,y,0):x^2+y^2lt1}$ is not closed because $$left(frac12,0,0right),left(frac23,0,0right),left(frac34,0,0right),dots,left(frac n{n+1},0,0right),dots$$ is a sequence of points in $S$ converging to the point $(1,0,0)$ which is not in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:41
1
1
$begingroup$
And it’s not open because every neighborhood of the origin contains points not in the set.
$endgroup$
– Lubin
Dec 5 '18 at 3:45
$begingroup$
And it’s not open because every neighborhood of the origin contains points not in the set.
$endgroup$
– Lubin
Dec 5 '18 at 3:45
2
2
$begingroup$
The set $S={(x,y,0):x^2+y^2lt1}$ is not open because $$(0,0,1),(0,0,1/2),(0,0,1/3),dots,(0,0,1/n),dots$$ is a sequence of points outside of $S$ congerging to the point $(0,0,0)$ in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:45
$begingroup$
The set $S={(x,y,0):x^2+y^2lt1}$ is not open because $$(0,0,1),(0,0,1/2),(0,0,1/3),dots,(0,0,1/n),dots$$ is a sequence of points outside of $S$ congerging to the point $(0,0,0)$ in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:45
$begingroup$
Is using the definition of open/closed not an option at all? I am frankly baffled why anyone would try and use the concept of locally compact here, when all you need to do is to check the definition?
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 9:46
$begingroup$
Is using the definition of open/closed not an option at all? I am frankly baffled why anyone would try and use the concept of locally compact here, when all you need to do is to check the definition?
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 9:46
add a comment |
2 Answers
2
active
oldest
votes
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Denote $S:={(x,y,z)inmathbb R^3:z=0,~x^2+y^2<1}$.
The set is clearly not closed because one can easily produce a sequence in $S$ which converges to a point outside of $S$ (indeed, see bof's comment).
To see why it is not open, consider the definition of an open set. Take a point $(x,y,0)in S$. Then any open ball of radius $r>0$ around $(x,y,0)$ contains points of the form $(x,y,z)$ where $z>0$, and these points are not in $S$.
answered Dec 5 '18 at 3:45
DaveDave
8,76711033
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add a comment |
$begingroup$
I will call the set mentioned in the question $A$.
To show a set is not closed, it suffices to show that a set does not contain all of its limit points. So try to find a point not in $A$ such that all of its neighborhoods intersect $A$.
To show a set is not open, one can show that no basis element is contained within it. Note that basis elements of $mathbb{R}^3$ can be of the form $(a_1,b_1)times (a_2,b_2)times (a_3,b_3)$ where $a_i < b_i$ for $i$ in ${1, 2, 3}$.
answered Dec 5 '18 at 3:51
John BJohn B
1766
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Denote $S:={(x,y,z)inmathbb R^3:z=0,~x^2+y^2<1}$.
The set is clearly not closed because one can easily produce a sequence in $S$ which converges to a point outside of $S$ (indeed, see bof's comment).
To see why it is not open, consider the definition of an open set. Take a point $(x,y,0)in S$. Then any open ball of radius $r>0$ around $(x,y,0)$ contains points of the form $(x,y,z)$ where $z>0$, and these points are not in $S$.
answered Dec 5 '18 at 3:45
DaveDave
8,76711033
$endgroup$
add a comment |
$begingroup$
Denote $S:={(x,y,z)inmathbb R^3:z=0,~x^2+y^2<1}$.
The set is clearly not closed because one can easily produce a sequence in $S$ which converges to a point outside of $S$ (indeed, see bof's comment).
To see why it is not open, consider the definition of an open set. Take a point $(x,y,0)in S$. Then any open ball of radius $r>0$ around $(x,y,0)$ contains points of the form $(x,y,z)$ where $z>0$, and these points are not in $S$.
answered Dec 5 '18 at 3:45
DaveDave
8,76711033
$endgroup$
add a comment |
$begingroup$
Denote $S:={(x,y,z)inmathbb R^3:z=0,~x^2+y^2<1}$.
The set is clearly not closed because one can easily produce a sequence in $S$ which converges to a point outside of $S$ (indeed, see bof's comment).
To see why it is not open, consider the definition of an open set. Take a point $(x,y,0)in S$. Then any open ball of radius $r>0$ around $(x,y,0)$ contains points of the form $(x,y,z)$ where $z>0$, and these points are not in $S$.
answered Dec 5 '18 at 3:45
DaveDave
8,76711033
$endgroup$
Denote $S:={(x,y,z)inmathbb R^3:z=0,~x^2+y^2<1}$.
The set is clearly not closed because one can easily produce a sequence in $S$ which converges to a point outside of $S$ (indeed, see bof's comment).
To see why it is not open, consider the definition of an open set. Take a point $(x,y,0)in S$. Then any open ball of radius $r>0$ around $(x,y,0)$ contains points of the form $(x,y,z)$ where $z>0$, and these points are not in $S$.
answered Dec 5 '18 at 3:45
DaveDave
8,76711033
answered Dec 5 '18 at 3:45
DaveDave
8,76711033
answered Dec 5 '18 at 3:45
DaveDave
8,76711033
answered Dec 5 '18 at 3:45
DaveDave
8,76711033
8,76711033
add a comment |
add a comment |
$begingroup$
I will call the set mentioned in the question $A$.
To show a set is not closed, it suffices to show that a set does not contain all of its limit points. So try to find a point not in $A$ such that all of its neighborhoods intersect $A$.
To show a set is not open, one can show that no basis element is contained within it. Note that basis elements of $mathbb{R}^3$ can be of the form $(a_1,b_1)times (a_2,b_2)times (a_3,b_3)$ where $a_i < b_i$ for $i$ in ${1, 2, 3}$.
answered Dec 5 '18 at 3:51
John BJohn B
1766
$endgroup$
add a comment |
$begingroup$
I will call the set mentioned in the question $A$.
To show a set is not closed, it suffices to show that a set does not contain all of its limit points. So try to find a point not in $A$ such that all of its neighborhoods intersect $A$.
To show a set is not open, one can show that no basis element is contained within it. Note that basis elements of $mathbb{R}^3$ can be of the form $(a_1,b_1)times (a_2,b_2)times (a_3,b_3)$ where $a_i < b_i$ for $i$ in ${1, 2, 3}$.
answered Dec 5 '18 at 3:51
John BJohn B
1766
$endgroup$
add a comment |
$begingroup$
I will call the set mentioned in the question $A$.
To show a set is not closed, it suffices to show that a set does not contain all of its limit points. So try to find a point not in $A$ such that all of its neighborhoods intersect $A$.
To show a set is not open, one can show that no basis element is contained within it. Note that basis elements of $mathbb{R}^3$ can be of the form $(a_1,b_1)times (a_2,b_2)times (a_3,b_3)$ where $a_i < b_i$ for $i$ in ${1, 2, 3}$.
answered Dec 5 '18 at 3:51
John BJohn B
1766
$endgroup$
I will call the set mentioned in the question $A$.
To show a set is not closed, it suffices to show that a set does not contain all of its limit points. So try to find a point not in $A$ such that all of its neighborhoods intersect $A$.
To show a set is not open, one can show that no basis element is contained within it. Note that basis elements of $mathbb{R}^3$ can be of the form $(a_1,b_1)times (a_2,b_2)times (a_3,b_3)$ where $a_i < b_i$ for $i$ in ${1, 2, 3}$.
answered Dec 5 '18 at 3:51
John BJohn B
1766
answered Dec 5 '18 at 3:51
John BJohn B
1766
answered Dec 5 '18 at 3:51
John BJohn B
1766
answered Dec 5 '18 at 3:51
John BJohn B
1766
1766
add a comment |
add a comment |
1
$begingroup$
Who says a locally compact set can't be closed? Every compact set is both locally compact and closed.
$endgroup$
– bof
Dec 5 '18 at 3:36
3
$begingroup$
The set $S={(x,y,0):x^2+y^2lt1}$ is not closed because $$left(frac12,0,0right),left(frac23,0,0right),left(frac34,0,0right),dots,left(frac n{n+1},0,0right),dots$$ is a sequence of points in $S$ converging to the point $(1,0,0)$ which is not in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:41
1
$begingroup$
And it’s not open because every neighborhood of the origin contains points not in the set.
$endgroup$
– Lubin
Dec 5 '18 at 3:45
2
$begingroup$
The set $S={(x,y,0):x^2+y^2lt1}$ is not open because $$(0,0,1),(0,0,1/2),(0,0,1/3),dots,(0,0,1/n),dots$$ is a sequence of points outside of $S$ congerging to the point $(0,0,0)$ in $S$.
$endgroup$
– bof
Dec 5 '18 at 3:45
$begingroup$
Is using the definition of open/closed not an option at all? I am frankly baffled why anyone would try and use the concept of locally compact here, when all you need to do is to check the definition?
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 9:46