Mixing
Consider ${(−1)^n + 1/n : n in Bbb Nsetminus {0}}subset Bbb R$. Is this set open or closed?
0
$begingroup$
Consider ${(−1)^n + 1/n : n in Bbb Nsetminus {0}}subset Bbb R$. Is this set open or closed?
As this is a union of discrete points, it should be closed but in the set limit of convergent sequence $1+frac{1}{2n}$, which is $1$, does not belong to this set, so is this set open, closed or neither of the two?
real-analysis general-topology
edited Jan 30 at 18:06
Shaun
10.3k113686
asked Jan 30 at 17:59
njjwnjjw
34
$endgroup$
add a comment |
0
$begingroup$
Consider ${(−1)^n + 1/n : n in Bbb Nsetminus {0}}subset Bbb R$. Is this set open or closed?
As this is a union of discrete points, it should be closed but in the set limit of convergent sequence $1+frac{1}{2n}$, which is $1$, does not belong to this set, so is this set open, closed or neither of the two?
real-analysis general-topology
edited Jan 30 at 18:06
Shaun
10.3k113686
asked Jan 30 at 17:59
njjwnjjw
34
$endgroup$
$begingroup$
You have a valid argument that the set is not closed. Now, is it open?
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
There is no rule that a set of discrete points is necessarily closed. A set consisting of a single point is closed, but closedness is not preserved by taking arbitrary unions.
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
complement of set of discrete points will be union of open sets so it should be closed as its complement is open i think
$endgroup$
– njjw
Jan 30 at 18:05
1
$begingroup$
x @njjw, no -- the complement of a union of singletons will be an intersection of open sets, and openness is not preserved by arbitrary intersections.
$endgroup$
– Henning Makholm
Jan 30 at 18:06
add a comment |
0
0
0
0
$begingroup$
Consider ${(−1)^n + 1/n : n in Bbb Nsetminus {0}}subset Bbb R$. Is this set open or closed?
As this is a union of discrete points, it should be closed but in the set limit of convergent sequence $1+frac{1}{2n}$, which is $1$, does not belong to this set, so is this set open, closed or neither of the two?
real-analysis general-topology
edited Jan 30 at 18:06
Shaun
10.3k113686
asked Jan 30 at 17:59
njjwnjjw
34
$endgroup$
Consider ${(−1)^n + 1/n : n in Bbb Nsetminus {0}}subset Bbb R$. Is this set open or closed?
As this is a union of discrete points, it should be closed but in the set limit of convergent sequence $1+frac{1}{2n}$, which is $1$, does not belong to this set, so is this set open, closed or neither of the two?
real-analysis general-topology
real-analysis general-topology
edited Jan 30 at 18:06
Shaun
10.3k113686
asked Jan 30 at 17:59
njjwnjjw
34
edited Jan 30 at 18:06
Shaun
10.3k113686
asked Jan 30 at 17:59
njjwnjjw
34
edited Jan 30 at 18:06
Shaun
10.3k113686
edited Jan 30 at 18:06
Shaun
10.3k113686
edited Jan 30 at 18:06
Shaun
10.3k113686
10.3k113686
asked Jan 30 at 17:59
njjwnjjw
34
asked Jan 30 at 17:59
njjwnjjw
34
asked Jan 30 at 17:59
njjwnjjw
34
34
$begingroup$
You have a valid argument that the set is not closed. Now, is it open?
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
There is no rule that a set of discrete points is necessarily closed. A set consisting of a single point is closed, but closedness is not preserved by taking arbitrary unions.
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
complement of set of discrete points will be union of open sets so it should be closed as its complement is open i think
$endgroup$
– njjw
Jan 30 at 18:05
1
$begingroup$
x @njjw, no -- the complement of a union of singletons will be an intersection of open sets, and openness is not preserved by arbitrary intersections.
$endgroup$
– Henning Makholm
Jan 30 at 18:06
add a comment |
$begingroup$
You have a valid argument that the set is not closed. Now, is it open?
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
There is no rule that a set of discrete points is necessarily closed. A set consisting of a single point is closed, but closedness is not preserved by taking arbitrary unions.
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
complement of set of discrete points will be union of open sets so it should be closed as its complement is open i think
$endgroup$
– njjw
Jan 30 at 18:05
1
$begingroup$
x @njjw, no -- the complement of a union of singletons will be an intersection of open sets, and openness is not preserved by arbitrary intersections.
$endgroup$
– Henning Makholm
Jan 30 at 18:06
$begingroup$
You have a valid argument that the set is not closed. Now, is it open?
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
You have a valid argument that the set is not closed. Now, is it open?
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
There is no rule that a set of discrete points is necessarily closed. A set consisting of a single point is closed, but closedness is not preserved by taking arbitrary unions.
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
There is no rule that a set of discrete points is necessarily closed. A set consisting of a single point is closed, but closedness is not preserved by taking arbitrary unions.
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
complement of set of discrete points will be union of open sets so it should be closed as its complement is open i think
$endgroup$
– njjw
Jan 30 at 18:05
$begingroup$
complement of set of discrete points will be union of open sets so it should be closed as its complement is open i think
$endgroup$
– njjw
Jan 30 at 18:05
1
1
$begingroup$
x @njjw, no -- the complement of a union of singletons will be an intersection of open sets, and openness is not preserved by arbitrary intersections.
$endgroup$
– Henning Makholm
Jan 30 at 18:06
$begingroup$
x @njjw, no -- the complement of a union of singletons will be an intersection of open sets, and openness is not preserved by arbitrary intersections.
$endgroup$
– Henning Makholm
Jan 30 at 18:06
add a comment |
1 Answer
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When $n=1$ the term is $0,$ yet the set doesn't contain any open interval around $0.$ So the set is not open. You already showed it isn't closed.
answered Jan 30 at 18:04
coffeemathcoffeemath
2,9171415
$endgroup$
$begingroup$
but why my first argument that this set is closed is wrong as it is a defination of closed set
$endgroup$
– njjw
Jan 30 at 18:09
1
$begingroup$
Your argument produced a point, namely $1,$ which is a limit point of the set but not a member of the set. That means the set is not closed, since a closed set is one which contains all of its limit points. So what you actually showed in your argument is that the set is not closed.
$endgroup$
– coffeemath
Jan 30 at 18:30
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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1
$begingroup$
When $n=1$ the term is $0,$ yet the set doesn't contain any open interval around $0.$ So the set is not open. You already showed it isn't closed.
answered Jan 30 at 18:04
coffeemathcoffeemath
2,9171415
$endgroup$
$begingroup$
but why my first argument that this set is closed is wrong as it is a defination of closed set
$endgroup$
– njjw
Jan 30 at 18:09
1
$begingroup$
Your argument produced a point, namely $1,$ which is a limit point of the set but not a member of the set. That means the set is not closed, since a closed set is one which contains all of its limit points. So what you actually showed in your argument is that the set is not closed.
$endgroup$
– coffeemath
Jan 30 at 18:30
add a comment |
1
$begingroup$
When $n=1$ the term is $0,$ yet the set doesn't contain any open interval around $0.$ So the set is not open. You already showed it isn't closed.
answered Jan 30 at 18:04
coffeemathcoffeemath
2,9171415
$endgroup$
$begingroup$
but why my first argument that this set is closed is wrong as it is a defination of closed set
$endgroup$
– njjw
Jan 30 at 18:09
1
$begingroup$
Your argument produced a point, namely $1,$ which is a limit point of the set but not a member of the set. That means the set is not closed, since a closed set is one which contains all of its limit points. So what you actually showed in your argument is that the set is not closed.
$endgroup$
– coffeemath
Jan 30 at 18:30
add a comment |
1
1
1
$begingroup$
When $n=1$ the term is $0,$ yet the set doesn't contain any open interval around $0.$ So the set is not open. You already showed it isn't closed.
answered Jan 30 at 18:04
coffeemathcoffeemath
2,9171415
$endgroup$
When $n=1$ the term is $0,$ yet the set doesn't contain any open interval around $0.$ So the set is not open. You already showed it isn't closed.
answered Jan 30 at 18:04
coffeemathcoffeemath
2,9171415
answered Jan 30 at 18:04
coffeemathcoffeemath
2,9171415
answered Jan 30 at 18:04
coffeemathcoffeemath
2,9171415
answered Jan 30 at 18:04
coffeemathcoffeemath
2,9171415
2,9171415
$begingroup$
but why my first argument that this set is closed is wrong as it is a defination of closed set
$endgroup$
– njjw
Jan 30 at 18:09
1
$begingroup$
Your argument produced a point, namely $1,$ which is a limit point of the set but not a member of the set. That means the set is not closed, since a closed set is one which contains all of its limit points. So what you actually showed in your argument is that the set is not closed.
$endgroup$
– coffeemath
Jan 30 at 18:30
add a comment |
$begingroup$
but why my first argument that this set is closed is wrong as it is a defination of closed set
$endgroup$
– njjw
Jan 30 at 18:09
1
$begingroup$
Your argument produced a point, namely $1,$ which is a limit point of the set but not a member of the set. That means the set is not closed, since a closed set is one which contains all of its limit points. So what you actually showed in your argument is that the set is not closed.
$endgroup$
– coffeemath
Jan 30 at 18:30
$begingroup$
but why my first argument that this set is closed is wrong as it is a defination of closed set
$endgroup$
– njjw
Jan 30 at 18:09
$begingroup$
but why my first argument that this set is closed is wrong as it is a defination of closed set
$endgroup$
– njjw
Jan 30 at 18:09
1
1
$begingroup$
Your argument produced a point, namely $1,$ which is a limit point of the set but not a member of the set. That means the set is not closed, since a closed set is one which contains all of its limit points. So what you actually showed in your argument is that the set is not closed.
$endgroup$
– coffeemath
Jan 30 at 18:30
$begingroup$
Your argument produced a point, namely $1,$ which is a limit point of the set but not a member of the set. That means the set is not closed, since a closed set is one which contains all of its limit points. So what you actually showed in your argument is that the set is not closed.
$endgroup$
– coffeemath
Jan 30 at 18:30
add a comment |
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$begingroup$
You have a valid argument that the set is not closed. Now, is it open?
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
There is no rule that a set of discrete points is necessarily closed. A set consisting of a single point is closed, but closedness is not preserved by taking arbitrary unions.
$endgroup$
– Henning Makholm
Jan 30 at 18:03
$begingroup$
complement of set of discrete points will be union of open sets so it should be closed as its complement is open i think
$endgroup$
– njjw
Jan 30 at 18:05
1
$begingroup$
x @njjw, no -- the complement of a union of singletons will be an intersection of open sets, and openness is not preserved by arbitrary intersections.
$endgroup$
– Henning Makholm
Jan 30 at 18:06