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Find set $A$ whose limit points are the same as its boundary points
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I was reading a problem that asked to find a non-empty set $A$ such that set of limit points in $A$ is the same as the set of boundary points in $A$. The solution given was to pick the set of all $xinmathbbR^2$ such that $|x|=1$. I was wondering if there are any other such sets $A$? Below is the problem and solution in case you wanted to see the solution they had.
Edit
Also, I'm a bit confused by their solution. If we were to consider the entire circle, the boundary of the circle would be the outer points. but since we are only considering those outer points, what are they bounding?
Thanks
real-analysis analysis
edited Jul 18 at 2:56
Key Flex
4,346425
asked Jul 17 at 23:30
john fowles
1,093817
add a comment |Â
up vote
0
down vote
favorite
I was reading a problem that asked to find a non-empty set $A$ such that set of limit points in $A$ is the same as the set of boundary points in $A$. The solution given was to pick the set of all $xinmathbbR^2$ such that $|x|=1$. I was wondering if there are any other such sets $A$? Below is the problem and solution in case you wanted to see the solution they had.
Edit
Also, I'm a bit confused by their solution. If we were to consider the entire circle, the boundary of the circle would be the outer points. but since we are only considering those outer points, what are they bounding?
Thanks
real-analysis analysis
edited Jul 18 at 2:56
Key Flex
4,346425
asked Jul 17 at 23:30
john fowles
1,093817
1
Boundary points do not do any "bounding". The boundary of a subset $A$ of a space $S$ is $overline A cap overline Ssetminus A.$
– DanielWainfleet
Jul 18 at 0:33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was reading a problem that asked to find a non-empty set $A$ such that set of limit points in $A$ is the same as the set of boundary points in $A$. The solution given was to pick the set of all $xinmathbbR^2$ such that $|x|=1$. I was wondering if there are any other such sets $A$? Below is the problem and solution in case you wanted to see the solution they had.
Edit
Also, I'm a bit confused by their solution. If we were to consider the entire circle, the boundary of the circle would be the outer points. but since we are only considering those outer points, what are they bounding?
Thanks
real-analysis analysis
edited Jul 18 at 2:56
Key Flex
4,346425
asked Jul 17 at 23:30
john fowles
1,093817
I was reading a problem that asked to find a non-empty set $A$ such that set of limit points in $A$ is the same as the set of boundary points in $A$. The solution given was to pick the set of all $xinmathbbR^2$ such that $|x|=1$. I was wondering if there are any other such sets $A$? Below is the problem and solution in case you wanted to see the solution they had.
Edit
Also, I'm a bit confused by their solution. If we were to consider the entire circle, the boundary of the circle would be the outer points. but since we are only considering those outer points, what are they bounding?
Thanks
real-analysis analysis
edited Jul 18 at 2:56
Key Flex
4,346425
asked Jul 17 at 23:30
john fowles
1,093817
edited Jul 18 at 2:56
Key Flex
4,346425
edited Jul 18 at 2:56
Key Flex
4,346425
edited Jul 18 at 2:56
Key Flex
4,346425
4,346425
asked Jul 17 at 23:30
john fowles
1,093817
asked Jul 17 at 23:30
john fowles
1,093817
asked Jul 17 at 23:30
john fowles
1,093817
1,093817
1
Boundary points do not do any "bounding". The boundary of a subset $A$ of a space $S$ is $overline A cap overline Ssetminus A.$
– DanielWainfleet
Jul 18 at 0:33
add a comment |Â
1
Boundary points do not do any "bounding". The boundary of a subset $A$ of a space $S$ is $overline A cap overline Ssetminus A.$
– DanielWainfleet
Jul 18 at 0:33
1
1
Boundary points do not do any "bounding". The boundary of a subset $A$ of a space $S$ is $overline A cap overline Ssetminus A.$
– DanielWainfleet
Jul 18 at 0:33
Boundary points do not do any "bounding". The boundary of a subset $A$ of a space $S$ is $overline A cap overline Ssetminus A.$
– DanielWainfleet
Jul 18 at 0:33
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Almost any curve in $mathbbR^2$ will have this property: any conic section, graph of an elementary function, etc. (Every point in the curve is a limit point of the curve, and of its complement)
answered Jul 17 at 23:36
Spencer
1678
you mean $mathbbR^2$, right? and I'm just confused about the idea of the boundary. If we consider any appropriate curve in $mathbbR^2$ then why are those points boundary points?
– john fowles
Jul 17 at 23:54
The boundary is the set of points in both the closure of a set and the closure of its complement. In the case of a curve (unless it's space-filling or otherwise strange), the closure of the curve is the curve itself. On the other hand, there are points arbitrarily close to any point in the curve that are not in the curve.
– Spencer
Jul 17 at 23:57
add a comment |Â
up vote
2
down vote
Another less trivial example is the set of rational numbers.
The set of its boundary points is the set of real numbers which is also the set of its limit points.
answered Jul 17 at 23:49
Mohammad Riazi-Kermani
27.5k41852
I really like this answer, but I gave the correct answer to the other user because he replied first. Thanks
– john fowles
Jul 18 at 0:08
@johnfowles Thanks for the comment. You are the OP and it is your right to accept one solution. It does not mean that other answers are not correct.
– Mohammad Riazi-Kermani
Jul 18 at 0:19
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Almost any curve in $mathbbR^2$ will have this property: any conic section, graph of an elementary function, etc. (Every point in the curve is a limit point of the curve, and of its complement)
answered Jul 17 at 23:36
Spencer
1678
you mean $mathbbR^2$, right? and I'm just confused about the idea of the boundary. If we consider any appropriate curve in $mathbbR^2$ then why are those points boundary points?
– john fowles
Jul 17 at 23:54
The boundary is the set of points in both the closure of a set and the closure of its complement. In the case of a curve (unless it's space-filling or otherwise strange), the closure of the curve is the curve itself. On the other hand, there are points arbitrarily close to any point in the curve that are not in the curve.
– Spencer
Jul 17 at 23:57
add a comment |Â
up vote
3
down vote
accepted
Almost any curve in $mathbbR^2$ will have this property: any conic section, graph of an elementary function, etc. (Every point in the curve is a limit point of the curve, and of its complement)
answered Jul 17 at 23:36
Spencer
1678
you mean $mathbbR^2$, right? and I'm just confused about the idea of the boundary. If we consider any appropriate curve in $mathbbR^2$ then why are those points boundary points?
– john fowles
Jul 17 at 23:54
The boundary is the set of points in both the closure of a set and the closure of its complement. In the case of a curve (unless it's space-filling or otherwise strange), the closure of the curve is the curve itself. On the other hand, there are points arbitrarily close to any point in the curve that are not in the curve.
– Spencer
Jul 17 at 23:57
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Almost any curve in $mathbbR^2$ will have this property: any conic section, graph of an elementary function, etc. (Every point in the curve is a limit point of the curve, and of its complement)
answered Jul 17 at 23:36
Spencer
1678
Almost any curve in $mathbbR^2$ will have this property: any conic section, graph of an elementary function, etc. (Every point in the curve is a limit point of the curve, and of its complement)
answered Jul 17 at 23:36
Spencer
1678
edited Jul 17 at 23:54
answered Jul 17 at 23:36
Spencer
1678
answered Jul 17 at 23:36
Spencer
1678
answered Jul 17 at 23:36
Spencer
1678
1678
you mean $mathbbR^2$, right? and I'm just confused about the idea of the boundary. If we consider any appropriate curve in $mathbbR^2$ then why are those points boundary points?
– john fowles
Jul 17 at 23:54
The boundary is the set of points in both the closure of a set and the closure of its complement. In the case of a curve (unless it's space-filling or otherwise strange), the closure of the curve is the curve itself. On the other hand, there are points arbitrarily close to any point in the curve that are not in the curve.
– Spencer
Jul 17 at 23:57
add a comment |Â
you mean $mathbbR^2$, right? and I'm just confused about the idea of the boundary. If we consider any appropriate curve in $mathbbR^2$ then why are those points boundary points?
– john fowles
Jul 17 at 23:54
The boundary is the set of points in both the closure of a set and the closure of its complement. In the case of a curve (unless it's space-filling or otherwise strange), the closure of the curve is the curve itself. On the other hand, there are points arbitrarily close to any point in the curve that are not in the curve.
– Spencer
Jul 17 at 23:57
you mean $mathbbR^2$, right? and I'm just confused about the idea of the boundary. If we consider any appropriate curve in $mathbbR^2$ then why are those points boundary points?
– john fowles
Jul 17 at 23:54
you mean $mathbbR^2$, right? and I'm just confused about the idea of the boundary. If we consider any appropriate curve in $mathbbR^2$ then why are those points boundary points?
– john fowles
Jul 17 at 23:54
The boundary is the set of points in both the closure of a set and the closure of its complement. In the case of a curve (unless it's space-filling or otherwise strange), the closure of the curve is the curve itself. On the other hand, there are points arbitrarily close to any point in the curve that are not in the curve.
– Spencer
Jul 17 at 23:57
The boundary is the set of points in both the closure of a set and the closure of its complement. In the case of a curve (unless it's space-filling or otherwise strange), the closure of the curve is the curve itself. On the other hand, there are points arbitrarily close to any point in the curve that are not in the curve.
– Spencer
Jul 17 at 23:57
add a comment |Â
up vote
2
down vote
Another less trivial example is the set of rational numbers.
The set of its boundary points is the set of real numbers which is also the set of its limit points.
answered Jul 17 at 23:49
Mohammad Riazi-Kermani
27.5k41852
I really like this answer, but I gave the correct answer to the other user because he replied first. Thanks
– john fowles
Jul 18 at 0:08
@johnfowles Thanks for the comment. You are the OP and it is your right to accept one solution. It does not mean that other answers are not correct.
– Mohammad Riazi-Kermani
Jul 18 at 0:19
add a comment |Â
up vote
2
down vote
Another less trivial example is the set of rational numbers.
The set of its boundary points is the set of real numbers which is also the set of its limit points.
answered Jul 17 at 23:49
Mohammad Riazi-Kermani
27.5k41852
I really like this answer, but I gave the correct answer to the other user because he replied first. Thanks
– john fowles
Jul 18 at 0:08
@johnfowles Thanks for the comment. You are the OP and it is your right to accept one solution. It does not mean that other answers are not correct.
– Mohammad Riazi-Kermani
Jul 18 at 0:19
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Another less trivial example is the set of rational numbers.
The set of its boundary points is the set of real numbers which is also the set of its limit points.
answered Jul 17 at 23:49
Mohammad Riazi-Kermani
27.5k41852
Another less trivial example is the set of rational numbers.
The set of its boundary points is the set of real numbers which is also the set of its limit points.
answered Jul 17 at 23:49
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 23:49
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 23:49
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 23:49
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
I really like this answer, but I gave the correct answer to the other user because he replied first. Thanks
– john fowles
Jul 18 at 0:08
@johnfowles Thanks for the comment. You are the OP and it is your right to accept one solution. It does not mean that other answers are not correct.
– Mohammad Riazi-Kermani
Jul 18 at 0:19
add a comment |Â
I really like this answer, but I gave the correct answer to the other user because he replied first. Thanks
– john fowles
Jul 18 at 0:08
@johnfowles Thanks for the comment. You are the OP and it is your right to accept one solution. It does not mean that other answers are not correct.
– Mohammad Riazi-Kermani
Jul 18 at 0:19
I really like this answer, but I gave the correct answer to the other user because he replied first. Thanks
– john fowles
Jul 18 at 0:08
I really like this answer, but I gave the correct answer to the other user because he replied first. Thanks
– john fowles
Jul 18 at 0:08
@johnfowles Thanks for the comment. You are the OP and it is your right to accept one solution. It does not mean that other answers are not correct.
– Mohammad Riazi-Kermani
Jul 18 at 0:19
@johnfowles Thanks for the comment. You are the OP and it is your right to accept one solution. It does not mean that other answers are not correct.
– Mohammad Riazi-Kermani
Jul 18 at 0:19
add a comment |Â
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1
Boundary points do not do any "bounding". The boundary of a subset $A$ of a space $S$ is $overline A cap overline Ssetminus A.$
– DanielWainfleet
Jul 18 at 0:33