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Limit point of the closed set $[0,1]$
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I am trying to understand the concept of limit point in general.
For the open set $(0,1)$, earlier discussions on stackexchange showed that all point in the open set $(0,1)$ is a limit point. Similarly, we can say the same about the closed set $[0,1]$, right ?. For every point $x$ in the set $[0,1]$, we can find $epsilon$ neighborhood ($V_epsilon$) such that $V_epsilon(x) cap A neq emptyset,x$.
If the above comment is true, anyone can give an example of a continuous set containing isolated points ?
real-analysis general-topology
asked Jul 18 at 20:15
Shew
454411
 |Â
show 2 more comments
up vote
0
down vote
favorite
I am trying to understand the concept of limit point in general.
For the open set $(0,1)$, earlier discussions on stackexchange showed that all point in the open set $(0,1)$ is a limit point. Similarly, we can say the same about the closed set $[0,1]$, right ?. For every point $x$ in the set $[0,1]$, we can find $epsilon$ neighborhood ($V_epsilon$) such that $V_epsilon(x) cap A neq emptyset,x$.
If the above comment is true, anyone can give an example of a continuous set containing isolated points ?
real-analysis general-topology
asked Jul 18 at 20:15
Shew
454411
If by "continuous set" you mean an interval, the only possibility is a singleton $ a $.
– Crostul
Jul 18 at 20:19
1
Yes it's true. Continuous set is unfortunately a little bit ambiguous. But if I understood well, indeed, set as $[a,b]$ has no isolated point. But if you take the discrete topology and $xin [0,1]$, then $[0,1]capx, -3=x$ and thus $x$ is an isolated point of $[0,1]$
– Peter
Jul 18 at 20:21
1
You need to show that for every $varepsilon$-neighbourhood of $x$, $V_varepsilon(x) cap A$ is contains a point in $A setminus x$, not that there is one. But indeed all $xin [0,1]$ are limit points.
– Henno Brandsma
Jul 18 at 21:33
/= $emptyset$,x is a misunderstanding
– William Elliot
Jul 19 at 2:46
1
As an intersection of A and an open set will never have the empty set as a member, your statement is always true. You are misusing set notation. Not subset of x is correct. @Shew
– William Elliot
Jul 20 at 8:09
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to understand the concept of limit point in general.
For the open set $(0,1)$, earlier discussions on stackexchange showed that all point in the open set $(0,1)$ is a limit point. Similarly, we can say the same about the closed set $[0,1]$, right ?. For every point $x$ in the set $[0,1]$, we can find $epsilon$ neighborhood ($V_epsilon$) such that $V_epsilon(x) cap A neq emptyset,x$.
If the above comment is true, anyone can give an example of a continuous set containing isolated points ?
real-analysis general-topology
asked Jul 18 at 20:15
Shew
454411
I am trying to understand the concept of limit point in general.
For the open set $(0,1)$, earlier discussions on stackexchange showed that all point in the open set $(0,1)$ is a limit point. Similarly, we can say the same about the closed set $[0,1]$, right ?. For every point $x$ in the set $[0,1]$, we can find $epsilon$ neighborhood ($V_epsilon$) such that $V_epsilon(x) cap A neq emptyset,x$.
If the above comment is true, anyone can give an example of a continuous set containing isolated points ?
real-analysis general-topology
asked Jul 18 at 20:15
Shew
454411
asked Jul 18 at 20:15
Shew
454411
asked Jul 18 at 20:15
Shew
454411
asked Jul 18 at 20:15
Shew
454411
454411
If by "continuous set" you mean an interval, the only possibility is a singleton $ a $.
– Crostul
Jul 18 at 20:19
1
Yes it's true. Continuous set is unfortunately a little bit ambiguous. But if I understood well, indeed, set as $[a,b]$ has no isolated point. But if you take the discrete topology and $xin [0,1]$, then $[0,1]capx, -3=x$ and thus $x$ is an isolated point of $[0,1]$
– Peter
Jul 18 at 20:21
1
You need to show that for every $varepsilon$-neighbourhood of $x$, $V_varepsilon(x) cap A$ is contains a point in $A setminus x$, not that there is one. But indeed all $xin [0,1]$ are limit points.
– Henno Brandsma
Jul 18 at 21:33
/= $emptyset$,x is a misunderstanding
– William Elliot
Jul 19 at 2:46
1
As an intersection of A and an open set will never have the empty set as a member, your statement is always true. You are misusing set notation. Not subset of x is correct. @Shew
– William Elliot
Jul 20 at 8:09
 |Â
show 2 more comments
If by "continuous set" you mean an interval, the only possibility is a singleton $ a $.
– Crostul
Jul 18 at 20:19
1
Yes it's true. Continuous set is unfortunately a little bit ambiguous. But if I understood well, indeed, set as $[a,b]$ has no isolated point. But if you take the discrete topology and $xin [0,1]$, then $[0,1]capx, -3=x$ and thus $x$ is an isolated point of $[0,1]$
– Peter
Jul 18 at 20:21
1
You need to show that for every $varepsilon$-neighbourhood of $x$, $V_varepsilon(x) cap A$ is contains a point in $A setminus x$, not that there is one. But indeed all $xin [0,1]$ are limit points.
– Henno Brandsma
Jul 18 at 21:33
/= $emptyset$,x is a misunderstanding
– William Elliot
Jul 19 at 2:46
1
As an intersection of A and an open set will never have the empty set as a member, your statement is always true. You are misusing set notation. Not subset of x is correct. @Shew
– William Elliot
Jul 20 at 8:09
If by "continuous set" you mean an interval, the only possibility is a singleton $ a $.
– Crostul
Jul 18 at 20:19
If by "continuous set" you mean an interval, the only possibility is a singleton $ a $.
– Crostul
Jul 18 at 20:19
1
1
Yes it's true. Continuous set is unfortunately a little bit ambiguous. But if I understood well, indeed, set as $[a,b]$ has no isolated point. But if you take the discrete topology and $xin [0,1]$, then $[0,1]capx, -3=x$ and thus $x$ is an isolated point of $[0,1]$
– Peter
Jul 18 at 20:21
Yes it's true. Continuous set is unfortunately a little bit ambiguous. But if I understood well, indeed, set as $[a,b]$ has no isolated point. But if you take the discrete topology and $xin [0,1]$, then $[0,1]capx, -3=x$ and thus $x$ is an isolated point of $[0,1]$
– Peter
Jul 18 at 20:21
1
1
You need to show that for every $varepsilon$-neighbourhood of $x$, $V_varepsilon(x) cap A$ is contains a point in $A setminus x$, not that there is one. But indeed all $xin [0,1]$ are limit points.
– Henno Brandsma
Jul 18 at 21:33
You need to show that for every $varepsilon$-neighbourhood of $x$, $V_varepsilon(x) cap A$ is contains a point in $A setminus x$, not that there is one. But indeed all $xin [0,1]$ are limit points.
– Henno Brandsma
Jul 18 at 21:33
/= $emptyset$,x is a misunderstanding
– William Elliot
Jul 19 at 2:46
/= $emptyset$,x is a misunderstanding
– William Elliot
Jul 19 at 2:46
1
1
As an intersection of A and an open set will never have the empty set as a member, your statement is always true. You are misusing set notation. Not subset of x is correct. @Shew
– William Elliot
Jul 20 at 8:09
As an intersection of A and an open set will never have the empty set as a member, your statement is always true. You are misusing set notation. Not subset of x is correct. @Shew
– William Elliot
Jul 20 at 8:09
 |Â
show 2 more comments
1 Answer
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By "continuous", I think you mean connected. But a connected set might not have all of its limit points.
Here's an example. Consider the set $(0,1)$. It is true that every point in the set is a limit point. For example, let's look at the point $1/2$. Is this a limit point? To be a limit point, we need to create a sequence of points in the set that approach this point. The Sequence $1/2, 1/2, 1/2, ...$ is such a set of points.
What about the point $0$? Is $0$ a limit point of this set? It is! Consider the sequence $1/2, 1/4, 1/8, 1/16, 1/32, ...$ This sequence is getting closer and closer to the point $0$, and no point in this sequence will ever leave the set $(0,1)$. Therefore, the set $(0,1)$ does not contain all its limit points.
Any set that contains all of its limit points is called a "closed" set. So $(0,1)$ is not a compact set. But $[0,1]$ is.
answered Jul 18 at 20:50
NicNic8
3,7113922
A set that contains all it's limit points is closed, but not necessarily compact; for example, $[0,infty)$ contains all it's limit points but is not compact.
– User8128
Jul 18 at 20:57
@User8128 Thank you for correcting my mistake.
– NicNic8
Jul 18 at 20:58
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
By "continuous", I think you mean connected. But a connected set might not have all of its limit points.
Here's an example. Consider the set $(0,1)$. It is true that every point in the set is a limit point. For example, let's look at the point $1/2$. Is this a limit point? To be a limit point, we need to create a sequence of points in the set that approach this point. The Sequence $1/2, 1/2, 1/2, ...$ is such a set of points.
What about the point $0$? Is $0$ a limit point of this set? It is! Consider the sequence $1/2, 1/4, 1/8, 1/16, 1/32, ...$ This sequence is getting closer and closer to the point $0$, and no point in this sequence will ever leave the set $(0,1)$. Therefore, the set $(0,1)$ does not contain all its limit points.
Any set that contains all of its limit points is called a "closed" set. So $(0,1)$ is not a compact set. But $[0,1]$ is.
answered Jul 18 at 20:50
NicNic8
3,7113922
A set that contains all it's limit points is closed, but not necessarily compact; for example, $[0,infty)$ contains all it's limit points but is not compact.
– User8128
Jul 18 at 20:57
@User8128 Thank you for correcting my mistake.
– NicNic8
Jul 18 at 20:58
add a comment |Â
up vote
0
down vote
By "continuous", I think you mean connected. But a connected set might not have all of its limit points.
Here's an example. Consider the set $(0,1)$. It is true that every point in the set is a limit point. For example, let's look at the point $1/2$. Is this a limit point? To be a limit point, we need to create a sequence of points in the set that approach this point. The Sequence $1/2, 1/2, 1/2, ...$ is such a set of points.
What about the point $0$? Is $0$ a limit point of this set? It is! Consider the sequence $1/2, 1/4, 1/8, 1/16, 1/32, ...$ This sequence is getting closer and closer to the point $0$, and no point in this sequence will ever leave the set $(0,1)$. Therefore, the set $(0,1)$ does not contain all its limit points.
Any set that contains all of its limit points is called a "closed" set. So $(0,1)$ is not a compact set. But $[0,1]$ is.
answered Jul 18 at 20:50
NicNic8
3,7113922
A set that contains all it's limit points is closed, but not necessarily compact; for example, $[0,infty)$ contains all it's limit points but is not compact.
– User8128
Jul 18 at 20:57
@User8128 Thank you for correcting my mistake.
– NicNic8
Jul 18 at 20:58
add a comment |Â
up vote
0
down vote
up vote
0
down vote
By "continuous", I think you mean connected. But a connected set might not have all of its limit points.
Here's an example. Consider the set $(0,1)$. It is true that every point in the set is a limit point. For example, let's look at the point $1/2$. Is this a limit point? To be a limit point, we need to create a sequence of points in the set that approach this point. The Sequence $1/2, 1/2, 1/2, ...$ is such a set of points.
What about the point $0$? Is $0$ a limit point of this set? It is! Consider the sequence $1/2, 1/4, 1/8, 1/16, 1/32, ...$ This sequence is getting closer and closer to the point $0$, and no point in this sequence will ever leave the set $(0,1)$. Therefore, the set $(0,1)$ does not contain all its limit points.
Any set that contains all of its limit points is called a "closed" set. So $(0,1)$ is not a compact set. But $[0,1]$ is.
answered Jul 18 at 20:50
NicNic8
3,7113922
By "continuous", I think you mean connected. But a connected set might not have all of its limit points.
Here's an example. Consider the set $(0,1)$. It is true that every point in the set is a limit point. For example, let's look at the point $1/2$. Is this a limit point? To be a limit point, we need to create a sequence of points in the set that approach this point. The Sequence $1/2, 1/2, 1/2, ...$ is such a set of points.
What about the point $0$? Is $0$ a limit point of this set? It is! Consider the sequence $1/2, 1/4, 1/8, 1/16, 1/32, ...$ This sequence is getting closer and closer to the point $0$, and no point in this sequence will ever leave the set $(0,1)$. Therefore, the set $(0,1)$ does not contain all its limit points.
Any set that contains all of its limit points is called a "closed" set. So $(0,1)$ is not a compact set. But $[0,1]$ is.
answered Jul 18 at 20:50
NicNic8
3,7113922
edited Jul 18 at 20:58
answered Jul 18 at 20:50
NicNic8
3,7113922
answered Jul 18 at 20:50
NicNic8
3,7113922
answered Jul 18 at 20:50
NicNic8
3,7113922
3,7113922
A set that contains all it's limit points is closed, but not necessarily compact; for example, $[0,infty)$ contains all it's limit points but is not compact.
– User8128
Jul 18 at 20:57
@User8128 Thank you for correcting my mistake.
– NicNic8
Jul 18 at 20:58
add a comment |Â
A set that contains all it's limit points is closed, but not necessarily compact; for example, $[0,infty)$ contains all it's limit points but is not compact.
– User8128
Jul 18 at 20:57
@User8128 Thank you for correcting my mistake.
– NicNic8
Jul 18 at 20:58
A set that contains all it's limit points is closed, but not necessarily compact; for example, $[0,infty)$ contains all it's limit points but is not compact.
– User8128
Jul 18 at 20:57
A set that contains all it's limit points is closed, but not necessarily compact; for example, $[0,infty)$ contains all it's limit points but is not compact.
– User8128
Jul 18 at 20:57
@User8128 Thank you for correcting my mistake.
– NicNic8
Jul 18 at 20:58
@User8128 Thank you for correcting my mistake.
– NicNic8
Jul 18 at 20:58
add a comment |Â
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If by "continuous set" you mean an interval, the only possibility is a singleton $ a $.
– Crostul
Jul 18 at 20:19
1
Yes it's true. Continuous set is unfortunately a little bit ambiguous. But if I understood well, indeed, set as $[a,b]$ has no isolated point. But if you take the discrete topology and $xin [0,1]$, then $[0,1]capx, -3=x$ and thus $x$ is an isolated point of $[0,1]$
– Peter
Jul 18 at 20:21
1
You need to show that for every $varepsilon$-neighbourhood of $x$, $V_varepsilon(x) cap A$ is contains a point in $A setminus x$, not that there is one. But indeed all $xin [0,1]$ are limit points.
– Henno Brandsma
Jul 18 at 21:33
/= $emptyset$,x is a misunderstanding
– William Elliot
Jul 19 at 2:46
1
As an intersection of A and an open set will never have the empty set as a member, your statement is always true. You are misusing set notation. Not subset of x is correct. @Shew
– William Elliot
Jul 20 at 8:09