In an indiscrete topology, does the empty set have any limit points?
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Since the intersection of any set with $emptyset$ is $emptyset$, it does not seem like $phi$ has any limit point. Is my reasoning correct?
general-topology
asked Nov 1 '14 at 15:08
StudentStudent
7141520
$endgroup$
add a comment |
$begingroup$
Since the intersection of any set with $emptyset$ is $emptyset$, it does not seem like $phi$ has any limit point. Is my reasoning correct?
general-topology
asked Nov 1 '14 at 15:08
StudentStudent
7141520
$endgroup$
1
$begingroup$
Yes, it is correct. No matter what topology you put, $emptyset$ has no limit points.
$endgroup$
– Jonas Gomes
Nov 1 '14 at 15:10
4
$begingroup$
For $emptyset$, useemptyset; for $varnothing$ usevarnothing.
$endgroup$
– MJD
Nov 1 '14 at 15:14
$begingroup$
the empty set has a unique topology, and of course it has no limit points since $emptyset$ has no points at all.
$endgroup$
– matiasdata
Nov 1 '14 at 15:17
add a comment |
$begingroup$
Since the intersection of any set with $emptyset$ is $emptyset$, it does not seem like $phi$ has any limit point. Is my reasoning correct?
general-topology
asked Nov 1 '14 at 15:08
StudentStudent
7141520
$endgroup$
Since the intersection of any set with $emptyset$ is $emptyset$, it does not seem like $phi$ has any limit point. Is my reasoning correct?
general-topology
general-topology
asked Nov 1 '14 at 15:08
StudentStudent
7141520
asked Nov 1 '14 at 15:08
StudentStudent
7141520
edited Jul 21 '16 at 13:42
Student
asked Nov 1 '14 at 15:08
StudentStudent
7141520
asked Nov 1 '14 at 15:08
StudentStudent
7141520
asked Nov 1 '14 at 15:08
StudentStudent
7141520
7141520
1
$begingroup$
Yes, it is correct. No matter what topology you put, $emptyset$ has no limit points.
$endgroup$
– Jonas Gomes
Nov 1 '14 at 15:10
4
$begingroup$
For $emptyset$, useemptyset; for $varnothing$ usevarnothing.
$endgroup$
– MJD
Nov 1 '14 at 15:14
$begingroup$
the empty set has a unique topology, and of course it has no limit points since $emptyset$ has no points at all.
$endgroup$
– matiasdata
Nov 1 '14 at 15:17
add a comment |
1
$begingroup$
Yes, it is correct. No matter what topology you put, $emptyset$ has no limit points.
$endgroup$
– Jonas Gomes
Nov 1 '14 at 15:10
4
$begingroup$
For $emptyset$, useemptyset; for $varnothing$ usevarnothing.
$endgroup$
– MJD
Nov 1 '14 at 15:14
$begingroup$
the empty set has a unique topology, and of course it has no limit points since $emptyset$ has no points at all.
$endgroup$
– matiasdata
Nov 1 '14 at 15:17
1
1
$begingroup$
Yes, it is correct. No matter what topology you put, $emptyset$ has no limit points.
$endgroup$
– Jonas Gomes
Nov 1 '14 at 15:10
$begingroup$
Yes, it is correct. No matter what topology you put, $emptyset$ has no limit points.
$endgroup$
– Jonas Gomes
Nov 1 '14 at 15:10
4
4
$begingroup$
For $emptyset$, use
emptyset; for $varnothing$ use varnothing.$endgroup$
– MJD
Nov 1 '14 at 15:14
$begingroup$
For $emptyset$, use
emptyset; for $varnothing$ use varnothing.$endgroup$
– MJD
Nov 1 '14 at 15:14
$begingroup$
the empty set has a unique topology, and of course it has no limit points since $emptyset$ has no points at all.
$endgroup$
– matiasdata
Nov 1 '14 at 15:17
$begingroup$
the empty set has a unique topology, and of course it has no limit points since $emptyset$ has no points at all.
$endgroup$
– matiasdata
Nov 1 '14 at 15:17
add a comment |
3 Answers
3
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oldest
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Empty set has no limit points since no set can have non-empty intersection with $varnothing$.
answered Nov 1 '14 at 15:41
Mathew GeorgeMathew George
513212
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add a comment |
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Suppose $emptyset$ has at least one limit point, say $x$. Then there exist a sequence $(x_n)$ in $emptyset$ which converge to $x.$ That is each $x_ninemptyset.$ This gives if empty set has a limit point it is not empty which contradicts to its definition. Hence??
answered Nov 1 '14 at 15:22
BumblebeeBumblebee
9,72412551
$endgroup$
3
$begingroup$
That's not the definition of limit point. A limit point of a set $S$ is a point $x$ such that $Ucap (Ssetminus {x}) neq varnothing$ for every neighbourhood $U$ of $x$. It does not follow that there is a sequence in $S$ converging to $x$ - and sometimes, such a sequence does not exist. If we were dealing with metric spaces, such a sequence would necessarily exist.
$endgroup$
– Daniel Fischer
Nov 1 '14 at 15:48
1
$begingroup$
After a long time now I can understand your comment :)
$endgroup$
– Bumblebee
Dec 30 '16 at 20:05
add a comment |
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Yes you would appear to be correct. And since it contains all its limit points (vacuously), it's closed. Additionally, this appears to be true in any topology.
answered Jan 1 at 5:06
Chris CusterChris Custer
14.2k3827
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3 Answers
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active
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3 Answers
3
active
oldest
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active
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active
oldest
votes
$begingroup$
Empty set has no limit points since no set can have non-empty intersection with $varnothing$.
answered Nov 1 '14 at 15:41
Mathew GeorgeMathew George
513212
$endgroup$
add a comment |
$begingroup$
Empty set has no limit points since no set can have non-empty intersection with $varnothing$.
answered Nov 1 '14 at 15:41
Mathew GeorgeMathew George
513212
$endgroup$
add a comment |
$begingroup$
Empty set has no limit points since no set can have non-empty intersection with $varnothing$.
answered Nov 1 '14 at 15:41
Mathew GeorgeMathew George
513212
$endgroup$
Empty set has no limit points since no set can have non-empty intersection with $varnothing$.
answered Nov 1 '14 at 15:41
Mathew GeorgeMathew George
513212
edited Jan 1 at 4:52
answered Nov 1 '14 at 15:41
Mathew GeorgeMathew George
513212
answered Nov 1 '14 at 15:41
Mathew GeorgeMathew George
513212
answered Nov 1 '14 at 15:41
Mathew GeorgeMathew George
513212
513212
add a comment |
add a comment |
$begingroup$
Suppose $emptyset$ has at least one limit point, say $x$. Then there exist a sequence $(x_n)$ in $emptyset$ which converge to $x.$ That is each $x_ninemptyset.$ This gives if empty set has a limit point it is not empty which contradicts to its definition. Hence??
answered Nov 1 '14 at 15:22
BumblebeeBumblebee
9,72412551
$endgroup$
3
$begingroup$
That's not the definition of limit point. A limit point of a set $S$ is a point $x$ such that $Ucap (Ssetminus {x}) neq varnothing$ for every neighbourhood $U$ of $x$. It does not follow that there is a sequence in $S$ converging to $x$ - and sometimes, such a sequence does not exist. If we were dealing with metric spaces, such a sequence would necessarily exist.
$endgroup$
– Daniel Fischer
Nov 1 '14 at 15:48
1
$begingroup$
After a long time now I can understand your comment :)
$endgroup$
– Bumblebee
Dec 30 '16 at 20:05
add a comment |
$begingroup$
Suppose $emptyset$ has at least one limit point, say $x$. Then there exist a sequence $(x_n)$ in $emptyset$ which converge to $x.$ That is each $x_ninemptyset.$ This gives if empty set has a limit point it is not empty which contradicts to its definition. Hence??
answered Nov 1 '14 at 15:22
BumblebeeBumblebee
9,72412551
$endgroup$
3
$begingroup$
That's not the definition of limit point. A limit point of a set $S$ is a point $x$ such that $Ucap (Ssetminus {x}) neq varnothing$ for every neighbourhood $U$ of $x$. It does not follow that there is a sequence in $S$ converging to $x$ - and sometimes, such a sequence does not exist. If we were dealing with metric spaces, such a sequence would necessarily exist.
$endgroup$
– Daniel Fischer
Nov 1 '14 at 15:48
1
$begingroup$
After a long time now I can understand your comment :)
$endgroup$
– Bumblebee
Dec 30 '16 at 20:05
add a comment |
$begingroup$
Suppose $emptyset$ has at least one limit point, say $x$. Then there exist a sequence $(x_n)$ in $emptyset$ which converge to $x.$ That is each $x_ninemptyset.$ This gives if empty set has a limit point it is not empty which contradicts to its definition. Hence??
answered Nov 1 '14 at 15:22
BumblebeeBumblebee
9,72412551
$endgroup$
Suppose $emptyset$ has at least one limit point, say $x$. Then there exist a sequence $(x_n)$ in $emptyset$ which converge to $x.$ That is each $x_ninemptyset.$ This gives if empty set has a limit point it is not empty which contradicts to its definition. Hence??
answered Nov 1 '14 at 15:22
BumblebeeBumblebee
9,72412551
answered Nov 1 '14 at 15:22
BumblebeeBumblebee
9,72412551
answered Nov 1 '14 at 15:22
BumblebeeBumblebee
9,72412551
answered Nov 1 '14 at 15:22
BumblebeeBumblebee
9,72412551
9,72412551
3
$begingroup$
That's not the definition of limit point. A limit point of a set $S$ is a point $x$ such that $Ucap (Ssetminus {x}) neq varnothing$ for every neighbourhood $U$ of $x$. It does not follow that there is a sequence in $S$ converging to $x$ - and sometimes, such a sequence does not exist. If we were dealing with metric spaces, such a sequence would necessarily exist.
$endgroup$
– Daniel Fischer
Nov 1 '14 at 15:48
1
$begingroup$
After a long time now I can understand your comment :)
$endgroup$
– Bumblebee
Dec 30 '16 at 20:05
add a comment |
3
$begingroup$
That's not the definition of limit point. A limit point of a set $S$ is a point $x$ such that $Ucap (Ssetminus {x}) neq varnothing$ for every neighbourhood $U$ of $x$. It does not follow that there is a sequence in $S$ converging to $x$ - and sometimes, such a sequence does not exist. If we were dealing with metric spaces, such a sequence would necessarily exist.
$endgroup$
– Daniel Fischer
Nov 1 '14 at 15:48
1
$begingroup$
After a long time now I can understand your comment :)
$endgroup$
– Bumblebee
Dec 30 '16 at 20:05
3
3
$begingroup$
That's not the definition of limit point. A limit point of a set $S$ is a point $x$ such that $Ucap (Ssetminus {x}) neq varnothing$ for every neighbourhood $U$ of $x$. It does not follow that there is a sequence in $S$ converging to $x$ - and sometimes, such a sequence does not exist. If we were dealing with metric spaces, such a sequence would necessarily exist.
$endgroup$
– Daniel Fischer
Nov 1 '14 at 15:48
$begingroup$
That's not the definition of limit point. A limit point of a set $S$ is a point $x$ such that $Ucap (Ssetminus {x}) neq varnothing$ for every neighbourhood $U$ of $x$. It does not follow that there is a sequence in $S$ converging to $x$ - and sometimes, such a sequence does not exist. If we were dealing with metric spaces, such a sequence would necessarily exist.
$endgroup$
– Daniel Fischer
Nov 1 '14 at 15:48
1
1
$begingroup$
After a long time now I can understand your comment :)
$endgroup$
– Bumblebee
Dec 30 '16 at 20:05
$begingroup$
After a long time now I can understand your comment :)
$endgroup$
– Bumblebee
Dec 30 '16 at 20:05
add a comment |
$begingroup$
Yes you would appear to be correct. And since it contains all its limit points (vacuously), it's closed. Additionally, this appears to be true in any topology.
answered Jan 1 at 5:06
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Yes you would appear to be correct. And since it contains all its limit points (vacuously), it's closed. Additionally, this appears to be true in any topology.
answered Jan 1 at 5:06
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Yes you would appear to be correct. And since it contains all its limit points (vacuously), it's closed. Additionally, this appears to be true in any topology.
answered Jan 1 at 5:06
Chris CusterChris Custer
14.2k3827
$endgroup$
Yes you would appear to be correct. And since it contains all its limit points (vacuously), it's closed. Additionally, this appears to be true in any topology.
answered Jan 1 at 5:06
Chris CusterChris Custer
14.2k3827
answered Jan 1 at 5:06
Chris CusterChris Custer
14.2k3827
answered Jan 1 at 5:06
Chris CusterChris Custer
14.2k3827
answered Jan 1 at 5:06
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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$begingroup$
Yes, it is correct. No matter what topology you put, $emptyset$ has no limit points.
$endgroup$
– Jonas Gomes
Nov 1 '14 at 15:10
4
$begingroup$
For $emptyset$, use
emptyset; for $varnothing$ usevarnothing.$endgroup$
– MJD
Nov 1 '14 at 15:14
$begingroup$
the empty set has a unique topology, and of course it has no limit points since $emptyset$ has no points at all.
$endgroup$
– matiasdata
Nov 1 '14 at 15:17