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Why do we require that a perfect set is closed?
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Definition of Perfect Set says
$E$ is perfect if $E$ is closed and if every point of $E$ is limit point of $E$.
Is it possible that every point of $E$ is limit point of $E$ but $E$ is not closed ?
I'am unable to find an example. Because Closed set is defined as if every limit point of $E$ is a point of $E$.
If not possible. Then what is need of including in Definition of Perfect set ? ( that If $E$ is closed )
Help would be appreciated !
real-analysis general-topology metric-spaces
edited Aug 14 at 15:27
Asaf Karagila♦
294k31409736
asked Aug 14 at 15:23
MathsforSS
427
add a comment |Â
up vote
4
down vote
favorite
Definition of Perfect Set says
$E$ is perfect if $E$ is closed and if every point of $E$ is limit point of $E$.
Is it possible that every point of $E$ is limit point of $E$ but $E$ is not closed ?
I'am unable to find an example. Because Closed set is defined as if every limit point of $E$ is a point of $E$.
If not possible. Then what is need of including in Definition of Perfect set ? ( that If $E$ is closed )
Help would be appreciated !
real-analysis general-topology metric-spaces
edited Aug 14 at 15:27
Asaf Karagila♦
294k31409736
asked Aug 14 at 15:23
MathsforSS
427
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Definition of Perfect Set says
$E$ is perfect if $E$ is closed and if every point of $E$ is limit point of $E$.
Is it possible that every point of $E$ is limit point of $E$ but $E$ is not closed ?
I'am unable to find an example. Because Closed set is defined as if every limit point of $E$ is a point of $E$.
If not possible. Then what is need of including in Definition of Perfect set ? ( that If $E$ is closed )
Help would be appreciated !
real-analysis general-topology metric-spaces
edited Aug 14 at 15:27
Asaf Karagila♦
294k31409736
asked Aug 14 at 15:23
MathsforSS
427
Definition of Perfect Set says
$E$ is perfect if $E$ is closed and if every point of $E$ is limit point of $E$.
Is it possible that every point of $E$ is limit point of $E$ but $E$ is not closed ?
I'am unable to find an example. Because Closed set is defined as if every limit point of $E$ is a point of $E$.
If not possible. Then what is need of including in Definition of Perfect set ? ( that If $E$ is closed )
Help would be appreciated !
real-analysis general-topology metric-spaces
edited Aug 14 at 15:27
Asaf Karagila♦
294k31409736
asked Aug 14 at 15:23
MathsforSS
427
edited Aug 14 at 15:27
Asaf Karagila♦
294k31409736
edited Aug 14 at 15:27
Asaf Karagila♦
294k31409736
edited Aug 14 at 15:27
Asaf Karagila♦
294k31409736
294k31409736
asked Aug 14 at 15:23
MathsforSS
427
asked Aug 14 at 15:23
MathsforSS
427
asked Aug 14 at 15:23
MathsforSS
427
427
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
11
down vote
accepted
Yes, of course. Note that $Bbb Q$ is not closed in $Bbb R$, but every rational number is the limit of a sequence of rational numbers, all different than your proposed limit.
In other words, $Bbb Q$ has no isolated points, but it's not closed as a subset of $Bbb R$.
answered Aug 14 at 15:25
Asaf Karagila♦
294k31409736
add a comment |Â
up vote
5
down vote
take $(a,b)subsetmathbbR$. If $xin(a,b)$ then $exists a_nin(a,b):a_nto x$
answered Aug 14 at 15:27
giannispapav
1,284222
3
$xin(a,b)$ then $exists a_nin(a,b):a_nto x$ isn't enough for $x$ to be a limit point.
– JiK
Aug 14 at 17:15
add a comment |Â
up vote
2
down vote
Others have already given nice examples.
About the notion: $E$ is closed if $E$ contains its limit points, or equivalently $E' subseteq E$.
A set is perfect when moreover $E subseteq E'$ (all points are limit points, or there are no isolated points) so iff $E' = E$. So it's a fixed point of the derived set operator $E to E'$, so to speak. This indeed implies closedness by the above criterion for closedness.
This notion is natural to consider when we start with some closed subset $E$ and form the succesive derived sets $E, E', E'', ldots$ or more formally define $E^(0) =E$, $E^(alpha+1) = (E^(alpha))'$ for all succesor ordinals $alpha+1$, and $E^(alpha) = bigcap_beta < alpha E^(beta)$ when $alpha$ is a limit ordinal. One can show that for some ordinal $gamma$, $E^(gamma) = E^(gamma+1)=: K$, and this $K$ is called the perfect kernel of $E$. Such considerations led Cantor to define ordinals and to consider perfect subsets of the reals.
answered Aug 14 at 18:28
Henno Brandsma
92.5k342100
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
Yes, of course. Note that $Bbb Q$ is not closed in $Bbb R$, but every rational number is the limit of a sequence of rational numbers, all different than your proposed limit.
In other words, $Bbb Q$ has no isolated points, but it's not closed as a subset of $Bbb R$.
answered Aug 14 at 15:25
Asaf Karagila♦
294k31409736
add a comment |Â
up vote
11
down vote
accepted
Yes, of course. Note that $Bbb Q$ is not closed in $Bbb R$, but every rational number is the limit of a sequence of rational numbers, all different than your proposed limit.
In other words, $Bbb Q$ has no isolated points, but it's not closed as a subset of $Bbb R$.
answered Aug 14 at 15:25
Asaf Karagila♦
294k31409736
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
Yes, of course. Note that $Bbb Q$ is not closed in $Bbb R$, but every rational number is the limit of a sequence of rational numbers, all different than your proposed limit.
In other words, $Bbb Q$ has no isolated points, but it's not closed as a subset of $Bbb R$.
answered Aug 14 at 15:25
Asaf Karagila♦
294k31409736
Yes, of course. Note that $Bbb Q$ is not closed in $Bbb R$, but every rational number is the limit of a sequence of rational numbers, all different than your proposed limit.
In other words, $Bbb Q$ has no isolated points, but it's not closed as a subset of $Bbb R$.
answered Aug 14 at 15:25
Asaf Karagila♦
294k31409736
answered Aug 14 at 15:25
Asaf Karagila♦
294k31409736
answered Aug 14 at 15:25
Asaf Karagila♦
294k31409736
answered Aug 14 at 15:25
Asaf Karagila♦
294k31409736
294k31409736
add a comment |Â
add a comment |Â
up vote
5
down vote
take $(a,b)subsetmathbbR$. If $xin(a,b)$ then $exists a_nin(a,b):a_nto x$
answered Aug 14 at 15:27
giannispapav
1,284222
3
$xin(a,b)$ then $exists a_nin(a,b):a_nto x$ isn't enough for $x$ to be a limit point.
– JiK
Aug 14 at 17:15
add a comment |Â
up vote
5
down vote
take $(a,b)subsetmathbbR$. If $xin(a,b)$ then $exists a_nin(a,b):a_nto x$
answered Aug 14 at 15:27
giannispapav
1,284222
3
$xin(a,b)$ then $exists a_nin(a,b):a_nto x$ isn't enough for $x$ to be a limit point.
– JiK
Aug 14 at 17:15
add a comment |Â
up vote
5
down vote
up vote
5
down vote
take $(a,b)subsetmathbbR$. If $xin(a,b)$ then $exists a_nin(a,b):a_nto x$
answered Aug 14 at 15:27
giannispapav
1,284222
take $(a,b)subsetmathbbR$. If $xin(a,b)$ then $exists a_nin(a,b):a_nto x$
answered Aug 14 at 15:27
giannispapav
1,284222
answered Aug 14 at 15:27
giannispapav
1,284222
answered Aug 14 at 15:27
giannispapav
1,284222
answered Aug 14 at 15:27
giannispapav
1,284222
1,284222
3
$xin(a,b)$ then $exists a_nin(a,b):a_nto x$ isn't enough for $x$ to be a limit point.
– JiK
Aug 14 at 17:15
add a comment |Â
3
$xin(a,b)$ then $exists a_nin(a,b):a_nto x$ isn't enough for $x$ to be a limit point.
– JiK
Aug 14 at 17:15
3
3
$xin(a,b)$ then $exists a_nin(a,b):a_nto x$ isn't enough for $x$ to be a limit point.
– JiK
Aug 14 at 17:15
$xin(a,b)$ then $exists a_nin(a,b):a_nto x$ isn't enough for $x$ to be a limit point.
– JiK
Aug 14 at 17:15
add a comment |Â
up vote
2
down vote
Others have already given nice examples.
About the notion: $E$ is closed if $E$ contains its limit points, or equivalently $E' subseteq E$.
A set is perfect when moreover $E subseteq E'$ (all points are limit points, or there are no isolated points) so iff $E' = E$. So it's a fixed point of the derived set operator $E to E'$, so to speak. This indeed implies closedness by the above criterion for closedness.
This notion is natural to consider when we start with some closed subset $E$ and form the succesive derived sets $E, E', E'', ldots$ or more formally define $E^(0) =E$, $E^(alpha+1) = (E^(alpha))'$ for all succesor ordinals $alpha+1$, and $E^(alpha) = bigcap_beta < alpha E^(beta)$ when $alpha$ is a limit ordinal. One can show that for some ordinal $gamma$, $E^(gamma) = E^(gamma+1)=: K$, and this $K$ is called the perfect kernel of $E$. Such considerations led Cantor to define ordinals and to consider perfect subsets of the reals.
answered Aug 14 at 18:28
Henno Brandsma
92.5k342100
add a comment |Â
up vote
2
down vote
Others have already given nice examples.
About the notion: $E$ is closed if $E$ contains its limit points, or equivalently $E' subseteq E$.
A set is perfect when moreover $E subseteq E'$ (all points are limit points, or there are no isolated points) so iff $E' = E$. So it's a fixed point of the derived set operator $E to E'$, so to speak. This indeed implies closedness by the above criterion for closedness.
This notion is natural to consider when we start with some closed subset $E$ and form the succesive derived sets $E, E', E'', ldots$ or more formally define $E^(0) =E$, $E^(alpha+1) = (E^(alpha))'$ for all succesor ordinals $alpha+1$, and $E^(alpha) = bigcap_beta < alpha E^(beta)$ when $alpha$ is a limit ordinal. One can show that for some ordinal $gamma$, $E^(gamma) = E^(gamma+1)=: K$, and this $K$ is called the perfect kernel of $E$. Such considerations led Cantor to define ordinals and to consider perfect subsets of the reals.
answered Aug 14 at 18:28
Henno Brandsma
92.5k342100
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Others have already given nice examples.
About the notion: $E$ is closed if $E$ contains its limit points, or equivalently $E' subseteq E$.
A set is perfect when moreover $E subseteq E'$ (all points are limit points, or there are no isolated points) so iff $E' = E$. So it's a fixed point of the derived set operator $E to E'$, so to speak. This indeed implies closedness by the above criterion for closedness.
This notion is natural to consider when we start with some closed subset $E$ and form the succesive derived sets $E, E', E'', ldots$ or more formally define $E^(0) =E$, $E^(alpha+1) = (E^(alpha))'$ for all succesor ordinals $alpha+1$, and $E^(alpha) = bigcap_beta < alpha E^(beta)$ when $alpha$ is a limit ordinal. One can show that for some ordinal $gamma$, $E^(gamma) = E^(gamma+1)=: K$, and this $K$ is called the perfect kernel of $E$. Such considerations led Cantor to define ordinals and to consider perfect subsets of the reals.
answered Aug 14 at 18:28
Henno Brandsma
92.5k342100
Others have already given nice examples.
About the notion: $E$ is closed if $E$ contains its limit points, or equivalently $E' subseteq E$.
A set is perfect when moreover $E subseteq E'$ (all points are limit points, or there are no isolated points) so iff $E' = E$. So it's a fixed point of the derived set operator $E to E'$, so to speak. This indeed implies closedness by the above criterion for closedness.
This notion is natural to consider when we start with some closed subset $E$ and form the succesive derived sets $E, E', E'', ldots$ or more formally define $E^(0) =E$, $E^(alpha+1) = (E^(alpha))'$ for all succesor ordinals $alpha+1$, and $E^(alpha) = bigcap_beta < alpha E^(beta)$ when $alpha$ is a limit ordinal. One can show that for some ordinal $gamma$, $E^(gamma) = E^(gamma+1)=: K$, and this $K$ is called the perfect kernel of $E$. Such considerations led Cantor to define ordinals and to consider perfect subsets of the reals.
answered Aug 14 at 18:28
Henno Brandsma
92.5k342100
answered Aug 14 at 18:28
Henno Brandsma
92.5k342100
answered Aug 14 at 18:28
Henno Brandsma
92.5k342100
answered Aug 14 at 18:28
Henno Brandsma
92.5k342100
92.5k342100
add a comment |Â
add a comment |Â
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