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Is any uncountable, scattered subset of [0,1] dense?
The Next CEO of Stack OverflowDoes every uncountable real set touch a rational number?Valid Proof that the Irrationals are Uncountable?countable dense subsetsIf two lower semicontinuous functions agree on a dense subset of $[0,1]$, are they equal?Infinitely uncountable set with only isolated points.If $Bsubset Asubset mathbbR, A,B$ are uncountable, then $A-B$ is Uncountable.Dense sets in complement-countable topologyHow to obtain a countable dense subset disjoint from the closure of a free sequence$(X,d)$ is separable if and only if $X$ does not contain an uncountable closed discrete subsetSequence in $(0,1)$ without accumulation points
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Just out of curiosity, I am wondering if any uncountable, scattered subset $Usubset[0,1]$ must be dense in $[0,1]$ (endowed with the Euclidean topology). It's not necessarily true if $U$ is countable, since we can take $U=frac1n : ninmathbbN$, and there are countably many open sets in $[0,1]$ which $U$ does not intersect.
But I'm having a hard time figuring out whether it's true for uncountable $U$. Any help would be appreciated. (I added the "scattered" hypothesis) since any other interval contained in $[0,1]$ would suffice as a counterexample).
real-analysis general-topology real-numbers
asked Mar 28 at 3:41
RlosRlos
1498
$endgroup$
|
show 2 more comments
$begingroup$
Just out of curiosity, I am wondering if any uncountable, scattered subset $Usubset[0,1]$ must be dense in $[0,1]$ (endowed with the Euclidean topology). It's not necessarily true if $U$ is countable, since we can take $U=frac1n : ninmathbbN$, and there are countably many open sets in $[0,1]$ which $U$ does not intersect.
But I'm having a hard time figuring out whether it's true for uncountable $U$. Any help would be appreciated. (I added the "scattered" hypothesis) since any other interval contained in $[0,1]$ would suffice as a counterexample).
real-analysis general-topology real-numbers
asked Mar 28 at 3:41
RlosRlos
1498
$endgroup$
2
$begingroup$
What does "scattered" mean?
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– Michael
Mar 28 at 3:46
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@Michael en.wikipedia.org/wiki/Glossary_of_topology#S
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– Alex Kruckman
Mar 28 at 3:47
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There are interesting counterexamples, like the Cantor set. But there are also completely trivial counterexamples: take any uncountable scattered $Usubset [0,1]$ and let $U' = (1/2)U = x/2mid xin U$. Then $U'$ is uncountable and scattered, but not dense in $[0,1]$, since it's contained in $[0,1/2]$.
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– Alex Kruckman
Mar 28 at 3:49
2
$begingroup$
How can a scattered set be dense? Assume that $Xsubset [0,1]$ is scattered. Then, by definition, $X$ contains a point $x$ that is isolated from $X$. Thus, there exists a nbh $U$ of $x$ such that $Ucap X = x$. Now, how can points in $Usetminusx$ be approximated by points in $X$?
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– amsmath
Mar 28 at 3:56
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@DavidMitra The Cantor set is not scattered, but dense in itself.
$endgroup$
– Henno Brandsma
Mar 28 at 5:14
|
show 2 more comments
$begingroup$
Just out of curiosity, I am wondering if any uncountable, scattered subset $Usubset[0,1]$ must be dense in $[0,1]$ (endowed with the Euclidean topology). It's not necessarily true if $U$ is countable, since we can take $U=frac1n : ninmathbbN$, and there are countably many open sets in $[0,1]$ which $U$ does not intersect.
But I'm having a hard time figuring out whether it's true for uncountable $U$. Any help would be appreciated. (I added the "scattered" hypothesis) since any other interval contained in $[0,1]$ would suffice as a counterexample).
real-analysis general-topology real-numbers
asked Mar 28 at 3:41
RlosRlos
1498
$endgroup$
Just out of curiosity, I am wondering if any uncountable, scattered subset $Usubset[0,1]$ must be dense in $[0,1]$ (endowed with the Euclidean topology). It's not necessarily true if $U$ is countable, since we can take $U=frac1n : ninmathbbN$, and there are countably many open sets in $[0,1]$ which $U$ does not intersect.
But I'm having a hard time figuring out whether it's true for uncountable $U$. Any help would be appreciated. (I added the "scattered" hypothesis) since any other interval contained in $[0,1]$ would suffice as a counterexample).
real-analysis general-topology real-numbers
real-analysis general-topology real-numbers
asked Mar 28 at 3:41
RlosRlos
1498
asked Mar 28 at 3:41
RlosRlos
1498
edited Mar 28 at 3:43
Rlos
asked Mar 28 at 3:41
RlosRlos
1498
asked Mar 28 at 3:41
RlosRlos
1498
asked Mar 28 at 3:41
RlosRlos
1498
1498
2
$begingroup$
What does "scattered" mean?
$endgroup$
– Michael
Mar 28 at 3:46
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@Michael en.wikipedia.org/wiki/Glossary_of_topology#S
$endgroup$
– Alex Kruckman
Mar 28 at 3:47
$begingroup$
There are interesting counterexamples, like the Cantor set. But there are also completely trivial counterexamples: take any uncountable scattered $Usubset [0,1]$ and let $U' = (1/2)U = x/2mid xin U$. Then $U'$ is uncountable and scattered, but not dense in $[0,1]$, since it's contained in $[0,1/2]$.
$endgroup$
– Alex Kruckman
Mar 28 at 3:49
2
$begingroup$
How can a scattered set be dense? Assume that $Xsubset [0,1]$ is scattered. Then, by definition, $X$ contains a point $x$ that is isolated from $X$. Thus, there exists a nbh $U$ of $x$ such that $Ucap X = x$. Now, how can points in $Usetminusx$ be approximated by points in $X$?
$endgroup$
– amsmath
Mar 28 at 3:56
$begingroup$
@DavidMitra The Cantor set is not scattered, but dense in itself.
$endgroup$
– Henno Brandsma
Mar 28 at 5:14
|
show 2 more comments
2
$begingroup$
What does "scattered" mean?
$endgroup$
– Michael
Mar 28 at 3:46
$begingroup$
@Michael en.wikipedia.org/wiki/Glossary_of_topology#S
$endgroup$
– Alex Kruckman
Mar 28 at 3:47
$begingroup$
There are interesting counterexamples, like the Cantor set. But there are also completely trivial counterexamples: take any uncountable scattered $Usubset [0,1]$ and let $U' = (1/2)U = x/2mid xin U$. Then $U'$ is uncountable and scattered, but not dense in $[0,1]$, since it's contained in $[0,1/2]$.
$endgroup$
– Alex Kruckman
Mar 28 at 3:49
2
$begingroup$
How can a scattered set be dense? Assume that $Xsubset [0,1]$ is scattered. Then, by definition, $X$ contains a point $x$ that is isolated from $X$. Thus, there exists a nbh $U$ of $x$ such that $Ucap X = x$. Now, how can points in $Usetminusx$ be approximated by points in $X$?
$endgroup$
– amsmath
Mar 28 at 3:56
$begingroup$
@DavidMitra The Cantor set is not scattered, but dense in itself.
$endgroup$
– Henno Brandsma
Mar 28 at 5:14
2
2
$begingroup$
What does "scattered" mean?
$endgroup$
– Michael
Mar 28 at 3:46
$begingroup$
What does "scattered" mean?
$endgroup$
– Michael
Mar 28 at 3:46
$begingroup$
@Michael en.wikipedia.org/wiki/Glossary_of_topology#S
$endgroup$
– Alex Kruckman
Mar 28 at 3:47
$begingroup$
@Michael en.wikipedia.org/wiki/Glossary_of_topology#S
$endgroup$
– Alex Kruckman
Mar 28 at 3:47
$begingroup$
There are interesting counterexamples, like the Cantor set. But there are also completely trivial counterexamples: take any uncountable scattered $Usubset [0,1]$ and let $U' = (1/2)U = x/2mid xin U$. Then $U'$ is uncountable and scattered, but not dense in $[0,1]$, since it's contained in $[0,1/2]$.
$endgroup$
– Alex Kruckman
Mar 28 at 3:49
$begingroup$
There are interesting counterexamples, like the Cantor set. But there are also completely trivial counterexamples: take any uncountable scattered $Usubset [0,1]$ and let $U' = (1/2)U = x/2mid xin U$. Then $U'$ is uncountable and scattered, but not dense in $[0,1]$, since it's contained in $[0,1/2]$.
$endgroup$
– Alex Kruckman
Mar 28 at 3:49
2
2
$begingroup$
How can a scattered set be dense? Assume that $Xsubset [0,1]$ is scattered. Then, by definition, $X$ contains a point $x$ that is isolated from $X$. Thus, there exists a nbh $U$ of $x$ such that $Ucap X = x$. Now, how can points in $Usetminusx$ be approximated by points in $X$?
$endgroup$
– amsmath
Mar 28 at 3:56
$begingroup$
How can a scattered set be dense? Assume that $Xsubset [0,1]$ is scattered. Then, by definition, $X$ contains a point $x$ that is isolated from $X$. Thus, there exists a nbh $U$ of $x$ such that $Ucap X = x$. Now, how can points in $Usetminusx$ be approximated by points in $X$?
$endgroup$
– amsmath
Mar 28 at 3:56
$begingroup$
@DavidMitra The Cantor set is not scattered, but dense in itself.
$endgroup$
– Henno Brandsma
Mar 28 at 5:14
$begingroup$
@DavidMitra The Cantor set is not scattered, but dense in itself.
$endgroup$
– Henno Brandsma
Mar 28 at 5:14
|
show 2 more comments
3 Answers
3
active
oldest
votes
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There are no uncountable scattered subsets of $[0,1]$. This follows from the theory of Cantor-Bendixson rank, for instance. Given any scattered $Asubset[0,1]$, there must be some ordinal $alpha$ such that the $alpha$th Cantor-Bendixson derivative $A^alpha$ of $A$ is empty. The least such $alpha$ must be countable, since the Cantor-Bendixson derivatives are a descending chain of closed subsets of $A$ and $A$ is second-countable. Since every subset of $A$ can have only countably many isolated points (again by second-countability), this means $A$ is countable.
answered Mar 28 at 3:59
Eric WofseyEric Wofsey
191k14216349
$endgroup$
1
$begingroup$
So logically speaking the question is true: all uncountable scattered subsets of $[0,1]$ are dense.
$endgroup$
– Henno Brandsma
Mar 28 at 5:13
$begingroup$
Well now I see why I couldn't find an example. Where can I learn more about the Cantor-Bendixson derivatives? Never heard of such concept before.
$endgroup$
– Rlos
Mar 28 at 5:46
$begingroup$
You can look at en.wikipedia.org/wiki/Derived_set_(mathematics) for a start. The standard basic application of them (which is closely related to how I use them here) is to prove the Cantor-Bendixson theorem which I'm sure you can find written up many places on the internet.
$endgroup$
– Eric Wofsey
Mar 28 at 6:04
add a comment |
$begingroup$
The argument in a comment is worth restating: if $X$ is a set without isolated points (a crowded space) that is $T_1$ then no scattered subset (countable or not) can be dense in $X$:
Let $C$ be scattered. So it has an isolated point $p in C$, so there is an open set $U$ of $X$ such that $U cap C =p$. But then $Usetminus p$ is non-empty (as $X$ is crowded) and open (as $X$ is $T_1$, $p$ is closed) and misses $C$. So $C$ is not dense.
This certainly applies to $X=[0,1]$.
answered Mar 28 at 17:26
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
add a comment |
$begingroup$
An uncountable set does not have to be dense. A simple example would be to take the half interval $[0,frac12]$ which is uncountable but not dense in $[0,1]$. But we can do even better, because we can find a set that is uncountable and not dense in any open interval. The Cantor set is uncountable and dense nowhere. https://en.wikipedia.org/wiki/Cantor_set
answered Mar 28 at 3:47
A. KriegmanA. Kriegman
133
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The Cantor set is not scattered, though.
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– Eric Wofsey
Mar 28 at 3:59
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@EricWofsey Could you please answer my (maybe stupid) question in the comments above?
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– amsmath
Mar 28 at 4:07
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@amsmath: Your argument is correct.
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– Eric Wofsey
Mar 28 at 4:08
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@EricWofsey Thank you very much!
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– amsmath
Mar 28 at 4:10
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@EricWofsey If your defining "scattered" as each element of the set is in an open interval containing no other points in the set, then you are correct that there are no uncountable scattered sets. However, proving this is not very hard and the "Cantor-Bendixson derivative" that you used is overkill. If you define scattered as not dense in any open set, then the cantor set is scattered.
$endgroup$
– A. Kriegman
Mar 28 at 14:32
|
show 1 more comment
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3 Answers
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3 Answers
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$begingroup$
There are no uncountable scattered subsets of $[0,1]$. This follows from the theory of Cantor-Bendixson rank, for instance. Given any scattered $Asubset[0,1]$, there must be some ordinal $alpha$ such that the $alpha$th Cantor-Bendixson derivative $A^alpha$ of $A$ is empty. The least such $alpha$ must be countable, since the Cantor-Bendixson derivatives are a descending chain of closed subsets of $A$ and $A$ is second-countable. Since every subset of $A$ can have only countably many isolated points (again by second-countability), this means $A$ is countable.
answered Mar 28 at 3:59
Eric WofseyEric Wofsey
191k14216349
$endgroup$
1
$begingroup$
So logically speaking the question is true: all uncountable scattered subsets of $[0,1]$ are dense.
$endgroup$
– Henno Brandsma
Mar 28 at 5:13
$begingroup$
Well now I see why I couldn't find an example. Where can I learn more about the Cantor-Bendixson derivatives? Never heard of such concept before.
$endgroup$
– Rlos
Mar 28 at 5:46
$begingroup$
You can look at en.wikipedia.org/wiki/Derived_set_(mathematics) for a start. The standard basic application of them (which is closely related to how I use them here) is to prove the Cantor-Bendixson theorem which I'm sure you can find written up many places on the internet.
$endgroup$
– Eric Wofsey
Mar 28 at 6:04
add a comment |
$begingroup$
There are no uncountable scattered subsets of $[0,1]$. This follows from the theory of Cantor-Bendixson rank, for instance. Given any scattered $Asubset[0,1]$, there must be some ordinal $alpha$ such that the $alpha$th Cantor-Bendixson derivative $A^alpha$ of $A$ is empty. The least such $alpha$ must be countable, since the Cantor-Bendixson derivatives are a descending chain of closed subsets of $A$ and $A$ is second-countable. Since every subset of $A$ can have only countably many isolated points (again by second-countability), this means $A$ is countable.
answered Mar 28 at 3:59
Eric WofseyEric Wofsey
191k14216349
$endgroup$
1
$begingroup$
So logically speaking the question is true: all uncountable scattered subsets of $[0,1]$ are dense.
$endgroup$
– Henno Brandsma
Mar 28 at 5:13
$begingroup$
Well now I see why I couldn't find an example. Where can I learn more about the Cantor-Bendixson derivatives? Never heard of such concept before.
$endgroup$
– Rlos
Mar 28 at 5:46
$begingroup$
You can look at en.wikipedia.org/wiki/Derived_set_(mathematics) for a start. The standard basic application of them (which is closely related to how I use them here) is to prove the Cantor-Bendixson theorem which I'm sure you can find written up many places on the internet.
$endgroup$
– Eric Wofsey
Mar 28 at 6:04
add a comment |
$begingroup$
There are no uncountable scattered subsets of $[0,1]$. This follows from the theory of Cantor-Bendixson rank, for instance. Given any scattered $Asubset[0,1]$, there must be some ordinal $alpha$ such that the $alpha$th Cantor-Bendixson derivative $A^alpha$ of $A$ is empty. The least such $alpha$ must be countable, since the Cantor-Bendixson derivatives are a descending chain of closed subsets of $A$ and $A$ is second-countable. Since every subset of $A$ can have only countably many isolated points (again by second-countability), this means $A$ is countable.
answered Mar 28 at 3:59
Eric WofseyEric Wofsey
191k14216349
$endgroup$
There are no uncountable scattered subsets of $[0,1]$. This follows from the theory of Cantor-Bendixson rank, for instance. Given any scattered $Asubset[0,1]$, there must be some ordinal $alpha$ such that the $alpha$th Cantor-Bendixson derivative $A^alpha$ of $A$ is empty. The least such $alpha$ must be countable, since the Cantor-Bendixson derivatives are a descending chain of closed subsets of $A$ and $A$ is second-countable. Since every subset of $A$ can have only countably many isolated points (again by second-countability), this means $A$ is countable.
answered Mar 28 at 3:59
Eric WofseyEric Wofsey
191k14216349
answered Mar 28 at 3:59
Eric WofseyEric Wofsey
191k14216349
answered Mar 28 at 3:59
Eric WofseyEric Wofsey
191k14216349
answered Mar 28 at 3:59
Eric WofseyEric Wofsey
191k14216349
191k14216349
1
$begingroup$
So logically speaking the question is true: all uncountable scattered subsets of $[0,1]$ are dense.
$endgroup$
– Henno Brandsma
Mar 28 at 5:13
$begingroup$
Well now I see why I couldn't find an example. Where can I learn more about the Cantor-Bendixson derivatives? Never heard of such concept before.
$endgroup$
– Rlos
Mar 28 at 5:46
$begingroup$
You can look at en.wikipedia.org/wiki/Derived_set_(mathematics) for a start. The standard basic application of them (which is closely related to how I use them here) is to prove the Cantor-Bendixson theorem which I'm sure you can find written up many places on the internet.
$endgroup$
– Eric Wofsey
Mar 28 at 6:04
add a comment |
1
$begingroup$
So logically speaking the question is true: all uncountable scattered subsets of $[0,1]$ are dense.
$endgroup$
– Henno Brandsma
Mar 28 at 5:13
$begingroup$
Well now I see why I couldn't find an example. Where can I learn more about the Cantor-Bendixson derivatives? Never heard of such concept before.
$endgroup$
– Rlos
Mar 28 at 5:46
$begingroup$
You can look at en.wikipedia.org/wiki/Derived_set_(mathematics) for a start. The standard basic application of them (which is closely related to how I use them here) is to prove the Cantor-Bendixson theorem which I'm sure you can find written up many places on the internet.
$endgroup$
– Eric Wofsey
Mar 28 at 6:04
1
1
$begingroup$
So logically speaking the question is true: all uncountable scattered subsets of $[0,1]$ are dense.
$endgroup$
– Henno Brandsma
Mar 28 at 5:13
$begingroup$
So logically speaking the question is true: all uncountable scattered subsets of $[0,1]$ are dense.
$endgroup$
– Henno Brandsma
Mar 28 at 5:13
$begingroup$
Well now I see why I couldn't find an example. Where can I learn more about the Cantor-Bendixson derivatives? Never heard of such concept before.
$endgroup$
– Rlos
Mar 28 at 5:46
$begingroup$
Well now I see why I couldn't find an example. Where can I learn more about the Cantor-Bendixson derivatives? Never heard of such concept before.
$endgroup$
– Rlos
Mar 28 at 5:46
$begingroup$
You can look at en.wikipedia.org/wiki/Derived_set_(mathematics) for a start. The standard basic application of them (which is closely related to how I use them here) is to prove the Cantor-Bendixson theorem which I'm sure you can find written up many places on the internet.
$endgroup$
– Eric Wofsey
Mar 28 at 6:04
$begingroup$
You can look at en.wikipedia.org/wiki/Derived_set_(mathematics) for a start. The standard basic application of them (which is closely related to how I use them here) is to prove the Cantor-Bendixson theorem which I'm sure you can find written up many places on the internet.
$endgroup$
– Eric Wofsey
Mar 28 at 6:04
add a comment |
$begingroup$
The argument in a comment is worth restating: if $X$ is a set without isolated points (a crowded space) that is $T_1$ then no scattered subset (countable or not) can be dense in $X$:
Let $C$ be scattered. So it has an isolated point $p in C$, so there is an open set $U$ of $X$ such that $U cap C =p$. But then $Usetminus p$ is non-empty (as $X$ is crowded) and open (as $X$ is $T_1$, $p$ is closed) and misses $C$. So $C$ is not dense.
This certainly applies to $X=[0,1]$.
answered Mar 28 at 17:26
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
add a comment |
$begingroup$
The argument in a comment is worth restating: if $X$ is a set without isolated points (a crowded space) that is $T_1$ then no scattered subset (countable or not) can be dense in $X$:
Let $C$ be scattered. So it has an isolated point $p in C$, so there is an open set $U$ of $X$ such that $U cap C =p$. But then $Usetminus p$ is non-empty (as $X$ is crowded) and open (as $X$ is $T_1$, $p$ is closed) and misses $C$. So $C$ is not dense.
This certainly applies to $X=[0,1]$.
answered Mar 28 at 17:26
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
add a comment |
$begingroup$
The argument in a comment is worth restating: if $X$ is a set without isolated points (a crowded space) that is $T_1$ then no scattered subset (countable or not) can be dense in $X$:
Let $C$ be scattered. So it has an isolated point $p in C$, so there is an open set $U$ of $X$ such that $U cap C =p$. But then $Usetminus p$ is non-empty (as $X$ is crowded) and open (as $X$ is $T_1$, $p$ is closed) and misses $C$. So $C$ is not dense.
This certainly applies to $X=[0,1]$.
answered Mar 28 at 17:26
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
The argument in a comment is worth restating: if $X$ is a set without isolated points (a crowded space) that is $T_1$ then no scattered subset (countable or not) can be dense in $X$:
Let $C$ be scattered. So it has an isolated point $p in C$, so there is an open set $U$ of $X$ such that $U cap C =p$. But then $Usetminus p$ is non-empty (as $X$ is crowded) and open (as $X$ is $T_1$, $p$ is closed) and misses $C$. So $C$ is not dense.
This certainly applies to $X=[0,1]$.
answered Mar 28 at 17:26
Henno BrandsmaHenno Brandsma
115k348124
answered Mar 28 at 17:26
Henno BrandsmaHenno Brandsma
115k348124
answered Mar 28 at 17:26
Henno BrandsmaHenno Brandsma
115k348124
answered Mar 28 at 17:26
Henno BrandsmaHenno Brandsma
115k348124
115k348124
add a comment |
add a comment |
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An uncountable set does not have to be dense. A simple example would be to take the half interval $[0,frac12]$ which is uncountable but not dense in $[0,1]$. But we can do even better, because we can find a set that is uncountable and not dense in any open interval. The Cantor set is uncountable and dense nowhere. https://en.wikipedia.org/wiki/Cantor_set
answered Mar 28 at 3:47
A. KriegmanA. Kriegman
133
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The Cantor set is not scattered, though.
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– Eric Wofsey
Mar 28 at 3:59
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@EricWofsey Could you please answer my (maybe stupid) question in the comments above?
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– amsmath
Mar 28 at 4:07
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@amsmath: Your argument is correct.
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– Eric Wofsey
Mar 28 at 4:08
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@EricWofsey Thank you very much!
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– amsmath
Mar 28 at 4:10
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@EricWofsey If your defining "scattered" as each element of the set is in an open interval containing no other points in the set, then you are correct that there are no uncountable scattered sets. However, proving this is not very hard and the "Cantor-Bendixson derivative" that you used is overkill. If you define scattered as not dense in any open set, then the cantor set is scattered.
$endgroup$
– A. Kriegman
Mar 28 at 14:32
|
show 1 more comment
$begingroup$
An uncountable set does not have to be dense. A simple example would be to take the half interval $[0,frac12]$ which is uncountable but not dense in $[0,1]$. But we can do even better, because we can find a set that is uncountable and not dense in any open interval. The Cantor set is uncountable and dense nowhere. https://en.wikipedia.org/wiki/Cantor_set
answered Mar 28 at 3:47
A. KriegmanA. Kriegman
133
$endgroup$
$begingroup$
The Cantor set is not scattered, though.
$endgroup$
– Eric Wofsey
Mar 28 at 3:59
$begingroup$
@EricWofsey Could you please answer my (maybe stupid) question in the comments above?
$endgroup$
– amsmath
Mar 28 at 4:07
$begingroup$
@amsmath: Your argument is correct.
$endgroup$
– Eric Wofsey
Mar 28 at 4:08
$begingroup$
@EricWofsey Thank you very much!
$endgroup$
– amsmath
Mar 28 at 4:10
$begingroup$
@EricWofsey If your defining "scattered" as each element of the set is in an open interval containing no other points in the set, then you are correct that there are no uncountable scattered sets. However, proving this is not very hard and the "Cantor-Bendixson derivative" that you used is overkill. If you define scattered as not dense in any open set, then the cantor set is scattered.
$endgroup$
– A. Kriegman
Mar 28 at 14:32
|
show 1 more comment
$begingroup$
An uncountable set does not have to be dense. A simple example would be to take the half interval $[0,frac12]$ which is uncountable but not dense in $[0,1]$. But we can do even better, because we can find a set that is uncountable and not dense in any open interval. The Cantor set is uncountable and dense nowhere. https://en.wikipedia.org/wiki/Cantor_set
answered Mar 28 at 3:47
A. KriegmanA. Kriegman
133
$endgroup$
An uncountable set does not have to be dense. A simple example would be to take the half interval $[0,frac12]$ which is uncountable but not dense in $[0,1]$. But we can do even better, because we can find a set that is uncountable and not dense in any open interval. The Cantor set is uncountable and dense nowhere. https://en.wikipedia.org/wiki/Cantor_set
answered Mar 28 at 3:47
A. KriegmanA. Kriegman
133
answered Mar 28 at 3:47
A. KriegmanA. Kriegman
133
answered Mar 28 at 3:47
A. KriegmanA. Kriegman
133
answered Mar 28 at 3:47
A. KriegmanA. Kriegman
133
133
$begingroup$
The Cantor set is not scattered, though.
$endgroup$
– Eric Wofsey
Mar 28 at 3:59
$begingroup$
@EricWofsey Could you please answer my (maybe stupid) question in the comments above?
$endgroup$
– amsmath
Mar 28 at 4:07
$begingroup$
@amsmath: Your argument is correct.
$endgroup$
– Eric Wofsey
Mar 28 at 4:08
$begingroup$
@EricWofsey Thank you very much!
$endgroup$
– amsmath
Mar 28 at 4:10
$begingroup$
@EricWofsey If your defining "scattered" as each element of the set is in an open interval containing no other points in the set, then you are correct that there are no uncountable scattered sets. However, proving this is not very hard and the "Cantor-Bendixson derivative" that you used is overkill. If you define scattered as not dense in any open set, then the cantor set is scattered.
$endgroup$
– A. Kriegman
Mar 28 at 14:32
|
show 1 more comment
$begingroup$
The Cantor set is not scattered, though.
$endgroup$
– Eric Wofsey
Mar 28 at 3:59
$begingroup$
@EricWofsey Could you please answer my (maybe stupid) question in the comments above?
$endgroup$
– amsmath
Mar 28 at 4:07
$begingroup$
@amsmath: Your argument is correct.
$endgroup$
– Eric Wofsey
Mar 28 at 4:08
$begingroup$
@EricWofsey Thank you very much!
$endgroup$
– amsmath
Mar 28 at 4:10
$begingroup$
@EricWofsey If your defining "scattered" as each element of the set is in an open interval containing no other points in the set, then you are correct that there are no uncountable scattered sets. However, proving this is not very hard and the "Cantor-Bendixson derivative" that you used is overkill. If you define scattered as not dense in any open set, then the cantor set is scattered.
$endgroup$
– A. Kriegman
Mar 28 at 14:32
$begingroup$
The Cantor set is not scattered, though.
$endgroup$
– Eric Wofsey
Mar 28 at 3:59
$begingroup$
The Cantor set is not scattered, though.
$endgroup$
– Eric Wofsey
Mar 28 at 3:59
$begingroup$
@EricWofsey Could you please answer my (maybe stupid) question in the comments above?
$endgroup$
– amsmath
Mar 28 at 4:07
$begingroup$
@EricWofsey Could you please answer my (maybe stupid) question in the comments above?
$endgroup$
– amsmath
Mar 28 at 4:07
$begingroup$
@amsmath: Your argument is correct.
$endgroup$
– Eric Wofsey
Mar 28 at 4:08
$begingroup$
@amsmath: Your argument is correct.
$endgroup$
– Eric Wofsey
Mar 28 at 4:08
$begingroup$
@EricWofsey Thank you very much!
$endgroup$
– amsmath
Mar 28 at 4:10
$begingroup$
@EricWofsey Thank you very much!
$endgroup$
– amsmath
Mar 28 at 4:10
$begingroup$
@EricWofsey If your defining "scattered" as each element of the set is in an open interval containing no other points in the set, then you are correct that there are no uncountable scattered sets. However, proving this is not very hard and the "Cantor-Bendixson derivative" that you used is overkill. If you define scattered as not dense in any open set, then the cantor set is scattered.
$endgroup$
– A. Kriegman
Mar 28 at 14:32
$begingroup$
@EricWofsey If your defining "scattered" as each element of the set is in an open interval containing no other points in the set, then you are correct that there are no uncountable scattered sets. However, proving this is not very hard and the "Cantor-Bendixson derivative" that you used is overkill. If you define scattered as not dense in any open set, then the cantor set is scattered.
$endgroup$
– A. Kriegman
Mar 28 at 14:32
|
show 1 more comment
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2
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What does "scattered" mean?
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– Michael
Mar 28 at 3:46
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@Michael en.wikipedia.org/wiki/Glossary_of_topology#S
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– Alex Kruckman
Mar 28 at 3:47
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There are interesting counterexamples, like the Cantor set. But there are also completely trivial counterexamples: take any uncountable scattered $Usubset [0,1]$ and let $U' = (1/2)U = x/2mid xin U$. Then $U'$ is uncountable and scattered, but not dense in $[0,1]$, since it's contained in $[0,1/2]$.
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– Alex Kruckman
Mar 28 at 3:49
2
$begingroup$
How can a scattered set be dense? Assume that $Xsubset [0,1]$ is scattered. Then, by definition, $X$ contains a point $x$ that is isolated from $X$. Thus, there exists a nbh $U$ of $x$ such that $Ucap X = x$. Now, how can points in $Usetminusx$ be approximated by points in $X$?
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– amsmath
Mar 28 at 3:56
$begingroup$
@DavidMitra The Cantor set is not scattered, but dense in itself.
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– Henno Brandsma
Mar 28 at 5:14