3 dense uncountable pairwise disjoint subsets of real line
up vote
3
down vote
favorite
Can we find three dense uncountable pairwise disjoint subsets of $mathbb{R}$? If so, what are these three sets?
I feel like it's not possible.
I was trying to counter it using Baire category and trying to construct such sets using the idea of construction of Cantor set (setting elements in trinary system). But I failed in both ways. I'm not even sure if answer would be affirmative or not.
real-analysis general-topology real-numbers
edited Dec 4 at 16:36
amWhy
191k28224439
asked Dec 4 at 16:29
ChakSayantan
1206
add a comment |
up vote
3
down vote
favorite
Can we find three dense uncountable pairwise disjoint subsets of $mathbb{R}$? If so, what are these three sets?
I feel like it's not possible.
I was trying to counter it using Baire category and trying to construct such sets using the idea of construction of Cantor set (setting elements in trinary system). But I failed in both ways. I'm not even sure if answer would be affirmative or not.
real-analysis general-topology real-numbers
edited Dec 4 at 16:36
amWhy
191k28224439
asked Dec 4 at 16:29
ChakSayantan
1206
1
Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
– Not Mike
Dec 4 at 16:40
@NotMike can you explain a bit more.. some precise example would be appreciated
– ChakSayantan
Dec 4 at 16:43
You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
– Asaf Karagila♦
Dec 4 at 17:22
Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
– timtfj
Dec 4 at 18:07
@timtfj partition property is not needed as long as it serves the other conditions
– ChakSayantan
Dec 4 at 18:16
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Can we find three dense uncountable pairwise disjoint subsets of $mathbb{R}$? If so, what are these three sets?
I feel like it's not possible.
I was trying to counter it using Baire category and trying to construct such sets using the idea of construction of Cantor set (setting elements in trinary system). But I failed in both ways. I'm not even sure if answer would be affirmative or not.
real-analysis general-topology real-numbers
edited Dec 4 at 16:36
amWhy
191k28224439
asked Dec 4 at 16:29
ChakSayantan
1206
Can we find three dense uncountable pairwise disjoint subsets of $mathbb{R}$? If so, what are these three sets?
I feel like it's not possible.
I was trying to counter it using Baire category and trying to construct such sets using the idea of construction of Cantor set (setting elements in trinary system). But I failed in both ways. I'm not even sure if answer would be affirmative or not.
real-analysis general-topology real-numbers
real-analysis general-topology real-numbers
edited Dec 4 at 16:36
amWhy
191k28224439
asked Dec 4 at 16:29
ChakSayantan
1206
edited Dec 4 at 16:36
amWhy
191k28224439
asked Dec 4 at 16:29
ChakSayantan
1206
edited Dec 4 at 16:36
amWhy
191k28224439
edited Dec 4 at 16:36
amWhy
191k28224439
edited Dec 4 at 16:36
amWhy
191k28224439
191k28224439
asked Dec 4 at 16:29
ChakSayantan
1206
asked Dec 4 at 16:29
ChakSayantan
1206
asked Dec 4 at 16:29
ChakSayantan
1206
1206
1
Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
– Not Mike
Dec 4 at 16:40
@NotMike can you explain a bit more.. some precise example would be appreciated
– ChakSayantan
Dec 4 at 16:43
You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
– Asaf Karagila♦
Dec 4 at 17:22
Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
– timtfj
Dec 4 at 18:07
@timtfj partition property is not needed as long as it serves the other conditions
– ChakSayantan
Dec 4 at 18:16
add a comment |
1
Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
– Not Mike
Dec 4 at 16:40
@NotMike can you explain a bit more.. some precise example would be appreciated
– ChakSayantan
Dec 4 at 16:43
You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
– Asaf Karagila♦
Dec 4 at 17:22
Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
– timtfj
Dec 4 at 18:07
@timtfj partition property is not needed as long as it serves the other conditions
– ChakSayantan
Dec 4 at 18:16
1
1
Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
– Not Mike
Dec 4 at 16:40
Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
– Not Mike
Dec 4 at 16:40
@NotMike can you explain a bit more.. some precise example would be appreciated
– ChakSayantan
Dec 4 at 16:43
@NotMike can you explain a bit more.. some precise example would be appreciated
– ChakSayantan
Dec 4 at 16:43
You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
– Asaf Karagila♦
Dec 4 at 17:22
You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
– Asaf Karagila♦
Dec 4 at 17:22
Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
– timtfj
Dec 4 at 18:07
Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
– timtfj
Dec 4 at 18:07
@timtfj partition property is not needed as long as it serves the other conditions
– ChakSayantan
Dec 4 at 18:16
@timtfj partition property is not needed as long as it serves the other conditions
– ChakSayantan
Dec 4 at 18:16
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Fix three disjoint dense subsets of $mathbb{Q}$, call them $D_0, D_1$ and $D_2$.
Let $I_0=(0,1)backslash mathbb{Q}$, $I_1=(1,2)backslash mathbb{Q}$, $I_2=(2,3)backslash mathbb{Q}$ and set $J_k = I_k cup D_k.$
Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.
answered Dec 4 at 16:53
Not Mike
31818
So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
– ChakSayantan
Dec 4 at 17:13
You take the union
– Not Mike
Dec 4 at 17:26
then is it dence in R? I can't view it properly
– ChakSayantan
Dec 4 at 18:17
Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
– Not Mike
Dec 4 at 19:02
ah! Got it. Thanks.
– ChakSayantan
Dec 4 at 19:05
add a comment |
up vote
1
down vote
Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${1,2,3}$.
Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${4,5,6}$.
Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${7,8,9}$.
Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $mathbb R$, and $D=mathbb Rsetminus(Acup Bcup C)$ is a fourth one.
answered Dec 4 at 22:33
bof
49.8k457119
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Fix three disjoint dense subsets of $mathbb{Q}$, call them $D_0, D_1$ and $D_2$.
Let $I_0=(0,1)backslash mathbb{Q}$, $I_1=(1,2)backslash mathbb{Q}$, $I_2=(2,3)backslash mathbb{Q}$ and set $J_k = I_k cup D_k.$
Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.
answered Dec 4 at 16:53
Not Mike
31818
So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
– ChakSayantan
Dec 4 at 17:13
You take the union
– Not Mike
Dec 4 at 17:26
then is it dence in R? I can't view it properly
– ChakSayantan
Dec 4 at 18:17
Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
– Not Mike
Dec 4 at 19:02
ah! Got it. Thanks.
– ChakSayantan
Dec 4 at 19:05
add a comment |
up vote
2
down vote
accepted
Fix three disjoint dense subsets of $mathbb{Q}$, call them $D_0, D_1$ and $D_2$.
Let $I_0=(0,1)backslash mathbb{Q}$, $I_1=(1,2)backslash mathbb{Q}$, $I_2=(2,3)backslash mathbb{Q}$ and set $J_k = I_k cup D_k.$
Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.
answered Dec 4 at 16:53
Not Mike
31818
So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
– ChakSayantan
Dec 4 at 17:13
You take the union
– Not Mike
Dec 4 at 17:26
then is it dence in R? I can't view it properly
– ChakSayantan
Dec 4 at 18:17
Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
– Not Mike
Dec 4 at 19:02
ah! Got it. Thanks.
– ChakSayantan
Dec 4 at 19:05
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Fix three disjoint dense subsets of $mathbb{Q}$, call them $D_0, D_1$ and $D_2$.
Let $I_0=(0,1)backslash mathbb{Q}$, $I_1=(1,2)backslash mathbb{Q}$, $I_2=(2,3)backslash mathbb{Q}$ and set $J_k = I_k cup D_k.$
Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.
answered Dec 4 at 16:53
Not Mike
31818
Fix three disjoint dense subsets of $mathbb{Q}$, call them $D_0, D_1$ and $D_2$.
Let $I_0=(0,1)backslash mathbb{Q}$, $I_1=(1,2)backslash mathbb{Q}$, $I_2=(2,3)backslash mathbb{Q}$ and set $J_k = I_k cup D_k.$
Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.
answered Dec 4 at 16:53
Not Mike
31818
answered Dec 4 at 16:53
Not Mike
31818
answered Dec 4 at 16:53
Not Mike
31818
answered Dec 4 at 16:53
Not Mike
31818
31818
So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
– ChakSayantan
Dec 4 at 17:13
You take the union
– Not Mike
Dec 4 at 17:26
then is it dence in R? I can't view it properly
– ChakSayantan
Dec 4 at 18:17
Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
– Not Mike
Dec 4 at 19:02
ah! Got it. Thanks.
– ChakSayantan
Dec 4 at 19:05
add a comment |
So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
– ChakSayantan
Dec 4 at 17:13
You take the union
– Not Mike
Dec 4 at 17:26
then is it dence in R? I can't view it properly
– ChakSayantan
Dec 4 at 18:17
Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
– Not Mike
Dec 4 at 19:02
ah! Got it. Thanks.
– ChakSayantan
Dec 4 at 19:05
So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
– ChakSayantan
Dec 4 at 17:13
So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
– ChakSayantan
Dec 4 at 17:13
You take the union
– Not Mike
Dec 4 at 17:26
You take the union
– Not Mike
Dec 4 at 17:26
then is it dence in R? I can't view it properly
– ChakSayantan
Dec 4 at 18:17
then is it dence in R? I can't view it properly
– ChakSayantan
Dec 4 at 18:17
Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
– Not Mike
Dec 4 at 19:02
Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
– Not Mike
Dec 4 at 19:02
ah! Got it. Thanks.
– ChakSayantan
Dec 4 at 19:05
ah! Got it. Thanks.
– ChakSayantan
Dec 4 at 19:05
add a comment |
up vote
1
down vote
Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${1,2,3}$.
Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${4,5,6}$.
Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${7,8,9}$.
Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $mathbb R$, and $D=mathbb Rsetminus(Acup Bcup C)$ is a fourth one.
answered Dec 4 at 22:33
bof
49.8k457119
add a comment |
up vote
1
down vote
Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${1,2,3}$.
Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${4,5,6}$.
Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${7,8,9}$.
Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $mathbb R$, and $D=mathbb Rsetminus(Acup Bcup C)$ is a fourth one.
answered Dec 4 at 22:33
bof
49.8k457119
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${1,2,3}$.
Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${4,5,6}$.
Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${7,8,9}$.
Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $mathbb R$, and $D=mathbb Rsetminus(Acup Bcup C)$ is a fourth one.
answered Dec 4 at 22:33
bof
49.8k457119
Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${1,2,3}$.
Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${4,5,6}$.
Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${7,8,9}$.
Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $mathbb R$, and $D=mathbb Rsetminus(Acup Bcup C)$ is a fourth one.
answered Dec 4 at 22:33
bof
49.8k457119
edited Dec 4 at 23:15
answered Dec 4 at 22:33
bof
49.8k457119
answered Dec 4 at 22:33
bof
49.8k457119
answered Dec 4 at 22:33
bof
49.8k457119
49.8k457119
add a comment |
add a comment |
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1
Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
– Not Mike
Dec 4 at 16:40
@NotMike can you explain a bit more.. some precise example would be appreciated
– ChakSayantan
Dec 4 at 16:43
You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
– Asaf Karagila♦
Dec 4 at 17:22
Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
– timtfj
Dec 4 at 18:07
@timtfj partition property is not needed as long as it serves the other conditions
– ChakSayantan
Dec 4 at 18:16