Mixing
Can the union of uncountable infinite sets be a countable infinite set?
$begingroup$
Can the union of uncountable infinite sets be a countable infinite set?
If there is such a set, I would be grateful if you answer the question by giving an example.
If the question is too simple, I apologize.
elementary-set-theory examples-counterexamples
edited Jan 27 at 23:26
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 9:51
ElementaryElementary
360111
$endgroup$
|
show 5 more comments
$begingroup$
Can the union of uncountable infinite sets be a countable infinite set?
If there is such a set, I would be grateful if you answer the question by giving an example.
If the question is too simple, I apologize.
elementary-set-theory examples-counterexamples
edited Jan 27 at 23:26
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 9:51
ElementaryElementary
360111
$endgroup$
$begingroup$
Oh, I am sorry for English. I fixed.
$endgroup$
– Elementary
Jan 27 at 10:00
2
$begingroup$
If you ask for all of them to be nonempty, no
$endgroup$
– Simone Ramello
Jan 27 at 10:00
1
$begingroup$
if their union is countable, then there is an injection from that Union to N, because that's what countable means by definition, so just restrict this injection to any of the unioned sets and that set would be countable also, which contradicts your premise of each of those sets being uncountable. So their union cannot be countable!
$endgroup$
– Zuhair
Jan 27 at 10:13
5
$begingroup$
Just checking we're not answering the wrong thing: you do mean a union of uncountable sets, not a union of uncountably many sets?
$endgroup$
– timtfj
Jan 27 at 12:10
2
$begingroup$
(For uncountably many countable sets, we could just point out that $mathbb N$ is the union of its uncountably many subsets.)
$endgroup$
– timtfj
Jan 27 at 13:08
|
show 5 more comments
$begingroup$
Can the union of uncountable infinite sets be a countable infinite set?
If there is such a set, I would be grateful if you answer the question by giving an example.
If the question is too simple, I apologize.
elementary-set-theory examples-counterexamples
edited Jan 27 at 23:26
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 9:51
ElementaryElementary
360111
$endgroup$
Can the union of uncountable infinite sets be a countable infinite set?
If there is such a set, I would be grateful if you answer the question by giving an example.
If the question is too simple, I apologize.
elementary-set-theory examples-counterexamples
elementary-set-theory examples-counterexamples
edited Jan 27 at 23:26
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 9:51
ElementaryElementary
360111
edited Jan 27 at 23:26
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 9:51
ElementaryElementary
360111
edited Jan 27 at 23:26
Andrés E. Caicedo
65.8k8160251
edited Jan 27 at 23:26
Andrés E. Caicedo
65.8k8160251
edited Jan 27 at 23:26
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Jan 27 at 9:51
ElementaryElementary
360111
asked Jan 27 at 9:51
ElementaryElementary
360111
asked Jan 27 at 9:51
ElementaryElementary
360111
360111
$begingroup$
Oh, I am sorry for English. I fixed.
$endgroup$
– Elementary
Jan 27 at 10:00
2
$begingroup$
If you ask for all of them to be nonempty, no
$endgroup$
– Simone Ramello
Jan 27 at 10:00
1
$begingroup$
if their union is countable, then there is an injection from that Union to N, because that's what countable means by definition, so just restrict this injection to any of the unioned sets and that set would be countable also, which contradicts your premise of each of those sets being uncountable. So their union cannot be countable!
$endgroup$
– Zuhair
Jan 27 at 10:13
5
$begingroup$
Just checking we're not answering the wrong thing: you do mean a union of uncountable sets, not a union of uncountably many sets?
$endgroup$
– timtfj
Jan 27 at 12:10
2
$begingroup$
(For uncountably many countable sets, we could just point out that $mathbb N$ is the union of its uncountably many subsets.)
$endgroup$
– timtfj
Jan 27 at 13:08
|
show 5 more comments
$begingroup$
Oh, I am sorry for English. I fixed.
$endgroup$
– Elementary
Jan 27 at 10:00
2
$begingroup$
If you ask for all of them to be nonempty, no
$endgroup$
– Simone Ramello
Jan 27 at 10:00
1
$begingroup$
if their union is countable, then there is an injection from that Union to N, because that's what countable means by definition, so just restrict this injection to any of the unioned sets and that set would be countable also, which contradicts your premise of each of those sets being uncountable. So their union cannot be countable!
$endgroup$
– Zuhair
Jan 27 at 10:13
5
$begingroup$
Just checking we're not answering the wrong thing: you do mean a union of uncountable sets, not a union of uncountably many sets?
$endgroup$
– timtfj
Jan 27 at 12:10
2
$begingroup$
(For uncountably many countable sets, we could just point out that $mathbb N$ is the union of its uncountably many subsets.)
$endgroup$
– timtfj
Jan 27 at 13:08
$begingroup$
Oh, I am sorry for English. I fixed.
$endgroup$
– Elementary
Jan 27 at 10:00
$begingroup$
Oh, I am sorry for English. I fixed.
$endgroup$
– Elementary
Jan 27 at 10:00
2
2
$begingroup$
If you ask for all of them to be nonempty, no
$endgroup$
– Simone Ramello
Jan 27 at 10:00
$begingroup$
If you ask for all of them to be nonempty, no
$endgroup$
– Simone Ramello
Jan 27 at 10:00
1
1
$begingroup$
if their union is countable, then there is an injection from that Union to N, because that's what countable means by definition, so just restrict this injection to any of the unioned sets and that set would be countable also, which contradicts your premise of each of those sets being uncountable. So their union cannot be countable!
$endgroup$
– Zuhair
Jan 27 at 10:13
$begingroup$
if their union is countable, then there is an injection from that Union to N, because that's what countable means by definition, so just restrict this injection to any of the unioned sets and that set would be countable also, which contradicts your premise of each of those sets being uncountable. So their union cannot be countable!
$endgroup$
– Zuhair
Jan 27 at 10:13
5
5
$begingroup$
Just checking we're not answering the wrong thing: you do mean a union of uncountable sets, not a union of uncountably many sets?
$endgroup$
– timtfj
Jan 27 at 12:10
$begingroup$
Just checking we're not answering the wrong thing: you do mean a union of uncountable sets, not a union of uncountably many sets?
$endgroup$
– timtfj
Jan 27 at 12:10
2
2
$begingroup$
(For uncountably many countable sets, we could just point out that $mathbb N$ is the union of its uncountably many subsets.)
$endgroup$
– timtfj
Jan 27 at 13:08
$begingroup$
(For uncountably many countable sets, we could just point out that $mathbb N$ is the union of its uncountably many subsets.)
$endgroup$
– timtfj
Jan 27 at 13:08
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Let $A$ be one of the uncountable sets, and let $B$ be their union.
Suppose $B$ is countable.
By the definition of set union, every element of $A$ is also an element of $B$: that is, $Asubseteq B$.
Therefore $B$ is a countable set with an uncountable subset, which is a contradiction.
Therefore $B$ cannot be countable.
Note: This relies on the theorem that every subset of a countable set is countable. The easiest way to see why that's true is to imagine counting the original set but skipping elements which aren't in the subset. A proof just involves expressing that more formally.
answered Jan 27 at 12:07
timtfjtimtfj
2,478420
$endgroup$
add a comment |
$begingroup$
Unions cannot decrease the cardinality of a set, so suppose $A_i$ are uncountable then the finite union has cardinality:
$$ |bigcuplimits_{i=0}^{N} A_i|geq |A_i|$$
for any of the $A_i$. Which means they are definitely uncountable.
answered Jan 27 at 10:11
Wesley StrikWesley Strik
2,209424
$endgroup$
$begingroup$
How are you just carrying the proof for finite sets and extending for infinite set? What is the definition of cardinality for infinite set whatever it is , but it is not number of an ordered set?
$endgroup$
– Bijayan Ray
Jan 27 at 12:13
$begingroup$
The pigeon hole principle is valid for finite sets but is not valid for infinite , an example is the infinite hotel paradox
$endgroup$
– Bijayan Ray
Jan 27 at 12:14
$begingroup$
en.wikipedia.org/wiki/Aleph_number
$endgroup$
– Wesley Strik
Jan 27 at 12:23
add a comment |
$begingroup$
Maybe the OP meant the intersection of uncountable sets. If so, then certainly it's possible with a countable infinity of input sets. The $p$-adic integers for any prime $p$ are uncountably infinite, but the intersection of all such sets for all primes $p$ is just the ordinary integers.
answered Jan 27 at 23:36
Oscar LanziOscar Lanzi
13.3k12136
$endgroup$
1
$begingroup$
Or even with just two sets, like $A=mathbb R_{>0}$ and $B=mathbb R_{<0} cup mathbb N$ (though this makes $B$ very much an invented-for-a-proof set!)
$endgroup$
– timtfj
Jan 28 at 16:51
$begingroup$
(I'm unsure of the usual notation. I mean positive reals in one set, and natural numbers and negative reals in the other, so they just have $mathbb N$ in common.)
$endgroup$
– timtfj
Jan 28 at 16:56
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $A$ be one of the uncountable sets, and let $B$ be their union.
Suppose $B$ is countable.
By the definition of set union, every element of $A$ is also an element of $B$: that is, $Asubseteq B$.
Therefore $B$ is a countable set with an uncountable subset, which is a contradiction.
Therefore $B$ cannot be countable.
Note: This relies on the theorem that every subset of a countable set is countable. The easiest way to see why that's true is to imagine counting the original set but skipping elements which aren't in the subset. A proof just involves expressing that more formally.
answered Jan 27 at 12:07
timtfjtimtfj
2,478420
$endgroup$
add a comment |
$begingroup$
Let $A$ be one of the uncountable sets, and let $B$ be their union.
Suppose $B$ is countable.
By the definition of set union, every element of $A$ is also an element of $B$: that is, $Asubseteq B$.
Therefore $B$ is a countable set with an uncountable subset, which is a contradiction.
Therefore $B$ cannot be countable.
Note: This relies on the theorem that every subset of a countable set is countable. The easiest way to see why that's true is to imagine counting the original set but skipping elements which aren't in the subset. A proof just involves expressing that more formally.
answered Jan 27 at 12:07
timtfjtimtfj
2,478420
$endgroup$
add a comment |
$begingroup$
Let $A$ be one of the uncountable sets, and let $B$ be their union.
Suppose $B$ is countable.
By the definition of set union, every element of $A$ is also an element of $B$: that is, $Asubseteq B$.
Therefore $B$ is a countable set with an uncountable subset, which is a contradiction.
Therefore $B$ cannot be countable.
Note: This relies on the theorem that every subset of a countable set is countable. The easiest way to see why that's true is to imagine counting the original set but skipping elements which aren't in the subset. A proof just involves expressing that more formally.
answered Jan 27 at 12:07
timtfjtimtfj
2,478420
$endgroup$
Let $A$ be one of the uncountable sets, and let $B$ be their union.
Suppose $B$ is countable.
By the definition of set union, every element of $A$ is also an element of $B$: that is, $Asubseteq B$.
Therefore $B$ is a countable set with an uncountable subset, which is a contradiction.
Therefore $B$ cannot be countable.
Note: This relies on the theorem that every subset of a countable set is countable. The easiest way to see why that's true is to imagine counting the original set but skipping elements which aren't in the subset. A proof just involves expressing that more formally.
answered Jan 27 at 12:07
timtfjtimtfj
2,478420
edited Jan 27 at 12:57
answered Jan 27 at 12:07
timtfjtimtfj
2,478420
answered Jan 27 at 12:07
timtfjtimtfj
2,478420
answered Jan 27 at 12:07
timtfjtimtfj
2,478420
2,478420
add a comment |
add a comment |
$begingroup$
Unions cannot decrease the cardinality of a set, so suppose $A_i$ are uncountable then the finite union has cardinality:
$$ |bigcuplimits_{i=0}^{N} A_i|geq |A_i|$$
for any of the $A_i$. Which means they are definitely uncountable.
answered Jan 27 at 10:11
Wesley StrikWesley Strik
2,209424
$endgroup$
$begingroup$
How are you just carrying the proof for finite sets and extending for infinite set? What is the definition of cardinality for infinite set whatever it is , but it is not number of an ordered set?
$endgroup$
– Bijayan Ray
Jan 27 at 12:13
$begingroup$
The pigeon hole principle is valid for finite sets but is not valid for infinite , an example is the infinite hotel paradox
$endgroup$
– Bijayan Ray
Jan 27 at 12:14
$begingroup$
en.wikipedia.org/wiki/Aleph_number
$endgroup$
– Wesley Strik
Jan 27 at 12:23
add a comment |
$begingroup$
Unions cannot decrease the cardinality of a set, so suppose $A_i$ are uncountable then the finite union has cardinality:
$$ |bigcuplimits_{i=0}^{N} A_i|geq |A_i|$$
for any of the $A_i$. Which means they are definitely uncountable.
answered Jan 27 at 10:11
Wesley StrikWesley Strik
2,209424
$endgroup$
$begingroup$
How are you just carrying the proof for finite sets and extending for infinite set? What is the definition of cardinality for infinite set whatever it is , but it is not number of an ordered set?
$endgroup$
– Bijayan Ray
Jan 27 at 12:13
$begingroup$
The pigeon hole principle is valid for finite sets but is not valid for infinite , an example is the infinite hotel paradox
$endgroup$
– Bijayan Ray
Jan 27 at 12:14
$begingroup$
en.wikipedia.org/wiki/Aleph_number
$endgroup$
– Wesley Strik
Jan 27 at 12:23
add a comment |
$begingroup$
Unions cannot decrease the cardinality of a set, so suppose $A_i$ are uncountable then the finite union has cardinality:
$$ |bigcuplimits_{i=0}^{N} A_i|geq |A_i|$$
for any of the $A_i$. Which means they are definitely uncountable.
answered Jan 27 at 10:11
Wesley StrikWesley Strik
2,209424
$endgroup$
Unions cannot decrease the cardinality of a set, so suppose $A_i$ are uncountable then the finite union has cardinality:
$$ |bigcuplimits_{i=0}^{N} A_i|geq |A_i|$$
for any of the $A_i$. Which means they are definitely uncountable.
answered Jan 27 at 10:11
Wesley StrikWesley Strik
2,209424
edited Jan 27 at 12:27
answered Jan 27 at 10:11
Wesley StrikWesley Strik
2,209424
answered Jan 27 at 10:11
Wesley StrikWesley Strik
2,209424
answered Jan 27 at 10:11
Wesley StrikWesley Strik
2,209424
2,209424
$begingroup$
How are you just carrying the proof for finite sets and extending for infinite set? What is the definition of cardinality for infinite set whatever it is , but it is not number of an ordered set?
$endgroup$
– Bijayan Ray
Jan 27 at 12:13
$begingroup$
The pigeon hole principle is valid for finite sets but is not valid for infinite , an example is the infinite hotel paradox
$endgroup$
– Bijayan Ray
Jan 27 at 12:14
$begingroup$
en.wikipedia.org/wiki/Aleph_number
$endgroup$
– Wesley Strik
Jan 27 at 12:23
add a comment |
$begingroup$
How are you just carrying the proof for finite sets and extending for infinite set? What is the definition of cardinality for infinite set whatever it is , but it is not number of an ordered set?
$endgroup$
– Bijayan Ray
Jan 27 at 12:13
$begingroup$
The pigeon hole principle is valid for finite sets but is not valid for infinite , an example is the infinite hotel paradox
$endgroup$
– Bijayan Ray
Jan 27 at 12:14
$begingroup$
en.wikipedia.org/wiki/Aleph_number
$endgroup$
– Wesley Strik
Jan 27 at 12:23
$begingroup$
How are you just carrying the proof for finite sets and extending for infinite set? What is the definition of cardinality for infinite set whatever it is , but it is not number of an ordered set?
$endgroup$
– Bijayan Ray
Jan 27 at 12:13
$begingroup$
How are you just carrying the proof for finite sets and extending for infinite set? What is the definition of cardinality for infinite set whatever it is , but it is not number of an ordered set?
$endgroup$
– Bijayan Ray
Jan 27 at 12:13
$begingroup$
The pigeon hole principle is valid for finite sets but is not valid for infinite , an example is the infinite hotel paradox
$endgroup$
– Bijayan Ray
Jan 27 at 12:14
$begingroup$
The pigeon hole principle is valid for finite sets but is not valid for infinite , an example is the infinite hotel paradox
$endgroup$
– Bijayan Ray
Jan 27 at 12:14
$begingroup$
en.wikipedia.org/wiki/Aleph_number
$endgroup$
– Wesley Strik
Jan 27 at 12:23
$begingroup$
en.wikipedia.org/wiki/Aleph_number
$endgroup$
– Wesley Strik
Jan 27 at 12:23
add a comment |
$begingroup$
Maybe the OP meant the intersection of uncountable sets. If so, then certainly it's possible with a countable infinity of input sets. The $p$-adic integers for any prime $p$ are uncountably infinite, but the intersection of all such sets for all primes $p$ is just the ordinary integers.
answered Jan 27 at 23:36
Oscar LanziOscar Lanzi
13.3k12136
$endgroup$
1
$begingroup$
Or even with just two sets, like $A=mathbb R_{>0}$ and $B=mathbb R_{<0} cup mathbb N$ (though this makes $B$ very much an invented-for-a-proof set!)
$endgroup$
– timtfj
Jan 28 at 16:51
$begingroup$
(I'm unsure of the usual notation. I mean positive reals in one set, and natural numbers and negative reals in the other, so they just have $mathbb N$ in common.)
$endgroup$
– timtfj
Jan 28 at 16:56
add a comment |
$begingroup$
Maybe the OP meant the intersection of uncountable sets. If so, then certainly it's possible with a countable infinity of input sets. The $p$-adic integers for any prime $p$ are uncountably infinite, but the intersection of all such sets for all primes $p$ is just the ordinary integers.
answered Jan 27 at 23:36
Oscar LanziOscar Lanzi
13.3k12136
$endgroup$
1
$begingroup$
Or even with just two sets, like $A=mathbb R_{>0}$ and $B=mathbb R_{<0} cup mathbb N$ (though this makes $B$ very much an invented-for-a-proof set!)
$endgroup$
– timtfj
Jan 28 at 16:51
$begingroup$
(I'm unsure of the usual notation. I mean positive reals in one set, and natural numbers and negative reals in the other, so they just have $mathbb N$ in common.)
$endgroup$
– timtfj
Jan 28 at 16:56
add a comment |
$begingroup$
Maybe the OP meant the intersection of uncountable sets. If so, then certainly it's possible with a countable infinity of input sets. The $p$-adic integers for any prime $p$ are uncountably infinite, but the intersection of all such sets for all primes $p$ is just the ordinary integers.
answered Jan 27 at 23:36
Oscar LanziOscar Lanzi
13.3k12136
$endgroup$
Maybe the OP meant the intersection of uncountable sets. If so, then certainly it's possible with a countable infinity of input sets. The $p$-adic integers for any prime $p$ are uncountably infinite, but the intersection of all such sets for all primes $p$ is just the ordinary integers.
answered Jan 27 at 23:36
Oscar LanziOscar Lanzi
13.3k12136
answered Jan 27 at 23:36
Oscar LanziOscar Lanzi
13.3k12136
answered Jan 27 at 23:36
Oscar LanziOscar Lanzi
13.3k12136
answered Jan 27 at 23:36
Oscar LanziOscar Lanzi
13.3k12136
13.3k12136
1
$begingroup$
Or even with just two sets, like $A=mathbb R_{>0}$ and $B=mathbb R_{<0} cup mathbb N$ (though this makes $B$ very much an invented-for-a-proof set!)
$endgroup$
– timtfj
Jan 28 at 16:51
$begingroup$
(I'm unsure of the usual notation. I mean positive reals in one set, and natural numbers and negative reals in the other, so they just have $mathbb N$ in common.)
$endgroup$
– timtfj
Jan 28 at 16:56
add a comment |
1
$begingroup$
Or even with just two sets, like $A=mathbb R_{>0}$ and $B=mathbb R_{<0} cup mathbb N$ (though this makes $B$ very much an invented-for-a-proof set!)
$endgroup$
– timtfj
Jan 28 at 16:51
$begingroup$
(I'm unsure of the usual notation. I mean positive reals in one set, and natural numbers and negative reals in the other, so they just have $mathbb N$ in common.)
$endgroup$
– timtfj
Jan 28 at 16:56
1
1
$begingroup$
Or even with just two sets, like $A=mathbb R_{>0}$ and $B=mathbb R_{<0} cup mathbb N$ (though this makes $B$ very much an invented-for-a-proof set!)
$endgroup$
– timtfj
Jan 28 at 16:51
$begingroup$
Or even with just two sets, like $A=mathbb R_{>0}$ and $B=mathbb R_{<0} cup mathbb N$ (though this makes $B$ very much an invented-for-a-proof set!)
$endgroup$
– timtfj
Jan 28 at 16:51
$begingroup$
(I'm unsure of the usual notation. I mean positive reals in one set, and natural numbers and negative reals in the other, so they just have $mathbb N$ in common.)
$endgroup$
– timtfj
Jan 28 at 16:56
$begingroup$
(I'm unsure of the usual notation. I mean positive reals in one set, and natural numbers and negative reals in the other, so they just have $mathbb N$ in common.)
$endgroup$
– timtfj
Jan 28 at 16:56
add a comment |
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Oh, I am sorry for English. I fixed.
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– Elementary
Jan 27 at 10:00
2
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If you ask for all of them to be nonempty, no
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– Simone Ramello
Jan 27 at 10:00
1
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if their union is countable, then there is an injection from that Union to N, because that's what countable means by definition, so just restrict this injection to any of the unioned sets and that set would be countable also, which contradicts your premise of each of those sets being uncountable. So their union cannot be countable!
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– Zuhair
Jan 27 at 10:13
5
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Just checking we're not answering the wrong thing: you do mean a union of uncountable sets, not a union of uncountably many sets?
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– timtfj
Jan 27 at 12:10
2
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(For uncountably many countable sets, we could just point out that $mathbb N$ is the union of its uncountably many subsets.)
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– timtfj
Jan 27 at 13:08