Uncountability of the equivalence classes of $mathbb{R}/mathbb{Q}$
$begingroup$
Let $a,bin[0,1]$ and define the equivalence relation $sim$ by $asim biff a-binmathbb{Q}$. This relation partitions $[0,1]$ into equivalence classes where every class consists of a set of numbers which are equivalent under $sim$,
My textbook states (without proof):
The set $[0,1]/sim$ consists of uncountably many of these classes, where
each class consists of countably many members.
How can I formally prove this statement?
elementary-set-theory equivalence-relations
edited May 2 '13 at 19:52
Ross Millikan
301k24200375
asked May 2 '13 at 19:49
leoleo
261
$endgroup$
add a comment |
$begingroup$
Let $a,bin[0,1]$ and define the equivalence relation $sim$ by $asim biff a-binmathbb{Q}$. This relation partitions $[0,1]$ into equivalence classes where every class consists of a set of numbers which are equivalent under $sim$,
My textbook states (without proof):
The set $[0,1]/sim$ consists of uncountably many of these classes, where
each class consists of countably many members.
How can I formally prove this statement?
elementary-set-theory equivalence-relations
edited May 2 '13 at 19:52
Ross Millikan
301k24200375
asked May 2 '13 at 19:49
leoleo
261
$endgroup$
1
$begingroup$
The classes are disjoint and each class is countable. If the the set of classes were countable, then the union of all classes would be ....., but $[0,1]$ is ....
$endgroup$
– Amr
May 2 '13 at 19:52
add a comment |
$begingroup$
Let $a,bin[0,1]$ and define the equivalence relation $sim$ by $asim biff a-binmathbb{Q}$. This relation partitions $[0,1]$ into equivalence classes where every class consists of a set of numbers which are equivalent under $sim$,
My textbook states (without proof):
The set $[0,1]/sim$ consists of uncountably many of these classes, where
each class consists of countably many members.
How can I formally prove this statement?
elementary-set-theory equivalence-relations
edited May 2 '13 at 19:52
Ross Millikan
301k24200375
asked May 2 '13 at 19:49
leoleo
261
$endgroup$
Let $a,bin[0,1]$ and define the equivalence relation $sim$ by $asim biff a-binmathbb{Q}$. This relation partitions $[0,1]$ into equivalence classes where every class consists of a set of numbers which are equivalent under $sim$,
My textbook states (without proof):
The set $[0,1]/sim$ consists of uncountably many of these classes, where
each class consists of countably many members.
How can I formally prove this statement?
elementary-set-theory equivalence-relations
elementary-set-theory equivalence-relations
edited May 2 '13 at 19:52
Ross Millikan
301k24200375
asked May 2 '13 at 19:49
leoleo
261
edited May 2 '13 at 19:52
Ross Millikan
301k24200375
asked May 2 '13 at 19:49
leoleo
261
edited May 2 '13 at 19:52
Ross Millikan
301k24200375
edited May 2 '13 at 19:52
Ross Millikan
301k24200375
edited May 2 '13 at 19:52
Ross Millikan
301k24200375
301k24200375
asked May 2 '13 at 19:49
leoleo
261
asked May 2 '13 at 19:49
leoleo
261
asked May 2 '13 at 19:49
leoleo
261
261
1
$begingroup$
The classes are disjoint and each class is countable. If the the set of classes were countable, then the union of all classes would be ....., but $[0,1]$ is ....
$endgroup$
– Amr
May 2 '13 at 19:52
add a comment |
1
$begingroup$
The classes are disjoint and each class is countable. If the the set of classes were countable, then the union of all classes would be ....., but $[0,1]$ is ....
$endgroup$
– Amr
May 2 '13 at 19:52
1
1
$begingroup$
The classes are disjoint and each class is countable. If the the set of classes were countable, then the union of all classes would be ....., but $[0,1]$ is ....
$endgroup$
– Amr
May 2 '13 at 19:52
$begingroup$
The classes are disjoint and each class is countable. If the the set of classes were countable, then the union of all classes would be ....., but $[0,1]$ is ....
$endgroup$
– Amr
May 2 '13 at 19:52
add a comment |
2 Answers
2
active
oldest
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$begingroup$
If you know $Bbb Q$ is countable, that covers the second half. Then use the fact that a countable union of countable sets is again countable to show that there must be uncountably many classes.
answered May 2 '13 at 19:53
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Given $a in [0,1]$, the class of $a$ is ${a + q : a+qin[0,1], qinmathbb Q}$, so this class has the cardinality $aleph_0$. Now suppose there are countably many classes. The union of countably many countable sets is again countable, so you obtain that $[0,1]$ is countable, which is a contradiction. So there are uncountably many classes of equivalence under ~.
answered May 2 '13 at 19:55
zarathustrazarathustra
4,15511029
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
If you know $Bbb Q$ is countable, that covers the second half. Then use the fact that a countable union of countable sets is again countable to show that there must be uncountably many classes.
answered May 2 '13 at 19:53
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
If you know $Bbb Q$ is countable, that covers the second half. Then use the fact that a countable union of countable sets is again countable to show that there must be uncountably many classes.
answered May 2 '13 at 19:53
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
If you know $Bbb Q$ is countable, that covers the second half. Then use the fact that a countable union of countable sets is again countable to show that there must be uncountably many classes.
answered May 2 '13 at 19:53
Ross MillikanRoss Millikan
301k24200375
$endgroup$
If you know $Bbb Q$ is countable, that covers the second half. Then use the fact that a countable union of countable sets is again countable to show that there must be uncountably many classes.
answered May 2 '13 at 19:53
Ross MillikanRoss Millikan
301k24200375
answered May 2 '13 at 19:53
Ross MillikanRoss Millikan
301k24200375
answered May 2 '13 at 19:53
Ross MillikanRoss Millikan
301k24200375
answered May 2 '13 at 19:53
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
$begingroup$
Given $a in [0,1]$, the class of $a$ is ${a + q : a+qin[0,1], qinmathbb Q}$, so this class has the cardinality $aleph_0$. Now suppose there are countably many classes. The union of countably many countable sets is again countable, so you obtain that $[0,1]$ is countable, which is a contradiction. So there are uncountably many classes of equivalence under ~.
answered May 2 '13 at 19:55
zarathustrazarathustra
4,15511029
$endgroup$
add a comment |
$begingroup$
Given $a in [0,1]$, the class of $a$ is ${a + q : a+qin[0,1], qinmathbb Q}$, so this class has the cardinality $aleph_0$. Now suppose there are countably many classes. The union of countably many countable sets is again countable, so you obtain that $[0,1]$ is countable, which is a contradiction. So there are uncountably many classes of equivalence under ~.
answered May 2 '13 at 19:55
zarathustrazarathustra
4,15511029
$endgroup$
add a comment |
$begingroup$
Given $a in [0,1]$, the class of $a$ is ${a + q : a+qin[0,1], qinmathbb Q}$, so this class has the cardinality $aleph_0$. Now suppose there are countably many classes. The union of countably many countable sets is again countable, so you obtain that $[0,1]$ is countable, which is a contradiction. So there are uncountably many classes of equivalence under ~.
answered May 2 '13 at 19:55
zarathustrazarathustra
4,15511029
$endgroup$
Given $a in [0,1]$, the class of $a$ is ${a + q : a+qin[0,1], qinmathbb Q}$, so this class has the cardinality $aleph_0$. Now suppose there are countably many classes. The union of countably many countable sets is again countable, so you obtain that $[0,1]$ is countable, which is a contradiction. So there are uncountably many classes of equivalence under ~.
answered May 2 '13 at 19:55
zarathustrazarathustra
4,15511029
answered May 2 '13 at 19:55
zarathustrazarathustra
4,15511029
answered May 2 '13 at 19:55
zarathustrazarathustra
4,15511029
answered May 2 '13 at 19:55
zarathustrazarathustra
4,15511029
4,15511029
add a comment |
add a comment |
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The classes are disjoint and each class is countable. If the the set of classes were countable, then the union of all classes would be ....., but $[0,1]$ is ....
$endgroup$
– Amr
May 2 '13 at 19:52