Let $a,b\in[0,1]$ and define the equivalence relation $\sim$ by $a\sim b\iff a-b\in\mathbb{Q}$. 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?

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    $\begingroup$ 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

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.


Given $a \in [0,1]$, the class of $a$ is $\{a + q : a+q\in[0,1], q\in\mathbb Q\}$, 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 ~.


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