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Math Help - countability of the reals

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    countability of the reals

    We know that the real numbers  \bold{R} are uncountable. But what if we partitioned the real numbers into a infinite but countable number of closed intervals. So we would have for example:  [0,1] \cup [1,2] \cup [2,3], \dots . Do the same thing for the negative reals. Then the number of intervals are countable right?
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     [0,1] contains all the reals between  0 and  1 inclusive. Similarly,  [1,2] contains all the reals between  1 and  2 inclusive etc...
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    Quote Originally Posted by particlejohn View Post
    Then the number of intervals are countable right?
    Right. But each of these intervals is uncountable.
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    Hello,
    Quote Originally Posted by particlejohn View Post
    We know that the real numbers  \bold{R} are uncountable. But what if we partitioned the real numbers into a infinite but countable number of intervals. So we would have for example:  [0,1] \cup [1,2] \cup [2,3], \dots . Do the same thing for the negative reals. Then the number of intervals are countable right?
    Hmm I'd say yes, because there's an interval corresponding to an integer. And \mathbb{Z} is equipotent to \mathbb{N}, infinite and countable set.

    Let S=\{\dots \cup [-2,-1) \cup [-1,0) \cup [0,1) \cup \dots \}

    S=\{A_i ~|~ \forall x \in A_i, ~i \le x < i+1 \} (actually i is the least integer contained in A_i)

    Every element of S can be associated to an integer of \mathbb{Z}. Thus S is countable.


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    Quote Originally Posted by particlejohn View Post
    We know that the real numbers  \bold{R} are uncountable. But what if we partitioned the real numbers into a infinite but countable number of closed intervals. So we would have for example:  [0,1] \cup [1,2] \cup [2,3], \dots . Do the same thing for the negative reals. Then the number of intervals are countable right?
    You cannot find a countable partition of \mathbb{R} by countable sets.
    If that is what you are asking.
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