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Math Help - countable

  1. #1
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    countable

    Show that the subset  A of terminating sequences is countable (e.g.  a_n = 0 for  n >N ).

    So suppose  E is a countable subset of  A . Then  E = \{s_1, s_2, \cdots, s_n \} where  s_k (  k=1 \ldots n ) are terminating sequences. Is it possible to just say that is is impossible to construct a sequence  s such that  s \notin E . Thus not every subset  E of  A is proper. Hence  A is countable.

    Or is the only way to do it is to say that  A is equivalent to  \mathbb{Q}^{\mathbb{N}} ?
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Show that the subset  A of terminating sequences is countable (e.g.  a_n = 0 for  n >N ).

    So suppose  E is a countable subset of  A . Then  E = \{s_1, s_2, \cdots, s_n \} where  s_k (  k=1 \ldots n ) are terminating sequences. Is it possible to just say that is is impossible to construct a sequence  s such that  s \notin E . Thus not every subset  E of  A is proper. Hence  A is countable.

    Or is the only way to do it is to say that  A is equivalent to  \mathbb{Q}^{\mathbb{N}} ?
    How would you show that it is "impossible to construct a sequence s such that s \notin E? Wouldn't that depend on exactly what E was?

    I would do this: let [tex]E_N[tex] be the set of all sequences such that s_N\ne 0 but that s_n= 0 for all n> N. It should be easy to show that [tex]E_N[tex] is countable. And E is the union of all [tex]E_N[tex] for all positive integers, N. That is, E is the countable union of countable sets.
    Last edited by HallsofIvy; April 27th 2009 at 04:54 PM.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    How would you show that it is "impossible to construct a sequence s such that s \notin E? Wouldn't that depend on exactly what E was?

    I would do this: let EN be the set of all sequences such that s_N\ne 0 but that s_n= 0 for all n> N. It should be easy to show that EN is countable. And E is the union of all EN for all positive integers, N. That is, E is the countable union of countable sets.
    Well that is the traditional way of doing it. But to show a set is uncountable, for any countable set  E of  A you can construct a sequence  s such that  s \notin E right? So the choice of  E doesn't really matter. But is it possible to explicitly show that no such  s exists in this case?
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