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  1. #1
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    countable

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

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

    Or is the only way to do it is to say that $\displaystyle A $ is equivalent to $\displaystyle \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 $\displaystyle A $ of terminating sequences is countable (e.g. $\displaystyle a_n = 0 $ for $\displaystyle n >N $).

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

    Or is the only way to do it is to say that $\displaystyle A $ is equivalent to $\displaystyle \mathbb{Q}^{\mathbb{N}} $?
    How would you show that it is "impossible to construct a sequence $\displaystyle s$ such that $\displaystyle s \notin E$? 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 $\displaystyle s_N\ne 0$ but that $\displaystyle s_n= 0$ 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; Apr 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 $\displaystyle s$ such that $\displaystyle s \notin E$? Wouldn't that depend on exactly what E was?

    I would do this: let EN be the set of all sequences such that $\displaystyle s_N\ne 0$ but that $\displaystyle s_n= 0$ 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 $\displaystyle E $ of $\displaystyle A $ you can construct a sequence $\displaystyle s $ such that $\displaystyle s \notin E $ right? So the choice of $\displaystyle E $ doesn't really matter. But is it possible to explicitly show that no such $\displaystyle s $ exists in this case?
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