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I'm having trouble finding a proof (without the need for cases) for this statement: If A is at most countable, then there is a sequence $(a_n)_n$ such that $ A = \{a_n : n \in \Bbb N \} $.

I know that there exists a surjection from the naturals onto A, but can we then define that surjection as a sequence?

I know this should be simple. But I just don't know how to write a nice, technically sound proof.

Thanks

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    Thanks Kevin. I'm sorry if this seemed trivial.2012-09-30

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By definition there must be a injection to the set of natural numbers. So there is a uniq natural number $k_{n}$ asigned to each element of the set. Taking this numbers from smallest to biggest $k_{1} as index $n=k_{n}$ for $a_{n}$ you are done.

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    $a_{1}=f^{-1}(2011)$. If the set is finite you could asign the last element of the set to all naturals that are left. So sequence would have infinite equal elements $k_{n}$ for $n$ bigger then the cardinality of the set2012-10-06