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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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That surjection is a sequence. A sequence is nothing but a function with domain $\mathbb{N}$. For the sake of pedantry: Only at most cuntable, nonempty sets can be arranged as a sequence. –  Michael Greinecker Sep 30 '12 at 18:40
    
A sequence is an application from $\mathbb{N}$ to something else. If you have a surjection, that surjection can be used as your sequence... –  xavierm02 Sep 30 '12 at 18:40
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Isn't a surjection from ${\Bbb N}$ to $A$ exactly the same thing as a sequence whose values are the whole of $A$? –  user22805 Sep 30 '12 at 18:42
    
What definition have you learned for "at most countable"? –  vesszabo Sep 30 '12 at 18:48
    
The definition I learned for at most countable is this: A is at most countable is A is finite or countable. –  mkeachie Sep 30 '12 at 19:04
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1 Answer 1

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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 $k_{1}<k_{2}<\cdots<k_{n}$ as index $n=k_{n}$ for $a_{n}$ you are done.

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You are not done yet, there are two small problems to face... For example, if $k_1=2011$, what is $a_1$?Also, if the set is finite, you don't get a sequence, because you only get finitely many indices... –  N. S. Oct 6 '12 at 19:44
    
$a_{1}=f^{-1}(2011)$. 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 set –  Mykolas Oct 6 '12 at 19:50
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