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While I was reading Enderton's "A mathematical introduction to Logic", I came across the proof of the following sentence: "The set of all finite sequences of members of the countable set A is also countable".

Proof: The set S of all such finite sequences can be characterized by the equation $$S=\bigcup_{n \in N} A^{n+1}$$ Since A is countable, we have a function f mapping A one-to-one into N. The basic idea is to map S one-to-one into N by assigning to $(a_0,a_1,...,a_m)$ the number $2^{f(a_0)+1}3^{f(a_1)+1}\cdot ... \cdot p_m^{f(a_m)+1}$, where $p_m$ is the $(m+1)$st prime. This suffers from the defect that this assignment might not be well-defined. For conceivably there could be $(a_0,a_1,...,a_m)=(b_0,b_1,...,b_n)$, with $a_i$ and $b_j$ in A but with $m\neq n$. But this is not serious; just assign to each member of S the smallest number obtainable in the above fashion. This gives us a well-defined map; it is easy to see that it is one-to-one.

Note: P is a finite sequence of members of A iff for some positive integer $n$, we have $P=(x_1,...,x_n)$, where each $x_i \in A$.

First of all, I cannot understand why the former assignment might not be well-defined and the latter assignment is well-defined. Secondly, I cannot understand what Enderton means by "just assign to each member of S the smallest number obtainable in the above fashion". By the way, is $(a,b,c,d) = ((a,b),(c,d))$ true? Also, in which cases can I omit/add parentheses in a tuple so as to have an equal tuple?

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I don't see how two sequences with different length can be identical. How are $(x_1,\ldots,x_n)$ defined? –  Apostolos Jun 24 '12 at 16:01
    
Some $a_i$ might be itself a finite sequence of $b_j$'s, or the other way around. For example, if $n>m$, then $(a_1,...,a_m) = (b_1,...,b_m,...,b_n)$, where $n=m+k$. That is, $a_1 = (b_1,...,b_{k+1})$. –  Stavros Jun 24 '12 at 16:26
    
The elements of $S$ are of the form $(x_1,\ldots,x_n)$ where $x_i\in A$. Then, I don't see the point of considering sequences of sequences, it is irrelevant to the definition. Am I wrong? –  Apostolos Jun 24 '12 at 16:35
    
Honestly! I don't know. Both definition and proof are somewhat ambiguous. $A$ might be the set whose members are 1/2/.../n-tuples. –  Stavros Jun 24 '12 at 16:43
    
Apostolos, do you know in which cases I can omit/add parentheses in a tuple so as to have an equal tuple? –  Stavros Jun 24 '12 at 17:15

2 Answers 2

up vote 1 down vote accepted

So maybe it is a good idea to move part of the comments here, since they are relevant to answering the question:

First of all, there is no unique definition of $(a,b,c,d)$, but the standard one is $(a,(b,(c,d)))$. My guess is that depending on the definition it may be impossible that a problem can arise, but using the definition I provided it is possible to have a conflict.

The problem arises in the case some tuple of elements of $A$ is also in $A$. Let for example let's assume that $a,b\in A$ but at the same time $(a,b)\in A$. Furthermore let's assume that $f(a)=1,f(b)=2,f((a,b))=3$. Then the triple $(a,a,b)$ by the standard definition is $(a,(a,b))$. Then it is not clear whether to send $(a,(a,b))$ to $2^2\cdot3^4$ or to $2^2\cdot 3^2\cdot 5^3$.

What Enderton suggests is to pick the least $n$ such that the sequence $P$ lies in $A^n$. Hence in our previous example we should choose to denote $(a,(a,b))$ with $2^2\cdot 3^4$ since this element is in $A^2$ and in $A^3$ but $2<3$.

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Thanks! Your explanation was quite straightforward and detailed. Ι think that Ross Millikan's assignment is not well-defined due to the fact that $A$ might be equal to $\mathbb{Q}$. Then it could be $\prod_{k=0}^m p_k^{a_k} \notin \mathbb{N}$ for some $(a_0,a_1,...,a_m)$. –  Stavros Jun 24 '12 at 20:33

You don't define what $f$ is, but as long as it is injective and positive I don't see how you get a conflict. In particular, if $n \gt m$, one of the numbers will be divisible by $p_{m+1}$ and one will not. It looks simpler to me to take $(a_0,a_1,\ldots,a_m)$ to $2^{a_0}3^{a_1}\ldots p_m^{a_m}$. Unique factorization shows that each tuple goes to a distinct image.

For your last we haven't defined tuples of tuples, which is what $((a,b),(c,d))$ is. If we map it recursively using my function $(a,b)\to 2^a3^b$, so $(a,b,c,d) \to 2^a3^b5^c7^d$ while $((a,b),(c,d)) \to 2^{2^a3^b}3^{2^c3^d}$, which are not equal.

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First of, thanks for your answer. Enderton only says that f is injective. If A is the set whose members are n-tuples, then your assignment is not well-defined. As for my last question, I think that we have defined recursively tuples of tuples. For example, $(a,b,c,d) = ((a,b,c),d) = (((a,b),c),d)$. However, I don't know if there are any other cases in which I can add/omit the parentheses. –  Stavros Jun 24 '12 at 16:21
    
I was taking A to be a countable set, and in particular $\mathbb N$ which we can using the bijection that witnesses countability. Then I think this shows that the finite sequences from A are countable. If you want to show that the sequences of sequences are countable, we can just apply the theorem twice. –  Ross Millikan Jun 24 '12 at 16:29
    
By the way, if $A=Q$, then your assignment is well-defined? And in which case there could be $(a_0,...,a_m)=(b_0,...,b_m)$? Please enlighten me! Furthermore, I haven't understood what Enderton means by "just assign to each member of S the smallest number obtainable in the above fashion". What is the smallest number? –  Stavros Jun 24 '12 at 18:22
    
@Stavros: for the first, if one tuple is longer than the other, then the primes in the expansion for the short one only go up to $p_m$. For the second, just inject $\mathbb Q$ into $\mathbb N$ and choose the primes according to the injection. –  Ross Millikan Jun 24 '12 at 21:07

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