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I have a question about the convergence properties of a sequence in $\hat{\mathbb{Z}}$, the completion of $\mathbb{Z}$. It is part of an exercise is due to this syllabus. I got confused somewhere.

Choose $b \in \mathbb{Z}$. Let us define the sequence $(a_n)_{n \in \mathbb{N}}$ on $\mathbb{Z}$ as follows: $$ a_0 \ := \ b, \quad \quad \text{ and } \quad \quad a_{n+1} \ := \ 2^{a_n} \ \text{ for } n \geq 1 $$ The sequence $(a_n)_{n \in \mathbb{N}}$ can be seen as a sequence on $\hat{\mathbb{Z}}$ as $$ \hat{a}_n \ := \ (a_n + 1!\mathbb{Z}, \ a_n + 2! \mathbb{Z}, \ a_n + 3!\mathbb{Z}, \dots )_n \quad \quad \text{ for } n \in \mathbb{N}. $$ Show that

  • the sequence $(\hat{a}_n)_{n \in \mathbb{N}}$ converges in $\hat{\mathbb{Z}}$
  • the limit $\hat{a}$ does not depend on the choice of $b$.

My attempts

As I silently suggested $ \hat{\mathbb{Z}} \ := \ \varprojlim_{n \in \mathbb{N}}\mathbb{Z}/n\mathbb{Z} \ = \ \varprojlim_{n \in \mathbb{N}}\mathbb{Z}/n!\mathbb{Z}. $ The sequence $\hat{a}_n$ converges iff $$ \forall m \in \mathbb{Z}, \ \exists N \in \mathbb{N} \quad \text{ such that } \quad n \geq N \implies m! \ | \ a_n - a $$

I have no very little idea about what $a \in \mathbb{Z}$ could work. I only know that $2^M \ | \ a_n$ for arbitrarily large $M$ if $n$ is sufficiently large. That means that $a$ has to divide $2$ as many times as $m!$ does. This is valid for any $m$, so $2^M \ | \ a$ for any $M \in \mathbb{Z}$. That means that $a=0$ right? But $a_n$ can never become zero in for example $\mathbb{Z}/6\mathbb{Z}$, so I must be mistaking.

I know that usually we represent the sequence in $\hat{\mathbb{Z}}$ as $$ \left( \sum_{0 \leq n \leq 0} a_n n! + 1!\mathbb{Z}, \ \sum_{0 \leq n \leq 1} a_n n! + 2!\mathbb{Z}, \ \sum_{0 \leq n \leq 2} a_n n! + 3!\mathbb{Z}, \ \dots \right) $$ But the choice of the representation should not matter right? What am I doing wrong?

I would like to have some help to see what I am doing wrong and to make a step to the solution of this exercise. Please do not give entire answers, at least not yet.

Koenraad van Duin
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  • Why do you want to choose $a\in \mathbb Z$? This is certainly not possible. To show that $(a_n)_{n\in\mathbb N}$ converges in $\hat{\mathbb Z}$, it suffices to show that it is a Cauchy sequence (since $\hat{\mathbb Z}$ is complete). – Claudius Jul 14 '16 at 06:23
  • @user218931, I certainly didn't want to choose $a$, I wanted to find it in fact. – Koenraad van Duin Jul 14 '16 at 06:26
  • Moreover, I already tried to prove the Cauchy property. I know what it means and I understand that $\hat{\mathbb{Z}}$ is complete, but I didn't manage the prove the Cauchy property. And even if I could it wouldn't be enough that the limit doesn't depend on $b$. – Koenraad van Duin Jul 14 '16 at 07:26
  • But using (and proving) the Cauchy property is all you can do. If you have a sequence $(x_n)_{n}$ corresponding to another $b$, then all you need to show is that $(a_n-x_n)_n$ is a zero sequence, as then both $(a_n)_n$ and $(x_n)_n$ have necessarily the same limit. – Claudius Jul 14 '16 at 09:00
  • I can't prove that $(a_n-x_n)_n$ is a zero sequence right? The best I can show is that starting from some $N$ it will be. – Koenraad van Duin Jul 14 '16 at 11:01
  • With zero sequence I mean a sequence converging to zero. Otherwise even starting from some $N$ will not give a constantly-zero-sequence (in general). – Claudius Jul 14 '16 at 13:18

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