MHB Every Cauchy sequence converges

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The discussion emphasizes that the p-adic numbers are complete with respect to the p-norm, meaning every Cauchy sequence converges. A proof is provided, demonstrating that if a Cauchy sequence is bounded, it can be transformed into a sequence within the p-adic integers, which are shown to converge. The proof also establishes that the limit of the sequence lies in the p-adic integers. An alternative proof is suggested, utilizing the compactness of the space formed by the finite sets of p-adic integers, leading to the conclusion that the p-adic metric induces a complete topology. Overall, the thread effectively illustrates the completeness of p-adic numbers through various proofs.
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The p-adic numbers are complete with respect to the p-norm, ie every Cauchy sequence converges.

Proof:

Let $(x_i)_{i \in \mathbb{N}}$ a Cauchy-sequence in $\mathbb{Q}_p$.

We want to show that, without loss of generality, we can suppose that $x_i \in \mathbb{Z}_p$.Let the set $\{ |x_i|_p | i \in \mathbb{N}\} \subset \mathbb{R}$. We suppose that it is upper bounded. If not, there are $\forall m \in \mathbb{N}$ and $N \in \mathbb{N}$ two indices $i,j \geq N$ with $|x_i|_p> |x_j|_p \geq p^m$.

From the sentence: For $|x|_p \neq |y|_p$, it stands that $|x+y|_p=\max \{|x|_p, |y|_p \} $, we have that $|x_i-x_j|_p=|x_i|_p>p^m$. This is not possible for a Cauchy-sequence.

Now, we pick $m \in \mathbb{N}$ with $p^m \geq \max \{ |x_i|_p | i \in \mathbb{N}_0 \}$. Then, $(p^m x_i)_i$ is a Cauchy-sequence with $|p^mx_i|_p \leq 1 \Rightarrow p^mx_i \in \mathbb{Z}_p$. It converges exactly then, when $(x_i)_i$ converges.So, we just need to consider a Cauchy-sequence $(x_i)_i \in \mathbb{Z}_p$. We want to show that it converges and to determine its limit $z \in \mathbb{Z}_p$. For each $k \in \mathbb{N}$, we pick a $N_k \in \mathbb{N}$, so that:

$$|x_i-x_j|_p<p^{-k} ,\text{ for } i,j \geq N_k$$

We can suppose, that $N_k$ is an increasing sequence.The above inequality is equivalent to $v_p(x_i-x_j)>k$ or $x_i-x_j \in p^{k+1} \mathbb{Z}_p$. So, we find a $z_k \in \mathbb{Z}$, such that:

$$z_k \equiv x_i \equiv x_i \mod p^{k+1}\mathbb{Z}_p, \text{ for } i,j \geq N_k$$

Because of $N_{k+1} \geq N_k$, it stands that:
$$z_{k+1} \equiv x_{N_{k+1}} \equiv x_{N_k} \equiv z_k \mod p^{k+1} \mathbb{Z}_p$$

ie. $z=(\overline{z_k})_k \in \Pi \mathbb{Z}/p^{k+1}\mathbb{Z}$ is an element from $\mathbb{Z}_p$. Also, for $i \geq N_k$, it stands that:
$$x_i \equiv z_k \equiv \ \mod p^{k+1}\mathbb{Z}_p \Rightarrow |x_i-z|_p<p^{-p}$$

so $x_i$ converges to $z$.
Could you explain me step by step the above proof?
 
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Here is an easier proof that minimizes the use of $\varepsilon$'s . First define $X_n = \mathbb{Z}/p^n \mathbb{Z}$, and give these finite sets the discrete topology. The $p$-adic integers are the projective limit of the $X_n$.

Now consider the product,
$$ X = \prod_{n=1}^{\infty} X_n $$
This is a compact space because each $X_n$ is compact. Furthermore, $\mathbb{Z}_p$ is a closed subset of $X$. Thus, $\mathbb{Z}_p$ is a compact space.

This means that if you can find a metric which describes the topology on $\mathbb{Z}_p$ the metric must automatically be complete (since compact metric spaces are complete). You have to show, which is not so difficult, that the $p$-metric induces the same topology as $\mathbb{Z}_p$ inherits from being a subspace of $X$.

This will prove that $\mathbb{Z}_p$ is complete with respect to the $p$-adic metric. To show that $\mathbb{Q}_p$ is complete you "defractionize" it.
 
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