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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Sentence:

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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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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