Proof by hand that $\mathbf{Q}_p$ is complete

Question

We take the completion of $\mathbf{Q}$ with respect to the p-adic norm in the following way. We take the ring of all Cauchy sequences $C$ modulo the maximal ideal $M$ of all null sequences and we define $\mathbf{Q}_p:=C/M$. There is a non-archimedean norm on $\mathbf{Q}_p$: for an equivalence class $x\in \mathbf{Q}_p$ we take a representative Cauchy sequences $(a_n)$ and we define $|x|=\lim |a_n|$.

I proved that $\mathbf{Q}$ is dense in $\mathbf{Q}_p$. Indeed if $x\in \mathbf{Q}_p$, then we take a representative Cauchy sequence $(a_n)$ in $\mathbf{Q}$. Let $\epsilon>0$, then there exists $N\in \mathbf{N}$ such that for all $m\geq N$ we have $|a_m-a_N|<\epsilon$. Now take the element in $y\in \mathbf{Q}_p$ represented by the constant sequence $(a_N,a_N,a_N,\ldots)$. Then $|x-y|=\lim |a_N-a_m|\leq \epsilon$. $\square$

I can find nowhere in literature a straight forward proof by hand that $\mathbf{Q}_p$ is complete. It is always left as an exercise, which I understand because it is very involved to work with "Cauchy sequences of Cauchy sequences of cauchy sequences etc.". So I decided to try it myself, but I didn't succeed..

My try. Let $(a^{(n)})$ be a Cauchy sequence in $\mathbf{Q}_p$. That means that for every $n$, $a^{(n)}\in \mathbf{Q}_p$, so we can find a representative Cauchy sequence in $\mathbf{Q}$ which we denote $(a^{(n)}_j)_j$. I have to find an element $b\in \mathbf{Q}_p$ such that $0=\lim_{n\to\infty }|a^{(n)}-b|=\lim_{n\to\infty} \lim_{j\to\infty} |a^{(n)}_j-b_j|$...

I am looking for a clear proof (it can use density which I already proved) that $\mathbf{Q}_p$ is complete wrt this norm.

Answer 1

This can be approached from the point of view of metric spaces in general. As mentioned in the comments, by using the completion of a metric space by the Cauchy sequence space construction. The density result you proved can be used as it is used in the lemma below.

Let $X$ is a metric space with metric $d$. Let $\widehat{X}$ denote the set of all Cauchy sequences $\widehat{x}=(x_1,x_2,\dots)$ of points of $X$. Define $\widehat{x}\sim \widehat{y}$ if $d(x_n,y_n)\to 0$. The relation $\sim$ is an equivalence relation.

Let $[\widehat{x}]$ denote the equivalence class of $\widehat{x}$, and let $Y$ denote the set of these equivalence classes. Define a metric on $D$ on $Y$ by the equation $$D([\widehat{x}],[\widehat{y}])=\lim_{n\to \infty}d(x_n,y_n).$$ The map $h:X\to Y$ defined by $h(x)=[(x,x,\dots)]$ is an isometric embedding.

The set $h(X)$ is dense in $Y$, in fact, given $\widehat{x}=(x_1,x_2,\dots)\in \widehat{X}$, the sequence $h(x_n)$ of points in $Y$ converges to the point $[\widehat{x}]$.

The metric space $(Y,D)$ is complete because of the following lemma.

Lemma: If $A$ is a dense subset of a metric space $(Z,\rho)$, and if every Cauchy sequence in $A$ converges in $Z$, then $Z$ is complete.

Proof: Let $(z_n)$ be a Cauchy sequence in $Z$. Since $A$ is dense in $Z$, for each $n$ we can choose $x_n\in B\left(z_n,\frac{1}{2^n}\right)\cap A$.

Given $\varepsilon>0$, let $N$ be such that $m,n\geq N\implies \rho(z_m,z_n)<\frac{\varepsilon}{3}$ and that $\frac{1}{2^N}<\frac{\varepsilon}{3}$.

Then for $m,n\geq N$ we have: $$\rho(x_m,x_n)\leq\rho(x_m,z_m)+\rho(z_m,z_n)+\rho(z_n,x_n)<\varepsilon$$ Therefore, $(x_n)$ is a Cauchy sequence in $A$ and then it's convergent to some $x\in Z$.

Given $\varepsilon>0$, there exists $N$ be such that $n\geq N\implies \rho(x_n,x)<\frac{\varepsilon}{2}$ and that $\frac{1}{2^N}<\frac{\varepsilon}{2}$.

Then $n\geq N\implies \rho(z_n,x)\leq\rho(z_n,x_n)+\rho(x_n,x)<\varepsilon$.

Thus $(z_n)$ is convergent to $x$ and $Z$ is complete. $\blacksquare$

Answer 2

This is only a sketch, ask me for any particular details. So as in the comments, we write out a table of entries where the rows correspond to terms in our Cauchy sequence (edit: p-adic "decimal representation" which js to say we pick a very specific sequence for each number) in $\mathbb Q_p$ and each row is a sequence in $\mathbb Q$ that converges to the number corresponding to that row.

The limit point will be the following: Note that for any $k$ and a large enough row index, the first $k$ entries in each row don't change as we move down the rows. This is equivalent to the saying that the original sequence was Cauchy. So we pick the first $k$ entries of our limit point to be defined by these constant $k$ entries and let $k\to \infty$.

This is clearly a Cauchy sequence of rationals so converges to something in $\mathbb Q_p$ that will be our limit point. To show that our original sequence converges to this limit point, let us subtract off this sequence from each row. Then for large enough row indices, the first $k$ entries will be $0$ so it is indeed a limit point.

Edit: Here's an attempt at a more detailed answer:

Given a Cauchy sequence $x_n$ of elements in $\mathbb Q_p$ we can assume all but finitely many of them belong to $p^{-k}\mathbb Z_p$ since this is compact and rescaling, we can assume $k = 0$, let us write $x_n = \sum_{m\geq 0}y_{nm}p^m$.

Lemma: For any $k > 0$, we can pick $N > 0$ so that $y_{nl}$ is constant in $n$ for $n \geq N$ and $l \leq k$.

Proof: To say that $x_n$ is Cauchy is to say that $|x_n - x_{n+1}| \to 0$ as $n\geq \infty$ or equivalently, in the difference, the initial terms are zero for longer and longer sequences.

Now let us define the limit $y_k = \lim_{k\to \infty} y_{nk}$ and this is just eventually equal to $y_{nk}$ for $n \gg 0$ and $y = \sum_{k\geq 0}y_kp^k$.

Claim: Then $\lim_k x_k = y$.

Proof: To show this, we need to show that $y-x_k$ has longer and longer initial sequences of zeroes. This also follows from the previous lemma and $y_{nk}$ being an eventually constant sequence.