Dtyjlui

Let $p$ be a prime number. For any ratinoal number $x$, define $$|x|_p =
begin{cases}
0 ,, & mbox{if } ,x=0 \
p^{-alpha},, & mbox{if },x=p^alphafrac{n}{m} ,,,mbox{in which }m,ninmathbb{Z},,mbox{and},,(p,mn)=1
end{cases}$$

We claim that ${a_n}_{n=1}^infty$ is a Cauchy sequence iff $,forall epsilon>0$ , $exists N>0$ s.t. $,forall m,n>N$ we have $|a_m-a_n|_p<epsilon$

We claim that ${a_n}_{n=1}^infty$ $p$-converges to $A$, in which $A$ is a rational number, iff $,forall epsilon>0$ , $exists N>0$ s.t. $,forall n>N$ we have $|a_n-A|_p<epsilon$.

OUR AIM:
Find a Cauchy sequence that doesn't $p$-converge to any rational number.

My thought

I found out one thing that might help as follows,

For rational numbers $x_1,x_2,cdots,x_n$,
$$|x_1+x_2+cdots+x_n|_ple max{|x_1|_p,|x_2|_p,cdots,|x_n|_p}$$

Proof: We only need to prove $|x+y|_ple max{|x|_p,|y|_p}$.

If one of $x$ and $y$ is $0$ , it's obvious.

If $xne 0$ and $y ne 0$ , without loss of generality we let $|x|_p ge |y|_p$ , $x=p{^{{alpha}_{1}}}frac{n_1}{m_1}$ and $y=p{^{{alpha}_{2}}}frac{n_2}{m_2}$.

So $|x+y|_p=|p^{alpha_1}frac{n_1 m_2 + p^{alpha_2-alpha_1} n_2 m_1}{m_1 m_2}|_p le p^{-alpha_1}$ , which yields the conclusion. (Considering $n_1 m_2 + p^{alpha_2-alpha_1} n_2 m_1 in mathbb{Z}$ and $(p,m_1m_2)=1$)

Through this conclusion we can easily get that ${a_n}_{n=1}^infty$ is a Cauchy sequence iff ${a_{n+1}-a_n}_{n=1}^infty$ $p$-converges to $0$.

Then I tried some sequences like $a_n=displaystylesum_{k=1}^nfrac{1}{k!}$, which is Cauchy sequence obviously, but I got stuck on how to prove it doesn't $p$-converges to a rational number $A$.

Any helps or ideas would be highly appreciated!

To answer your question you need to invoke the completion $mathbb{Q}_p$ of $(mathbb{Q},|;;|_p)$. More specifically, you need to find a Cauchy sequence of rational numbers with a limit in $mathbb{Q}_psetminusmathbb{Q}$. In order to identify the elements of $mathbb{Q}_psetminusmathbb{Q}$, we need the following theorem.

Theorem: Let
$x=sum_{i=v}^infty r_i p^iinmathbb{Q}_p$ $(vinmathbb{Z},; 0leq r_ileq
p-1)$. Then $x$ is a rational number if and only if the sequence
$(r_i)_i$ of digits of $x$ is eventually periodic, i.e. there exists
$ninmathbb{N}$ such that the subsequence $(r_i)_{igeq n}$ is periodic.

Proof: Result 5.3 in Robert, Alain M., A course in $p$-adic analysis, Graduate Texts in Mathematics. 198. New York, NY: Springer. xv, 437 p. (2000). ZBL0947.11035.

We can use the theorem above to answer your question (and
to prove the incompleteness of $(mathbb{Q},|;;|_p)$).
Consider $x=sum_{i=0}^infty p^{i^2}inmathbb{Q}_p$.
Note that $x$ is the limit of a convergent sequence
of rational numbers, say $a_n=sum_{i=0}^n p^{i^2}$.
In fact, $|x-a_n|_p=e^{-(n+1)^2}$ for every $ninmathbb{N}$.
Thus $(a_n)_n$ is a Cauchy sequence in $(mathbb{Q},|;;|_p)$
but the theorem implies that $xnotinmathbb{Q}$,
i.e. $(a_n)_n$ is not convergent in $(mathbb{Q},|;;|_p)$.

For examples of Cauchy sequences that $p$-converge to rational numbers you can check that
$$frac{1}{1-p}=sum_{i=0}^infty p^{i}hspace{1cm}mbox{ and }hspace{1cm}-1=sum_{i=0}^infty (p-1)p^{i}$$

I don't quite understand your question. You put on $mathbf Q$ the $p$-adic valuation and you show that a sequence of rationals is $p$-adically Cauchy iff the $p$-adic distance between 2 consecutive terms tends to $0$ (because of the "ultrametric inequality"). Then you ask for a Cauchy sequence that doesn't converge $p$-adically in $mathbf Q$. For this, just construct the $p$-adic completion of $mathbf Q$, which is the field $mathbf Q_p$ of $p$-adic numbers (in the same way as the archimedean completion of $mathbf Q$ is $mathbf R$)! Your question is then equivalent to the $p$-adic non completeness of $mathbf Q$. This has nothing to do with the characterization of Cauchy sequences. The usual reason invoked for the archimedean non completeness of $mathbf Q$ is that $mathbf R$ is not countable (in fact card $mathbf R=aleph_1$). The unique expansion of any non null $p$-adic number as the sum of a polynomial in $1/p$ and a power series in $p$ shows that card $mathbf Q_p$ is also $aleph_1$ .

Addendum. If you ask for concrete examples, here are two parallel illustrations. In the archimedean case, the classical proof of the irrationality of $sqrt 2$ uses the unique factorization in $mathbf Z$: if $sqrt 2=m/n$, where $m, n$ are two coprime integers, it would follow that $2n^2=m^2$, and then unique factorization (up to a sign) would be contradicted. In the $p$-adic case, the same reasoning works at the beginning: $2n^2=m^2$ in the ring $mathbf Z_p$ of $p$-adic integers, but here unique factorization is up to units (=invertible elements) of $mathbf Z_p$ (whereas the only units of $mathbf Z$ are $pm 1$), and $2$ is a unit if $pneq 2$. A more elaborate calculation is needed. Using the binomial function $X(X-1)...(X-n+1)/n!$ , one can show that $sqrt 2in mathbf Q_p$ iff $pequiv pm 1$ mod $8$ .