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

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 $x_1,x_2,\cdots,x_n$, $$|x_1+x_2+\cdots+x_n|_p\le \max\{|x_1|_p,|x_2|_p,\cdots,|x_n|_p\}$$

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

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

If $x\ne 0$ 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 $a_n=\displaystyle\sum_{k=1}^n\frac{1}{k!}$, 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!

  • 1
    $\begingroup$ "which is a Cauchy sequence obviously". No it isn't. $\endgroup$ – Lorem Ipsum Jan 16 '19 at 19:22
  • $\begingroup$ How would you answer the analogous question with the usual norm, i.e. how would you write down a Cauchy sequence of rational numbers whose limit is not a rational number? $\endgroup$ – Lorem Ipsum Jan 16 '19 at 19:24
  • $\begingroup$ Have you looked at the $p$-adic norm of $p^n$? $\endgroup$ – Paul Sinclair Jan 17 '19 at 0:12
  • $\begingroup$ An interesting related result. A $p$-adic number is rational iff its $p$-adic expansion is eventually periodic. $\endgroup$ – GEdgar Jan 17 '19 at 14:02

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 $\mathbb{Q}_p\setminus\mathbb{Q}$. In order to identify the elements of $\mathbb{Q}_p\setminus\mathbb{Q}$, we need the following theorem.

Theorem: Let $x=\sum_{i=v}^\infty r_i p^i\in\mathbb{Q}_p$ $(v\in\mathbb{Z},\; 0\leq r_i\leq p-1)$. 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 $n\in\mathbb{N}$ such that the subsequence $(r_i)_{i\geq n}$ 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 $x=\sum_{i=0}^\infty p^{i^2}\in\mathbb{Q}_p$. 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 $n\in\mathbb{N}$. Thus $(a_n)_n$ is a Cauchy sequence in $(\mathbb{Q},|\;\;|_p)$ but the theorem implies that $x\not\in\mathbb{Q}$, 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 $$\frac{1}{1-p}=\sum_{i=0}^\infty p^{i}\hspace{1cm}\mbox{ and }\hspace{1cm}-1=\sum_{i=0}^\infty (p-1)p^{i}$$

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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 $p\neq 2$. A more elaborate calculation is needed. Using the binomial function $X(X-1)...(X-n+1)/n!$ , one can show that $\sqrt 2\in \mathbf Q_p$ iff $p\equiv \pm 1$ mod $8$ .

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