1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ "which is a Cauchy sequence obviously". No it isn't. $\endgroup$ – Lorem Ipsum Jan 16 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 at 19:24
  • $\begingroup$ Have you looked at the $p$-adic norm of $p^n$? $\endgroup$ – Paul Sinclair Jan 17 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 at 14:02
1
$\begingroup$

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 $$\frac{1}{1-p}=\sum_{i=0}^\infty p^{i}\hspace{1cm}\mbox{ and }\hspace{1cm}-1=\sum_{i=0}^\infty (p-1)p^{i}$$

$\endgroup$
0
$\begingroup$

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 $8$ .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.