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

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!

• "which is a Cauchy sequence obviously". No it isn't. – Lorem Ipsum Jan 16 '19 at 19:22
• 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? – Lorem Ipsum Jan 16 '19 at 19:24
• Have you looked at the $p$-adic norm of $p^n$? – Paul Sinclair Jan 17 '19 at 0:12
• An interesting related result. A $p$-adic number is rational iff its $p$-adic expansion is eventually periodic. – 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 $$\frac{1}{1-p}=\sum_{i=0}^\infty p^{i}\hspace{1cm}\mbox{ and }\hspace{1cm}-1=\sum_{i=0}^\infty (p-1)p^{i}$$

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 $$8$$ .