Show that $(a_k)$ is a Cauchy sequence in this metric space

Question

Context

The metric space $(ℓ_0, d)$, where $$ℓ_0 = \left\{(a_n) : a_n ∈ \mathbb{R}\text{ for each } n,\text { and } (a_n)\text{ is eventually zero} \right\}$$ and

$d: ℓ_0 \times ℓ_0 \rightarrow \mathbb{R}$ is given by $$d(a, b)=\sum_{n=1}^{\infty}{|a_n-b_n|},\text{ where } a = (a_n) \text{ and } b=(b_n)$$

Let $p$ be a prime and let $(a_k)$ be the sequence of points in $ℓ_0$ given by $$a_k=\left\{\frac{1}{p},\frac{1}{p^2},\frac{1}{p^3},...,\frac{1}{p^k},0,0,0,...\right\}\text{ for } k\in \mathbb{N}$$

(i) Calculate $d(a_{10}, a_9)$

$$\frac{1}{p^{10}}$$

(ii) Find $d(a_{k+1}, a_k)$ for each $k \in \mathbb{N}$

$$\frac{1}{p^{k+1}}$$

My question and attempt

(iii) Hence show that $(a_k)$ is a Cauchy sequence.

$$|a_k-a_{k-n}|=|(a_{k-n}-a)+(a_{k-n+1}-a)+...+(a_{k+1}-a)+(a_k-a)|$$ $\text {then due to the triangle inequality for modulus:}$ $$|a_k-a_{k-n}|\leq|a_{k-n}-a|+|a_{k-n+1}-a|+...+|a_{k+1}-a|+|a_k-a|$$ $\text { then because the limit } a \text{ is zero}$ $$|a_k-a_{k-n}|\leq|a_{k-n}|+|a_{k-n+1}|+...+|a_{k+1}|+|a_k|$$ $$|a_k-a_{k-n}|\leq \sum_{n}{\frac{1}{p^n}}$$ $$|a_k-a_{k-n}|\leq \frac{p^{-1}(1-(p^{-1})^n)}{1-p^{-1}}$$ $$|a_k-a_{k-n}|\leq \frac{p^{-1}-p^{-(n+1)}}{1-p^{-1}}$$ $$|a_k-a_{k-n}|\leq \frac{p^n-1}{p-1}$$ $\text{From here I would say we can always choose a } k,m \geq \mathbb{N}\text{ where } m=k-n \text{ so that }$ $$|a_k-a_{k-n}|\leq \frac{p^n-1}{p-1} \leq \frac{p^{\mathbb{N}}-1}{p-1}$$

I'm struggling greatly with how to present this especially with notation, if the logic of my answer is even correct in the first place. I would appreciate any critiques or corrections.

Answer 1

For $(a,b)\in(\ell_0)^2\ $ let's define $\ d(a,b)=\sum\limits_{n=1}^\infty|a_n-b_n|$

Also let's have $N(a)=\inf\{ n\in\mathbb N\mid \forall m\gt n, a_m=0\}$, it always exists since sequence in $\ell_0$ are eventually null.

Let's call $M(a,b)=\max(N(a),N(b))$.

• $d(a,b)=\sum\limits_{n=1}^{M(a,b)}|a_n-b_n|$ is a finite sum, so it is well defined and positive.

• $d(a,b)=d(b,a)$ is trivially verified since $|a_n-b_n|=|b_n-a_n|$

• $d(a,b)=0\iff\forall\ n\le M(a,b),\ |a_n-b_n|=0$ since it is a sum of positive terms, so for $n\le M(a,b), a_n=b_n$ and for $n>M(a,b), a_n=b_n=0\iff a=b$.

• $|\cdot|$ verifies the triangular inequality, and $d(\cdot,\cdot)$ is a finite sum, thus linear, so $d(\cdot,\cdot)$ also verifies the triangular inequality.

Note: you can redact the fourth point in details if you want, but I think it is not necessary.

So $d$ is a distance in $\ell_0$.

Let's have $\displaystyle (a_k)_k=\{\frac 1p,\frac 1{p^2},..,\frac 1{p^k},0,0,...\}$ for $k\in\mathbb N^*$ and $p$ prime.

For $n\in\mathbb N,\ N(a_k)=k$ and $N(a_{k+n})=k+n$ so

$\begin{array}{l} \displaystyle d(a_k,a_{k+n})=\sum\limits_{i=1}^\infty|{a_k}_i-{a_{(k+n)}}_i|=\\ \displaystyle \phantom{d(a_k,a_{k+n})}=\sum\limits_{i=1}^k|{a_k}_i-{a_{(k+n)}}_i|+\sum\limits_{i=k+1}^{k+n}|{a_k}_i-{a_{(k+n)}}_i|+\sum\limits_{i=k+1+n}^\infty|{a_k}_i-{a_{(k+n)}}_i|\\ \displaystyle \phantom{d(a_k,a_{k+n})}=\sum\limits_{i=1}^k |\frac 1{p^i}-\frac 1{p^i}|+\sum\limits_{i=k+1}^{k+n}|0-\frac 1{p^i}|+\sum\limits_{i=k+1+n}^\infty |0-0|\\ \displaystyle \phantom{d(a_k,a_{k+n})}=\sum\limits_{i=k+1}^{k+n}\frac 1{p^i}=\frac 1{p^{k+1}}\sum\limits_{i=0}^{n-1}\frac 1{p^i}=\frac 1{p^{k+1}}\frac{1-{\frac 1p}^n}{1-\frac 1p}\\ \end{array}$

Since $p$ is a prime number $0\le \frac 1p\le\frac 12$ and $\bigg|\frac{1-{\frac 1p}^n}{1-\frac 1p}\bigg|\le 2$

So $\displaystyle d(a_k,a_{k+n})\le \frac 2{p^{k+1}}\to 0$ when $k\to+\infty$ independantly of the $n$ chosen.

So $(a_k)_k$ is a Cauchy sequence in $(\ell_0,d)$.