Show that Cauchy sequence $(a_k)$ isn't d-convergent in the set of eventually zero sequences

Question

I'm doing a bonus homework question in which I'm struggling greatly. It's much tougher than anything that I will be formally assessed on. Any help would be appreciated because I would like to understand this.

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 know the sequence is Cauchy thanks to help I received previously for this specific question here: Previous part of question

My question now is to prove that the metric space $(ℓ_0, d)$ is not complete using the fact that $(a_k)$ is Cauchy.

My attempt

I know that in a complete metric space, every $d$-Cauchy sequence must be $d$-convergent. Hence I need to show that $(a_k)$ isn't $d$-convergent. I want to do a proof by contradiction.

Suppose $(a_k)$ is $d$-convergent. The limit of $(a_k)$, which I will label $(b)$, needs to be in $ℓ_0$ and hence is also a sequence. So we have:

$$(d((a_k), (b)))\rightarrow 0$$

and so

$$(\sum_{i=1}^{\infty}{|(a_k)_i-b|})\rightarrow 0$$

For $(\sum_{i=1}^{\infty}{|(a_k)_i-b|})$ to be a null sequence,

$$\sum_{i=1}^{\infty}{|(a_k)_i-b|}\rightarrow 0 \text{ as }i\rightarrow \infty$$

But every term is non-negative and so it can't be dropping to zero. However, what if $(a_k)$ is equal to its limit $(b)$? Then every term is zero and with the constant terms of zeros, which should not only be in $ℓ_0$, but also $d$-convergent? I phrase these as questions because I've clearly gone wrong, so my understanding must be failing somewhere. Other than that it is pretty much contradicted I think.

Thanks

Answer 1

For clarity, let's denote the elements of your metric space with boldface symbols.

Fix a large integer $N$. For $k$ large enough, the $N$-th entry in $$ {\bf a}^{k} = \left(p^{-1}, p^{-2}, \ldots, p^{-k}, 0, 0, \ldots \right) $$ has to be $p^{-N}$, a strictly positive number.

Therefore, you can go by contradiction: suppose the sequence of the elements ${\bf a}^{k}$ in your metric space, $(l_{0}, d)$, converges to some ${\bf b} = (b_{1}, b_{2}, \ldots)$ in $(l_{0}, d)$. By the definition of your metric space, there has to be an index $M$ such that all the $b$'s with index $M$ or higher are zero. But this would be impossible, since if we choose $N > M$ and $k$ large enough, then $b_{N} = p^{-N} > 0$.

The main difficulty here is to keep track of two indices: the index $k$ of the element ${\bf a}^{k}$ in the metric space, and, regarding that element as a sequence, the index $N$ of an entry in that sequence.

Answer 2

You were right to try contradiction. The best way to approach these kinds of proofs, I feel, is to ask yourself what intuitively goes wrong, why shouldn't this work? For thus problem, it seems that the problem is that $(a_k)$ is converging to an element which is not eventually zero. Let's see if we can make this more rigorous.

I am going to change your notation slightly since it is a little confusing, in my opinion. Let $(b)$ be the hypothetical limit, which we will force to not exist. To show this I will take $n\in \Bbb N$ and show that $b_n>0$. To this end by the definition of convergence note that for some $N\in\Bbb N$, and all $k\geq N$, $$\sum_{t=0}^{\infty}|a_{k,t} - b_t|\leq \frac{1}{2p^n}$$ Now, this implies in particular that $|a_{k,n}-b_n|\leq \frac{1}{2p^n}$ since all the terms are positive. For large enough $k$ this then becomes $$|\frac{1}{p^n}-a_n|\leq \frac{1}{2p^n}$$ and the result follows.