Prove the following space is not complete.

Question

I am asked to prove the following space of sequences is not complete:

$$D = \{ a = \{a_n\}_{n\in\mathbb{N}};\ a_n\in\mathbb{C};\ a_n\neq 0\quad \text{for a limited number of elements}\}$$

I followed this reasoning: let's take a sequence of elements of $D$ like

$$a^{(N)} = \left(1, \frac{1}{2}, \ldots , \frac{1}{N}, 0, 0, 0,...\right)$$

I can prove this is a Cauchy sequence, indeed

$$||a^{(N)} - a^{(M)}|| = \max\left\{\dfrac{1}{N}, \frac{1}{M}\right\} \to 0\qquad \text{as} \quad N,M\to +\infty$$

Yet I cannot understand the hint in the notes to prove $D$ is not complete.

The hint reads: "choose the limit sequence $a = \left(1, \frac{1}{2}, \ldots, \frac{1}{N}, \frac{1}{N+1}, \ldots\right)$ and see this does not belong to $D$..."

Can someone explain me this? I tried to do something but I don't understand why it doesn't belong to $D$. It's a Cauchy sequence, and converges.

Than you

Answer 1

Bear in mind the elements of $D$ are not numbers but sequence of numbers. When you say "let's take a sequence of elements of D like..." not that $a^N = (1,\frac 12, ....,\frac 1N, 0,0,0...)$ that $a^N$ is NOT the sequence. $a^N$ is just a single elemnt of the sequence. The actual sequence is:

$\{a^1, a^2, a^3, ......., a^N, a^{N+1}......\}$

Now somehow it looks like a norm was defined. I don't know how it was defined but if $||a^N - a^M||=\min\{\frac 1N, \frac 1M\}$. So we can define a concept of the limit of a sequence of sequence, and $n\to \infty$ we have $a^{N} \to $ the sequence $\{1, \frac 12, \frac 13, ......., \frac 1N, \frac 1{N+1},......\}$ and this is an infinite sequence.

As it is an infinite sequence it is not an element of $D$. (The sequence in $D$ are all with finite non-zero terms). So $\{a^1, a^2, a^3,......\}$ is a sequence of sequences that converges to a sequence that is not in $D$.

Hence $D$ is not complete.

Answer 2

The hint is, strictly speaking, wrong, or perhaps a little incomplete. The "limit sequence" $a$ is not an element of $D$, so it does not make (pedantic) sense to even call $a$ a limit sequence without passing to some ambient space (from the eyes of $D$, there is no such thing as '$a$' and to say $(a^{(n)})\to a$ in $D$ is nonsense). To see $D$ is not complete, we really need to show that the $(a^{(n)})_n$ have no limit in $D$ (if you like, we want to show that there is a 'hole' where $a$ 'should be').

I assume the supremum norm-metric on $D$.

Here is a hint that you'll hopefully find more helpful:

• Suppose $\alpha\in D$, $\alpha=(\alpha_1,\alpha_2,\cdots,\alpha_K,0,0,0,\cdots)$ is a limit of the $(a^{(n)})_n$.
• For large enough $i$ and $n$, $\alpha_i$ is zero but $a^{(n)}_i$ is not. Can you find a contradiction to the condition $\|a^{(n)}-\alpha\|\to0$?

A further hint is below.

Fix an $N\in\Bbb N$ with $N>K$: there exists $M\in\Bbb N$ such that $\|\alpha-a^{(M)}\|<1/N$. But, what can you say about: $$\max_{i\in\Bbb N}|\alpha_i-a^{(M)}_i|$$Considering that $\alpha_N=0$?