Space of real sequences with finitely many nonzero elements is a Banach space?

Question

Let $X$ be a vector space (over $\mathbb{R}$) consisting of all real-valued sequences, which have almost all elements equal to $0$.

For $x:= (x_n)_{n\in \mathbb{N}}^{\infty} \in X$ we define its norm as $N(x):= \sup{ \left \{ \frac{\left | x_n \right |}{n}: n\in \mathbb{N} \right \}}$.

Is $(X, N)$ a Banach space?

A sequence $(y_n)_{n\in \mathbb{N}}^{\infty} \subset X$ is a Cauchy sequence, if for all $\varepsilon >0$ and $k, l$ large enough, $N(y_k - y_l) < \varepsilon$, but here I have problems with understanding what is a norm of such difference since I need to consider a sequence of sequences. I think this may not be Banach since the norm is defined with supremum of nonzero elements and it could possibly tend to something different than zero, but I am confused with double indices.

Answer 1

No the sequence $u_n$ such that $u_n(m)=1, m\leq n$ and $u_n(m)=0, m>n$ is a Cauch sequence which does not converge.

Answer 2

Consider the sequence $\{x_n\}$ in $X$ defined by $x_n(k)=1$ if $1\leq k\leq n$, and $x_n(k)=0$ otherwise. If $n<m$, then $$ ||x_n-x_m||=\frac{1}{n+1} $$ hence $\{x_n\}$ is a Cauchy sequence in $X$. Can $\{x_n\}$ have a limit in $X$?

Answer 3

The answers posted above should suffice to answer your question.

More generally, if $X$ is an infinite dimensional vector space whose dimension is countable, then for any norm on $X$, $X$ is not complete and, thus, not a Banach space.

(To see that the space you have described has countable dimension, consider the sequence of elements ${\{e_n\}}_{n\in\mathbb{N}}$ in the space such that:

${e_n}(k)=0,$ for $k\neq n$ and $e_n(n)=1$)

As for the proof of this fact, it follows from the following:

Let $\{{x_n}\}_{n\in\mathbb{N}}$ be a countable Hamel basis of $X$ and for $n\in\mathbb{N}$ consider the subspace:

$F_n=span\{x_k:k\leq n\}$

Then, $F_n$ is a closed subspace of $X$ for each $n\in\mathbb{N}$ (as it is finite dimensional)

It is obvious that $X=\cup_{n\in\mathbb{N}}F_n$ thus since $X$ is a Banach space, by the Baire category theorem we would have that:

$intF_{n_0}\neq\emptyset$ for some $n_0 \in \mathbb{N}$

But this implies that $F_{n_0}=X$, thus $X$ is finite dimensional, a contradiction.