Prove that $V$ is a complete space

Question

Let $V$ be the space of real sequences with a finite number of elements $\neq 0$ ($ \exists N_x$ so that $x_k=0 \forall k>N_x$. Define $$||\vec x||_1=\sum_{k=1}^{N_x}|x_k|$$ Prove that $V$ is a complete space with that norm

I haven´t been able to concrete anything: I need to prove that every cauchy sequence converges.

Let $x$ be a cauchy sequence in $V$(note that $x=(x^1,x^2,...)$ where each $x^n$ is a sequence with a finite number of elements $\neq 0$)

so I need to prove that: $\forall \epsilon>0$ $\exists N(\epsilon)$ so that $\forall n>N$, $||x^n-y||_1<\epsilon$ where $y$ is the limit sequence

By hypothesis we know that $x$ is a cauchy sequence hence: $\exists N_1$ so that $\forall m,n>N_1$ $||x^n-x^m||_1<\epsilon$ that is $$\sum_{k=1}^{l}|x^n_k-x^m_k|<\epsilon$$

I took $l=$max{${p,q}$} where p is there number such that $\forall k>p$, $x^n_k=0$ and $q$ is the number such that $\forall k>q$, $x^m_k=0$

Bu I don´t know how to continue from here. Can you lend me a hand please? I would really appreciate it :)

Answer 1

This is not true because the sequence $$a_n =\sum_{j=1}^n \frac{1}{2^j} e_j $$ is Cauchy but it has no limit.

Answer 2

The space $V$ you're dealing with is a subspace of $l^1$, the set of all real sequences $x$ such that $$ \sum_{k=0}^{\infty}|x(k)| $$ is convergent (note the absolute value; I use function notation because of what follows).

The norm on this space is precisely $$ \|x\|_1=\sum_{k=0}^{\infty}|x(k)|. $$ This is a complete space because it's the set of Lebesgue integrable functions on $\mathbb{N}$ with the “counting measure” and there is no nonempty set with measure $0$, so equality almost everywhere is the same as equality.

Note that $V$ is a subspace of $l^1$ and that the norm induced on $V$ by $l^1$ is exactly the same as in your definition.

The subspace $V$ is dense in $l^1$. If $x\in l^1$, we can define $x_n$ as the “truncation of $x$ beyond $n$”: $$ x_n(k)=\begin{cases} x(k)&\text{if $k\le n$}\\ 0&\text{if $k>n$} \end{cases} $$ and it should be easy to see that $$ \lim_{n\to\infty}x_n=x $$ in the space $l^1$ with the $\|\cdot\|_1$ norm. Moreover, $x_n\in V$, for all $n$.

Therefore $V$ is not complete.