Is this space complete or is it incomplete?

Question

Show (if possible) that the space of all complex sequences $x=(x_n)$ with only a finite number of terms nonzero (the number of nonzero terms may be different for different members of the space) is complete. We are given a norm $\|x\|=\left(\sum|x_i|^2\right)^{1/2}$.

I have verified that this is a normed vector space since this norm satisfies the four axioms of a normed vector space.

We must show that every Cauchy sequence converges to a point in the space or exhibit a sequence that contradicts the claim.

I was given the following hint, but I'm not sure what to do.

Hint: Consider the sequence $(x_n)$ of sequences $x_n=(1,1/2,1/3,...,1/n,0,0,...)$.

Answer 1

First, an easy exercise: show that the sequence is Cauchy.

Now you know that, in order for the space to be complete, that sequence had better converge. If it does converge, what would it have to converge to? And is that "point" actually a point in your space?

Answer 2

This is a good hint. One useful idea would be to work in a space that we know is complete and see how it compares. So, observe that the space of all eventually zero complex sequences, call it $X$, is a subspace of $\ell_2$. That $\ell_2$ is complete is standard. See this wikipedia page for some details on sequence spaces. In particular, note that $X$ inherits the norm from $\ell_2$.

Now, only two details need to be checked. First, see that $(1,1/2,1/3,...) \in \ell_2$.

Second, check that $x_n \to (1,1/2,1/3,...)$ in $\ell_2$.

These two facts demonstrate that $X$ is not complete.

Answer 3

Let $x^{(n)}=\left(1,\frac12,\ldots,\frac1n,0,0,\ldots\right)$ for $n=1,2,\ldots$. Then if $n>m$, $$\left\|x^{(n)}-x^{(m)}\right\| = \left(\sum_{k=m+1}^n |x_k^{(n)}|^2\right)^{\frac12} = \left(\sum_{k=m+1}^n \frac1{k^2}\right)^{\frac12}. $$ Now, as $\sum_{k=1}^\infty\frac1{k^2}$ is a series with finite terms that converges, we know that $$\sum_{k=m+1}^n \frac1{k^2}\stackrel{n,m\to\infty}{\longrightarrow}0$$ and hence $$\left(\sum_{k=m+1}^n \frac1{k^2}\right)^{\frac12}\stackrel{n,m\to\infty}{\longrightarrow}0.$$ Therefore $x^{(n)}$ is a Cauchy sequence, so it has a limit $x$ in the space $$l^2(\mathbb C)=\left\{y^{(n)}\in\mathbb C^{\infty} : \sum_{k=1}^\infty |y_k^{(n)}|^2<\infty\right\}. $$ However, it is clear that $$x = \left(1,\frac12,\frac13, \ldots\right), $$ and so $x$ is not in our subspace (as it has an infinite number of nonzero terms). Therefore this subspace is not complete.