# Prove that the normed vector space $(S_F,\|\cdot\|_1)$ is not Banach.

$S_F$ is the space of real sequences $\mathbf a=(a_n)_{n=1}^{\infty}$ such that every sequence $\mathbf a\in S_F$ is eventually zero. $\|\cdot\|_1$ is the norm defined as $\|\mathbf a\|_1=\sum_{n=1}^{\infty}\lvert a_n\rvert$.

I know that a Banach space is one where every Cauchy sequence in $S_F$ converges to an element of $S_F$ but I don't know how to prove that a normed vector space is not Banach. Can anyone help?

• Hint: $S_F$ is a subspace of $\ell^1$. If it were complete, it would have to be closed... – user38355 Nov 19 '14 at 15:40

Let $$x^{(M)}\in S_F$$ be a sequence where $$x_n^{(N)} = 1/n^2$$ if $$n\leq N$$ and zero elsewhere. Note that $$\sum_n 1/n^2 < \infty$$. If $$M>N$$, then $$\lVert x^{(M)}-x^{(N)}\rVert = \sum_{k=N+1}^M 1/k^2\to 0$$ as $$N,M\to \infty$$, so the sequence $$(x^{(N)})_{N\in \mathbb{N}}$$ is cauchy.
Now we need to show that no point in $$S_F$$ can be the limit. Suppose $$y\in S_F$$ would be such that $$x^{(N)}\to y$$ as $$N\to\infty$$. Now there exists $$M$$ s.t. for all $$n>M$$, $$y_n=0$$. Now if $$N>M$$, $$\lVert y-x^{(N)}\rVert\geq 1/(M+1)^2$$, and thus $$x^{(N)}$$ cannot converge to $$y$$, which is a contradiction. This shows that $$S_F$$ cannot be complete.
You need to find a Cauchy sequence that is not convergent. In your case you could take an infinite sequence $\mathbf a$ such that $\|\mathbf a\|_1<\infty$, and approximate it by elements in $S_F$.