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$S_F$ 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?

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

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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 $S_F$.

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