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Our professor gave two examples of spaces of sequences one of which is Banach and the other not:

Consider the space of sequences $X$, where only finitely many terms are non-zero, with the norm defined as $||x|| = \sup \{\frac{|x_n|}{n} : n \in \mathbb{N}\}$. This is apparently a normed space but not Banach.

However, taking $Z$, the space of all sequences with norm defined as above, and requiring that the norm is finite - this is a Banach Space.

I know by definition that I would need to show that every cauchy sequence converges to something in the space to show that $Z$ is Banach, but the way it was stated in class - seemed like these were pretty obvious examples.

Is there an obvious reason or a quick way to see why $X$ is not banach and $Z$ is? Thanks in advance.

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For $X$, take $x_n=(1,1,1,\dots, 1, 0, 0, \dots)$, where the last $1$ is in the $n^{th}$ position. Then for any $n<m\in\mathbb{N}$, $||x_n-x_m||=\frac{1}{n}$ which goes to $0$ as $n\to\infty$, so $(x_n)$ is Cauchy. On the other hand, if $z\in X$ (thus only finitely many non-zero entries), you can find a $N\in\mathbb{N}$ such that $z_k=0$ for all $k\geq N$, hence $x_k-z$ has a $1$ in the $N^{th}$ position. Therefore, for any $k\geq N$, $||x_{k}-z||\geq\frac{1}{N}$. This means $(x_n)$ cannot converge to any $z\in X$.

For $Z$, the easiest is to notice that $(x_n)\in Z$ if and only if $(\frac{x_n}{n})\in l_{\infty}$, and $||(x_n)||_{Z}=||\frac{x_n}{n}||_{\infty}$. Since $l_{\infty}$ is a Banach space, so is $Z$. Alternatively, you can take the long route, proving that $Z$ is a Banach space the same way you presumably did for $l_{\infty}$ in class, with the obvious adjustments.

In fact, your example is a modified version, by changing the norm slightly, of the classical examples of $l_{\infty}$ (a Banach space) and $c_{00}$ (not a Banach space), both with the usual $\sup$ norm.

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  • $\begingroup$ Thanks for your great explanation. It is all much clearer to me now. Just to clarify for my own understanding, in showing that $X$ is not Banach, are you essentially saying that since any $z \in X$ only has finitely many non-zero terms - then there is really no candidate in $X$ which $(x_n)$ can converge to, since any such candidate, going far enough into the cauchy sequence shows that the norm is $> \frac{1}{N}$. $\endgroup$ – Kendrick Easley May 17 '16 at 21:49
  • $\begingroup$ Exactly, the distance to any element in the space is bounded from below. $\endgroup$ – Markus May 17 '16 at 22:20
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HINT.-Put $$a_n=\sum_{k=1}^n\frac 1k$$ Then $x_n=(a_1,a_2,...,a_n,0,0,0,...)\in X$ and the sequence $\{x_n\}$ is Cauchy but it is not convergent because$$\sum_{k=1}^{\infty}\frac 1k$$ is divergent. Hence $X$ is not Banach.

$Z$ is just a quite known Banach space of bounded sequences.

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  • $\begingroup$ Look at the norm, is not the classical sup norm on bounded sequences $\endgroup$ – Markus May 17 '16 at 16:28
  • $\begingroup$ That's right. Thanks. $\endgroup$ – Piquito May 19 '16 at 10:26

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