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Let $\mathbb{R}^\infty$ the vector space of sequences $(x_n)_{n\in\mathbb{N}}$ of real numbers, such that has a finitely many terms $x_n\neq 0$. Define on $\mathbb{R}^{\infty}$ the norm $\lVert x\rVert=\sqrt{\sum_{i=1}^\infty x_i^2}$, for $x=(x_n)_{n\in\mathbb{N}}$. Show that $(\mathbb{R}^\infty,\lVert\cdot\rVert)$ is not complete space.

The prove of this question -is similar to show that $\ell^{2}$ with the norm $\lVert x\rVert=\sqrt{\sum_{i=1}^\infty x_i^2}$, for $x=(x_n)_{n\in\mathbb{N}}$, is a complete metric space-. But, how the fact that "has a finitely many terms $\neq 0$" implies that space isn't complete?

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  • $\begingroup$ @CaptainLama It is not Cauchy, is it? $\endgroup$
    – Clement C.
    Mar 10 '16 at 3:41
  • $\begingroup$ Yes, it was stupid, I took the first thing that popped into mind. Anyway, to make up for it, a (hopefully) intelligent comment : if you know Baire's theorem, you can show that no countable-dimensional normed vector space can be complete. $\endgroup$ Mar 10 '16 at 3:42
  • $\begingroup$ @captain lama. $\mathbb{R}^{n}$ is countable dimensional normed space and it's complete. $\endgroup$
    – T. Eskin
    Mar 10 '16 at 4:13
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Consider first the sequence $u=(u_n)_{n\in\mathbb{N}}$ defined by $u_n = \frac{1}{n+1}$ (which is not in $\mathbb{R}^\infty$). From it, define the sequences $ u^{(k)} = (u^{(k)}_n)_{n\in\mathbb{N}} $ (for $k\in\mathbb{N}$) as the truncations of $u$ to the first terms: $$ u^{(k)}_n = \begin{cases} u_n & \text{ if } n \leq k \\ 0 & \text{ otherwise.} \\ \end{cases} $$ Then, you have $$\sum_{n=1}^\infty (u_n - u^{(k)}_n)^2 = \sum_{n=k+1}^\infty \frac{1}{(n+1)^2} \xrightarrow[k\to\infty]{} 0$$ so $(u^{(k)})_k$ converges to $u$ in $\ell^2$, and therefore is Cauchy. But then, it is also Cauchy in $(\mathbb{R}^\infty, \lVert\cdot\rVert)$ which is a subspace of $\ell^2$; yet it cannot converge in $(\mathbb{R}^\infty, \lVert\cdot\rVert)$, since $u\notin \mathbb{R}^\infty$.

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Let $e_m^*(e_k) = \delta_{mk}$, that is, the linear functional that picks out the $m$th component. It is easy to check that $e_m^*$ is continuous.

Suppose $x_n \to x$ in $\mathbb{R}^\infty$. Since $x \in \mathbb{R}^\infty$, there is some $N$ such that $e_N^*(x) = 0$, and hence $e_N^*(x_n) \to 0$.

Now let $x_n = \sum_{k=1}^n{1 \over 2^k} e_k$. It is easy to check that $x_n$ is Cauchy.

Choose some $m$, then we see that $e_m^*(x_n) = {1 \over 2^m}$ for all $n \ge m$.

Hence $x_n$ cannot converge to a point in $\mathbb{R}^\infty$.

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$$ \left( 1, \frac 1 2 , \frac 1 3, \ldots,\frac 1 n , 0 , 0, 0, 0, \ldots \right) \to\text{what} \in \ell^2 \text{ ?} $$

The point is you can have a sequence within the space of sequences with only finitely many nonzero terms, whose limit in $\ell^2$ requires infinitely many nonzero terms, and is thus not within that smaller space. The smaller space is therefore not complete.

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