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Is the completion of $\{x=(x_n)|x_n\in \mathbb R \text{ and  for a given } x,\text{ only finitely many } x_n\neq0\}$ equipped with the norm $\|x\|:= |x_1|+|x_2|+...$ simply the set of all real sequences?

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I think you want to say "only finitely many $x_n\ne0$"; otherwise, $\Vert x\Vert$ could be infinite. If this is what you meant, then the completion will be the set of absolutely summable sequences (that is, $\ell_1$). – David Mitra Dec 9 '11 at 9:48
@DavidMitra: Thanks! I misread my notes, you are right. – Volga Dec 9 '11 at 9:52

The "larger" space here is the set of all sequences $x$ such that $$\tag{1}\Vert x\Vert= \sum\limits_{n=1}^\infty |x(n)|$$ is finite. This is the space of absolutely summable sequences, denoted by $\ell_1$, with norm defined as in (1). One can show that $\ell_1$ is a complete normed linear space.

Let $F$ be your set of sequences. Then $F$ is the space of all sequences of finite support and sits inside $\ell_1$. Moreover, given an element $x$ in $\ell_1$, the sequence $\{y_n\}$ in $F$ with terms defined by $$y_n=(x(1),x(2),\ldots, x(n),0,0,\ldots)$$ converges in norm to $x$.

This shows that $F$ is a dense subset of $\ell_1$. As such, the completion of $F$ is $\ell_1$.

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Thanks, David! I have a question not directly related to this problem: in the definition of a "completion" it suffices that there exists a dense subset of the complete space which is isometric to the space to be completed. What does that mean? So we can change the metric? – Volga Dec 9 '11 at 12:38
Changing the space as well. However, there would be a 1-1, onto map between the spaces that preserves distances (so, the norm of an element in one space is the same as the norm of its image in the other space). – David Mitra Dec 9 '11 at 13:12

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