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?
Completing a normed space
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real-analysis
normed-spaces
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0@DavidMitra: Thanks! I misread my notes, you are right. – 2011-12-09
1 Answers
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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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0Changing 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). – 2011-12-09