Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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

1 Answer 1

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 $\ell_1$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.