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Let $X$ be the set of all real sequences, consider the subset $$S=\big\{x=\{x_n\}: x_n\in\mathbb{Q}~\forall n, x_n=0\text{ except for a finite number of n}\big\}.$$

We need to find out which of the followings are true.

  • $a.$ $S$ is dense in $l_1$, the space of absolutely summable sequences, provided with the metric $$d_1(x,y)=\sum_{n=1}^{\infty}|x_n-y_n|\;.$$

  • $b.$ $S$ is dense in $l_2$, the space of square summable sequences, provided with the metric $$d_2(x,y)=\left(\sum_{n=1}^{\infty}|x_n-y_n|^2\right)^{1/2}.$$

  • $c.$ $S$ is dense in $l_{\infty}$, the space of bounded sequences, provided with the metric $$d_{\infty}(x,y)=\sup_n\{|x_n-y_n|\}\;.$$

My Intution: I know that $l_p$ is seperable for $1\le p<\infty$, as $S$ is countable and dense so $a,b$ are true, $c$ is false as $l_{\infty}$ is not seperable. Am I right?

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You’re right that $S$ is dense in $\ell_1$ and $\ell_2$, but can you actually prove it? It’s not enough to know that they are separable. –  Brian M. Scott Jan 20 '13 at 11:53
    
No at the moment I can not prove it, I studied it many days before :(, is my guess for $c$ is also right? –  Une Femme Douce Jan 20 '13 at 11:55
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Here’s a hint for $\ell_1$: if $\epsilon>0$, there’s an $m$ such that $\sum_{n\ge m}|x_n|<\epsilon/2$, and you can choose rationals $q_0,\dots,q_{n-1}$ so that $\sum_{k=0}^{n-1}|x_k-q_k|<\epsilon/2$. –  Brian M. Scott Jan 20 '13 at 11:58

1 Answer 1

up vote 2 down vote accepted

Separability of $\ell_p$ by itself is not sufficient to state that some countable set is dense, but gives you a hint on how to construct an approximation sequence in $S$ for any sequences in $\ell_p$ when $p<\infty$. For the case $c$ consider the sequence consisting of $1$'s.

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could you please tell me about the $c$, if I take $x_n=1\forall n$ then? –  Une Femme Douce Jan 20 '13 at 12:33
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@Panu: yes. Then the sup-norm of the difference with a vector that has at least one zero component (in your case it is in $S$ and thus has even more) is always $1$ –  Ilya Jan 20 '13 at 12:35
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Hence $c$ is not dense. –  Une Femme Douce Jan 20 '13 at 12:37
    
@Panu: indeed, you're right - as well as you were right with answers for a and b. Your solution however shall be fixed in a and b –  Ilya Jan 20 '13 at 12:38
    
Thank you, pleased! –  Une Femme Douce Jan 20 '13 at 12:39

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