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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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? – La Belle Noiseuse Jan 20 '13 at 11:55
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
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.

share|cite|improve this answer
could you please tell me about the $c$, if I take $x_n=1\forall n$ then? – La Belle Noiseuse Jan 20 '13 at 12:33
@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
Hence $c$ is not dense. – La Belle Noiseuse 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! – La Belle Noiseuse Jan 20 '13 at 12:39

Your Answer


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.