this is my first question and I hope I don't make any relevant mistakes. For a little bit of context, in my real analysis homework I have the following problem.

Show that the subset D $\subset$ $l_2$ composed by all the sequences $(x_n)$ with $x_n\neq0$ being a rational number for a finite number of indexes, is countable and dense in $l_2$

Here the set $l_2$ is the set of all real number sequences $(x_n)$ that satisfy: $\sum_{n=1}^{\infty}|x_n|^2\leq \infty$ And the metric is given by:


My question in reality is, should I consider D as being the set of all sequences with a finite number of entries being rational numbers that are different than 0 and the rest of the sequence being all real numbers. Or should i consider D as the set of all sequences with finite entries beign rational numbers that are different than 0 and the rest of all entries being equal to 0 ?

Not sure if the point of the question is clear, in any case hit me up and we can dicuss the matter. Thanks in advance, any help would be appreciated

  • 3
    $\begingroup$ All zero except for finitely many that are rational numbers. $\endgroup$ – user647486 Apr 18 at 2:09
  • $\begingroup$ Any suggestions as to how to show that this set is dense in l2 ? $\endgroup$ – Guilherme takata Apr 18 at 14:39
  • $\begingroup$ If $x\in \ell^2$, then there is $N$ such that $\sum_{k>N}|x_k|^2<\epsilon^2/2$. Then take $y\in\ell^2$ such that $y_k=0$ for $k>N$ and $y_k\in\mathbb{Q}$ and $\sum_{k=1}^{N}|y_k-x_k|^2<\epsilon^2/2$. It follows that $\|y-x\|<\epsilon$. $\endgroup$ – user647486 Apr 18 at 14:43
  • $\begingroup$ Oh, got it now. Thank you $\endgroup$ – Guilherme takata Apr 18 at 18:25

$D$ is the set of all rational sequences with finitely many non-zero entries.

By a countability argument you can check that it's a countable set.

It's dense because you can truncate any sequence up to some term, the series that the norm must be convergent, and rationals are dense in $\mathbb{R}$.

  • $\begingroup$ I get that sequences in D have finitely many non zero entries, but the rest of the sequence is entierly composed of zeros or by any real numbers ? $\endgroup$ – Guilherme takata Apr 18 at 14:38
  • $\begingroup$ If the (infinitely many) other entries were allowed to be any real number, then they could be non-zero. So those are the sequences that, after some entry, are all zeroes. $\endgroup$ – AspiringMathematician Apr 22 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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