Prove that metric space $\left \langle \ell_p(\mathbb R), d_p(x,y)=(\sum_{i=1}^{\infty} |x_i|^p)^\frac{1}{p} \right \rangle$ is separable. Where $\ell_p(\mathbb R)=\left \{ (x_1,x_2,...,x_n,...):\sum_{i=1}^{\infty} |x_i|^p<\infty, p>1, x_i \in \mathbb R \right \}$

To show separability I need to find countable everywhere dense subset. I've already proved (with similar approach) that $\ell_p(\mathbb Q)$ is everywhere dense. However, I can't find a way to show $\ell_p(\mathbb Q)$ is countable.

  • 3
    $\begingroup$ I don't think $\ell_p(\mathbb Q)$ is countable. $\endgroup$
    – Tunococ
    Sep 8, 2013 at 12:28

2 Answers 2


To elaborate on above answer, let $M$ be the set of all sequences $(r_{0}, \ldots, r_{n}, 0,\ldots)$ with $r_{i}$ being rational and $n \in \mathbb{N}$. (This is the set of sequences with finite support and rational entries).

Since $\mathbb{Q}$ is countable and finite product of countable sets is countable and then countable union of countable sets is countable, we see that $M$ is countable.

Now let us see that $M$ is dense in $\ell_{p}(\mathbb{R})$. To do this, given any $x = (x_{n}) \in \ell_{p}(\mathbb{R})$, and for $\epsilon > 0$ we must find an element $y \in M$ such that $d(x,y) <\epsilon$. We have

$$ \sum\limits_{n=0}^{\infty} |x_{n}|^{p} < \infty $$

Hence, given $\epsilon > 0$, there exists $m \in \mathbb{N}$ such that

$$ \sum\limits_{n=m+1}^{\infty} |x_{n}|^{p} < \epsilon/2 $$

Now for $0\leq i \leq m$, choose $r_{i} \in \mathbb{Q}$ such that $|r_{i} - x_{i}| < \left(\frac{\epsilon}{2m}\right)^{\frac{1}{p}}$ (using that the rationals are dense in reals).

Then the element $y = (r_{0}, r_{1}, \ldots, r_{m}, 0, 0, \ldots) \in M$ is the required element.

  • $\begingroup$ Indeed! Thank you. $\endgroup$ Apr 9, 2020 at 4:07
  • $\begingroup$ You said countable union of countable set is countable,we see that M is countable.According to me, $M=\mathbb{Q} *\mathbb{Q} * ...* \mathbb{Q} *0 *...* 0$ so being finite product of countable set is countable.Why we need this extra statement:countable union of countable set is countable?M is finite product of countable set $\mathbb{Q}$ not countable union of? $\endgroup$ Mar 3, 2021 at 8:36
  • $\begingroup$ @mathstudent: $M$ is a union of the spaces of finite rational sequences of length $n$ for each $n \geq 1$, ie. $\mathbb{Q}\times 0...$ and $\mathbb{Q}\times\mathbb{Q}\times 0...$ and $\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}\times 0...$, etc. $M$ is the countable union of these spaces. $\endgroup$
    – ryan221b
    Oct 25, 2021 at 13:19

The sequences with finite support and rational entries are dense in $\ell_p(\mathbb R)$ for each $1\leqslant p\lt \infty$ (we can make the remainder of the series as small as we wish).


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