In my understanding, to prove that $\ell_p$ is a separable space, one may take the subspace of rational sequences, which satisfy the following condition:

$$\sum\limits_{k=1}^\infty |x_k|^p\in \mathbb Q\mbox{ with $\{x_k\}$ a rational sequence}$$

Then this space is countable and dense in $\ell_p$.

  • Countability:

Since $\mathbb Q$ is countable, there exists an injection $f:\mathbb Q\to \mathbb N$. Thus there exists an injection $g:(\ell_p)_\mathbb{Q}\to \mathbb N$, defined by $g\left[(x_k)_{k=1}^\infty\right]=f\left[\sum\limits_{k=1}^\infty |x_k|^p\right]\in \mathbb N$.

  • Density:

Let $(y_i)_{i=1}^\infty\in \ell_p$, then, since $|y_i|\in \mathbb R$ for all $i\in\mathbb{N}$, there is a sequence $(x_k)_{k=1}^\infty\in (\ell_p)_\mathbb{Q}$ (with $|x_k|\in\mathbb Q$) such that

$$\forall \varepsilon >0, |y_i-x_k|^p <\varepsilon, \mbox{ for some $k\in\mathbb{N}$ }$$

Do you think this proof is good enough, or am I missing some important point(s)?

  • $\begingroup$ You say $(x_i)_{i=1}^\infty\in \ell_p$, then $(x_k)_{k=1}^\infty\in (\ell_p)_\mathbb{Q}$. You should use a different symbol than $x$ for your sequence in $(\ell_p)_\mathbb{Q}$. Changing the letter for the index set, when it's still the same index set, doesn't change the sequence. $\endgroup$
    – Aweygan
    Commented Oct 4, 2017 at 16:15
  • $\begingroup$ Thanks, I'll edit my notation now. $\endgroup$
    – sequence
    Commented Oct 4, 2017 at 16:19
  • $\begingroup$ For the countability argument, why is $g$ an injection? For the density argument, how do you know that there is such a sequence $(x_k)$? $\endgroup$
    – user169852
    Commented Oct 4, 2017 at 16:31
  • $\begingroup$ Also, it's not clear to me that the set of rational sequences satisfying $\sum |x_k|^p \in \mathbb Q$ is actually a subspace. E.g. take $p = 3/2$ and define $x_1 = 1$, and $x_k = 0$ for $k > 1$. Then the sum of $(x_k)$ with itself is the sequence $(2, 0, 0, \ldots)$, and $\sum |x_k + x_k|^p = 2^p = 2^{3/2} \not\in \mathbb Q$. $\endgroup$
    – user169852
    Commented Oct 4, 2017 at 16:35
  • $\begingroup$ Do you mean that the sum $M:=\sum_{n=1}^\infty|x_n|^p$ is rational? Or that any $x_n$ in the seuqnce is rational? Beacuse in the first case the set obtained isn't countable. $\endgroup$
    – j3M
    Commented Oct 4, 2017 at 16:35

1 Answer 1


The set you describe, let's call it $X,$ is uncountable. Proof: Define $x_n = 1/2^{n/p}, n\in \mathbb N.$ Then $(x_n)\in X.$ Note that $x_m\ne x_n$ if $m\ne n.$ It follows that if $\sigma_1, \sigma_2$ are distinct bijections of $\mathbb N$ to $\mathbb N,$ then $x_{\sigma_1(n)}, x_{\sigma_2(n)}$ are distinct sequences in $X.$ Since the set of such bijections is uncountable, so is $X.$

I'm not sure why you're going to all this trouble. An easy countable dense subset of $l^p$ is the set of rational sequences in $l^p$ that are $0$ except for finitely many terms.

Added later: @DanielWainfleet pointed out that in my first answer I neglected the requirement that the sequences in $X$ should have all rational terms. To achieve that, inductively choose distinct rational $x_1,x_2,\dots$ such that

$$\frac{1-(x_1^p + \cdots + x_n^p)}{2} < x_{n+1}^p < 1-(x_1^p + \cdots + x_n^p).$$

for all $n.$ Then $\sum_{n=1}^{\infty}x_n^p = 1.$ The rest of the argument is the same.

  • $\begingroup$ If we consider the set of rational sequences in $\ell_p$, can we define such sequences as, say, $(x_k)=\{0,0,\dots, 0, f(n_1), f(n_2), \dots, f(n_l), \dots, 0,\dots\}$, where $f$ is an injection from $\mathbb{N}$ into $\mathbb{Q}$ (since $\mathbb{Q}$ is countable and $|\mathbb{N}|=|\mathbb{Q}|$)? Then all rationals would be covered by this set. $\endgroup$
    – sequence
    Commented Oct 4, 2017 at 19:48
  • 1
    $\begingroup$ Although your last sentence is well-said, it seems that in the Q we should have $x\in (x_k)_k \in X$ iff each $x_k\in \Bbb Q$ as well as $\|x\|^p\in \Bbb Q.$ If such $ x$ exists with $x_k\ne 0$ for infinitely many $k$ then by the methods of your answer, $X$ is uncountable. Such $x$ do exist, but it may not be obvious. $\endgroup$ Commented Oct 5, 2017 at 2:52
  • $\begingroup$ @DanielWainfleet Yes you are right. Thank you. I'll add to my answer. $\endgroup$
    – zhw.
    Commented Oct 6, 2017 at 17:30
  • $\begingroup$ @sequence I'm afraid I don't understand your comments. $\endgroup$
    – zhw.
    Commented Oct 6, 2017 at 17:36
  • $\begingroup$ To the proposer: The set of sequences in your last comment to this A is not a dense set. $\endgroup$ Commented Oct 6, 2017 at 19:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .