# Prove that sequence space $\ell_p(\mathbb R)$ is separable

Problem:

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.

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

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.

• Indeed! Thank you. Apr 9, 2020 at 4:07
• 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? Mar 3, 2021 at 8:36
• @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. 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).

I think it takes some more work to show that the set of all sequences with rational entries and finite non-zero element is countable (denoted $$A$$).

Because Countable infinitely cartesian product of countable set may not be countable, otherwise we shall be able to prove that $$\ell_p(Q)$$ is also countable, however, its cardinality is $$\mathbb{N}^{\mathbb{N}}=\mathcal{c}$$, which is the cardinality of $$\mathbb{R}$$ and shall be uncountable.

Also, the above set $$A$$ can not be expressed as a finite product of a countable set. Consider any finite product $$\underset{\text{M of Q here}}{\mathbb{Q} \times \mathbb{Q} \times \dots \mathbb{Q}}$$, then a rational sequence with (M+1) non-zero entries still has finite support but does not belong to the above product.

Instead, I think we can prove it by constructing a one-to-one mapping from $$A$$ to $$\mathbb{Q}$$.

Consider a mapping that encodes the denominator and numerator into decimal spaces separated by 0, e.g. $$f((1/3,3/5,\dots,100/101,\dots))=0.10303050\dots10001010\dots$$. Also, in order to take care of the negative entries, instead of using the irreducible fraction $$\frac{p}{q}$$, we use $$\frac{2p}{2q}$$ for a negative entry, so e.g. $$f((-1/3,3/5,\dots,-100/101,\dots))=0.20603050\dots20002020\dots$$.

Since each rational number can be expressed as its irreducible form $$\frac{p}{q}$$ where $$p,q \in \mathbb{N}$$ and since each sequence only has finite support, we know the image of $$f$$ is a subset of $$\mathbb{Q}$$. Suppose $$f(x)=f(y)\Rightarrow 0.p_1^x0q_1^x0p_2^x0q_2^x \dots p_n^x0q_n^x \dots =0.p_1^y0q_1^y0p_2^y0q_2^y \dots p_n^y0q_n^y \dots$$, then as there are only finite non-zero digits, we must have $$p_i^x=p_i^y$$ and $$q_i^x = q_i^y$$ for all i, hence $$x=y$$, and we know $$f$$ is one-to-one.

Since there exists an one-to-one mapping from $$A$$ to $$\mathbb{Q}$$, we know $$A$$ is also countable.