Exercise :

Show that the spaces $\ell^p, \; 1 \leq p < + \infty$ are separable.

Attempt :

In order to show that $\ell^p$ is separable for $\ell^p, \; 1 \leq p < + \infty$, we need to work over a set $D$ proving that it is countable and dense over $\ell^p$, while showing that for $x \in \ell^P$ and $y \in D$, it is $\|x-y\|<\varepsilon.$

Since $x \in \ell^p$, where $x=(x_n)_{n}$, this obviously means that : $$\sum_{n=1}^\infty |x_n|^p <\infty \implies \left( \sum_{n=1}^\infty |x_n|^p \right)^{1/p} < \infty \implies \left( \sum_{n=1}^\infty |x_n|^p \right)^{1/p}<\varepsilon$$ $$\Leftrightarrow$$ $$\sum_{n=1}^\infty |x_n|^p < \varepsilon^p, \quad \text{for some } \varepsilon >0$$

But that would mean that $\exists k \in \mathbb N$, such that :

$$\sum_{n=k+1}^\infty |x_n|^p < \left(\frac{\varepsilon}{2}\right)^p$$

Now, if I let $y=(y_n)_n$ be the sequence

$$y_n = \begin{cases} x_n& n \leq k \\ 0& n> k \end{cases}$$

then, one can easily see that :

$$\left\| x-y\right\|_p = \left( \sum_{n=k+1}^\infty |x_n|^p \right)^{1/p}<\frac{\varepsilon}{2}$$

Now, let $D_n$ be the set :

$$D_n = \left\{\sum_{i=1}^n q_ie_i : q_i \in \mathbb Q \right\}$$

This set is countable as $D_n$ can be correlated to $\mathbb Q^n$. Now, the union $$\bigcup_{n=1}^\infty D_n$$ is also countable as a union of countable sets. Now, it is :

$$\|x-z\|_p \leq \|x-y\|_p + \|y-z\|_p < \varepsilon/2 + \varepsilon/2 \equiv \varepsilon$$

This means that $D = \bigcup_{n=1}^\infty D_n$ is countable and dense over $\ell^p$ which means that $\ell^p$ is separable.

Question : How does one show that $\ell^\infty$ is not a separable space ?

  • 1
    $\begingroup$ Take $D$ to be all the finite sequences (i.e. they get only $0$ eventually) with rational values and argue almost as you did. $\endgroup$
    – Yanko
    Feb 9, 2019 at 20:12
  • 1
    $\begingroup$ For the second question. Consider all the $\{0,1\}$ sequences in $l^\infty$. This is a non-countable set of elements with distance $1$ from one another. [I believe you can answer your own question now, if you still struggle I can try to write a more detailed answer] $\endgroup$
    – Yanko
    Feb 9, 2019 at 20:14
  • $\begingroup$ @Yanko Kindly thanks for your inputs ! I just updated my answer based on your hint and some intuition. Is it correct ? Thanks for the hint over the final part, it's straighforward based on that intuition ! $\endgroup$
    – Rebellos
    Feb 9, 2019 at 20:17
  • $\begingroup$ It's good. but you have some index problems with the definition of $D_n$ (you want $\sum_{i=1}^n$ instead of $\sum_{n=1}^k$.) $\endgroup$
    – Yanko
    Feb 9, 2019 at 20:18
  • $\begingroup$ @Yanko Thanks, I messed up. Appreciate all your help ! $\endgroup$
    – Rebellos
    Feb 9, 2019 at 20:19

1 Answer 1


As Yanko suggested in the comments, you can do the following:

In order to show that $\ell_\infty$ is not separable it is enough to find $A \subset \ell_\infty$ that is not countable s.t. for each $x\ne y \in A$ , $||x-y|| =\delta \gt 0$. since then we can surround those points of $A$ with $\dfrac{1}{2}\delta$-Balls and then there couldn't be a countable set that is dense since it should've element in each such ball (and distinct ones in each ball).

So take $A = \{x=(x_n) : x_n \in \{0,1\} \}\subset \ell_\infty $. $A$ is uncountable.

Show that $||x-y||_{\infty} = 1$ iff $x\ne y$ and you are done .

  • $\begingroup$ That was Yanko's comment and hint from all along so I guess that he at least deserves some credits. Thanks for your input though. $\endgroup$
    – Rebellos
    Feb 9, 2019 at 20:30
  • $\begingroup$ Im sorry , i really didn't read the comments before i answered(i had this exact question in my problem set a while ago so i just answered) i will edit it. @Rebellos $\endgroup$
    – user335501
    Feb 9, 2019 at 20:32
  • $\begingroup$ Np, just mentioned ! $\endgroup$
    – Rebellos
    Feb 9, 2019 at 20:34

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