The spaces $\ell^p, \; 1 \leq p < + \infty$ are separable. On the other side, $\ell^\infty$ is not.

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 ?

• Take $D$ to be all the finite sequences (i.e. they get only $0$ eventually) with rational values and argue almost as you did. Feb 9 '19 at 20:12
• 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] Feb 9 '19 at 20:14
• @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 ! Feb 9 '19 at 20:17
• It's good. but you have some index problems with the definition of $D_n$ (you want $\sum_{i=1}^n$ instead of $\sum_{n=1}^k$.) Feb 9 '19 at 20:18
• @Yanko Thanks, I messed up. Appreciate all your help ! Feb 9 '19 at 20:19

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 .

• 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. Feb 9 '19 at 20:30
• 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
– user335501
Feb 9 '19 at 20:32
• Np, just mentioned ! Feb 9 '19 at 20:34