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

Question

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 ?

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 .