Size of $L^{\infty}$ I am confused about the 'size' of $L^\infty$: 
One the one hand, when we consider $L^\infty(\Omega)$ and $\mu(\Omega)$ is finite (and $\mu$ some measure), then we have the inclusion $L^\infty \subset L^p$ for some $p \geq 1$. So $L^\infty$ is the 'smallest' of the $L^p$ spaces.
On the other hand, we know for arbitrary open $\Omega \subset \mathbb{R^n}$ that $L^\infty$ is not separable, so it is in some sense 'too big' to be approximated by a countable subset, but $L^p$ for $1 \leq p < \infty$ are separable.
I find this somewhat confusing. What am I missing or getting wrong? 
Thanks!
 A: What's going on here, is that the separability doesn't just tell you about the size of sets in the sense of set theoretic inclusions but also tells you something about the topology on that set. On a finite measure space, $L^\infty(\Omega)$ is continuously embedded in $L^p(\Omega)$ but it gets a very different topology than the one we would give it if we just thought of it as a subset of $L^p(\Omega)$. Now let me give some details of this.
For simplicity lets fix a finite measure space $(\Omega, \mathcal{F}, \mu)$ and consider $L^p(\Omega)$.
In this case, when considered as sets, $L^\infty(\Omega) \subset L^p(\Omega)$. So we could consider $L^\infty(\Omega)$ with the subspace topology induced by the $L^p(\Omega)$-norm. For this topology, $L^\infty(\Omega)$ would be separable (as a subspace of a separable space). Of course, we don't want to think of $L^\infty(\Omega)$ with the subspace topology because it comes with its own natural topology, the one induced by its usual norm!
In fact, the embedding $L^\infty(\Omega) \hookrightarrow L^p(\Omega)$ is even continuous since if $f \in L^\infty(\Omega)$ then $$\int_\Omega |f|^p d\mu \leq \mu(\Omega) \|f\|_\infty^p. $$
However, we don't have such an inequality going the other way. So whilst the embedding is continuous, it is not a homeomorphic embedding. 
This means that the norm topology on $L^\infty(\Omega)$ is stronger than the one induced on $L^\infty(\Omega)$ by $L^p(\Omega)$ and so it is harder to approximate functions in this topology. The lack of separability of $L^\infty(\Omega)$ is a symptom of the fact that its inherent notion of approximation is much stronger, rather than of its 'size'.
