# A quantity measuring the separability of Banach spaces

Let $$X$$ be a Banach space. It is natural for us to introduce a quantity measuring the separability of sets as follows: for a subset $$A$$ of $$X$$, we set

$$\textrm{sep}(A)=\inf\{\epsilon>0: A\subseteq K+\epsilon B_{X}$$ for some countable subset $$K$$ of $$X\}$$.

Clearly, $$A$$ is separable if and only if $$\textrm{sep}(A)=0$$.

It is elementary that a Banach space $$X$$ is separable if $$X^{*}$$ is separable. My question is about a quantitative version of this result.

Question. Does there exist a universal constant $$C$$ such that $$\textrm{sep}(B_{X})\leq C\cdot \textrm{sep}(B_{X^{*}}) \, ?$$

For the unit ball $$B_X$$ of the Banach space there are only two possibilities:
sep$$(B_X)= 1$$, if $$B_X$$ is not separable, and sep$$(B_X)=0$$ if $$B_X$$ is separable. Indeed, if sep$$(B_X)<1$$ there are $$\varepsilon <1$$ and a countable subset $$K\subseteq B_X$$ with $$B_X\subseteq K + \varepsilon B_X$$. But this can be iterated, i.e., $$B_X\subseteq K +\varepsilon (K+\varepsilon B_X) \subseteq K_1+\varepsilon^2 B_X$$ where $$K_1=K+\varepsilon K$$ is again countable. Inductively, this implies $$B_X\subseteq K_n +\varepsilon^n B_X$$ for a countable set $$K_n$$ and hence sep$$(B_X)=0$$.
Therefore, $$C=1$$ satisfies the desired inequality sep$$(B_X)\le$$ sep$$(B_{X^*})$$.
• In particular, if $X$ is reflexive, then one has $\mathrm{sep}(B_X) = \mathrm{sep}(B_{X^\star})$, right ? Apr 12, 2022 at 14:44
• Sure, because sep$(B_{X^*})\le$ sep$(B_{X^{**}})=$ sep$(B_X)$. Apr 13, 2022 at 8:26