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.



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> **Question.** Does there exist a universal constant $C$ such that $$\textrm{sep}(B\_{X})\leq C\cdot \textrm{sep}(B\_{X^{\*}}) \, ?$$
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A quantity measuring the separability of Banach spaces
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$(B\_{X^\*})$.


