Let $X$ be a Banach space. Define the ordering $\leq$ on the subspaces of $X^*$ by $Y\leq Z$ iff $Y$ is a closed subspace of $Z.$ Then we define the following collections: $$\mathrm{Du}(X)=\{Y\leq X^*|Y\text{ weakly}^*\text{-dense in }X^*\}$$ the closed, weakly$^*$-dense subspaces of $X^*;$ $$\mathrm{Mu}(X)=\{Y\in\mathrm{Du}(X)|Z\in\mathrm{Du}(X),Z\leq Y\implies Z=Y\}$$ the minimal closed, weakly$^*$-dense subspaces of $X^*;$ $$\mathrm{Pre}(X)=\{Y\leq X^*|Y^{*}=X\}$$ the bipreduals of $X^*$ (which can always be identified as subspaces of $X^*$ by their canonical embedding).
As David Gao showed in the comments of this answer, $\mathrm{Pre}(X)\subseteq\mathrm{Mu}(X).$ Naturally, we ask the converse question:
Is it true that $\mathrm{Mu}(X)\subseteq\mathrm{Pre}(X)?$ Or is it ever the case that $\mathrm{Pre}(X)\subset\mathrm{Mu}(X)?$ My hope is that minimal closed, w$^*$-dense subspaces of $X^*$ serve as a natural generalization of preduals of $X.$ If the answer to the first question is yes, then this is not a generalization at all.
