3\mapsto 5,\, 4\mapsto 5,\, 5\mapsto 5\}$. Consider now the abstractions $\rho_1 \triangleq \{1,3,5\}$ and $\rho_2 \triangleq \{1,4,5\}$ and observe that $\rho_i {\circ}f {\circ}\rho_i = \mu{\circ}f {\circ}\mu$. However, we have that $\rho_1 \sqcap \rho_2 = \lambda x.x$, because the image of $\rho_1 \sqcap \rho_2$ is ${\mathcal{M}}(\rho_1 \cup \rho_2) = 
\{1,2,3,4,5\}$. Hence, $(\rho_1 \sqcap \rho_2) {\circ}f {\circ}(\rho_1 \sqcap \rho_2) = f \neq \mu{\circ}f {\circ}\mu$. Therefore, if we let $\rho_r = \sqcap \{ \rho \in \operatorname{uco}(C)~|~ \rho \circ f \circ 
 \rho = \mu \circ f \circ \mu\}$ then $\rho_r = \lambda x.x$. Consequently, the most concrete domain that induces the same best correct approximation of $f$ as $\mu$ does not exist.

The Solution
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Our key technical result is the following *constructive* characterization of the property of “having the same b.c.a.” for two comparable abstract domains. In the following, given a poset $A$ and any subset $S\subseteq A$, $\max (S) \triangleq \{
x\in S~|~ \forall y\in S.\; x\leq_A y \Rightarrow x = y\}$ denotes the set of maximal elements of $S$ in $A$.

\[key\] Let $f: C\rightarrow C$ and $A,B \in \operatorname{Abs}(C)$ such that $B\subseteq A$. Suppose that $f{\circ}\mu_A : C {\rightarrow}C$ is continuous (i.e., preserves lub’s of chains in $C$). Then, $$\textstyle
