Function of two sets Let $U$ be the set of all nonempty subsets of $[0,1]$ that are a union of finitely many  closed intervals (where an "interval" that is a single point does not count as an interval). Does there exist a function $f:U\times U\rightarrow U$ such that for any $A,B\in U$:
(a) $f(A,B) = f(B,A)$
(b) $f(A,B)$ has length (i.e. Lebesgue measure) less than $0.0001$.
(c) $f(A,B)\cap A$ has positive length.
(d)  The length of $f(X,B)\cap A$ is maximized at $X=A$.
This is a variant of this question with more restrictive conditions, so my guess would be that the answer is no.
 A: Note: This is not actually an answer, only a place to share some
preliminary observations in the hopes that it will be useful for others.
Idea for simplifying notation/proofs:
Instead of using closed intervals in the question, one could
use only half-open intervals (for example, of the form $[a,b)\subset [0,1)$).
The problem statements should be equivalent with closed intervals or half-open
intervals.
The formulation with half-open intervals has the advantage that
one does not have to deal with closed "intervals" which consist of a single point.
Moreover, this would have the advantage that
$U_0:=U\cup\{\emptyset\}$ is stable with respect to intersections, unions, complements,
i.e.
$$
A\cap B, A\cup B, [0,1)\setminus A \in U_0 = U\cup\{\emptyset\}
\qquad
\forall A,B\in U_0 = U\cup\{\emptyset\}
$$
holds.
Some consequences of (a)-(d):
Here, $|\cdot|$ refers to the Lebesgue measure
and we use the version with half-open intervals outlined above,
(but this can easily translated to the version with closed intervals).
The following should hold for all $A,B\in U$.
For now, I have not included details proofs.
(e) The length of $f(X,Y)\cap A$ is maximized at $(X,Y)=(A,A)$.
(f) We have $C\subset f(C,C)$, where $C=f(A,B)$.
(g) We have $C\subset f(C,B)$, where $C=f(A,B)$.
(h) We have $|f(A,B)|\leq \alpha := |f([0,1),[0,1))|$.
And also $\alpha\leq 0.0001$.
(i) We have $D_1 = f(D_1,D_1)$, where $D_1:=f([0,1),[0,1))$.
(j) We have $D_1 = f(D_1,[0,1))$, where $D_1:=f([0,1),[0,1))$.
some remarks on the new conditions:
(h) allows us to forget about $0.0001$, we just have to keep in mind that $\alpha$
is small.
It would be interesting to know if $f(C,C)=C$ can be shown for $C=f(A,B)$.
In the comments on MO it was first claimed that this is true, but then the user
said he does not remember the argument and does no longer make the claim.
I think one can maybe show something similar if we add intersections in the right places.
