# Function of two sets intersection

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) $$f(A,B)\neq [0,1]$$.

(b) $$f(A,B)\cap A\neq\emptyset$$ and $$f(A,B)\cap B\neq\emptyset$$.

(c) The length (i.e. Lebesgue measure) of $$f(X,B)\cap A$$ is maximized at $$X=A$$, and the length of $$f(A,X)\cap B$$ is maximized at $$X=B$$.

Any two of the three conditions can be satisfied:

• $$f(A,B)\equiv [0,1]$$ satisfy (b) and (c).

• $$f(A,B)\equiv Y$$ for any fixed $$Y\neq [0,1]$$ satisfy (a) and (c).

• $$f(A,B)$$ being any $$Y\neq [0,1]$$ that intersects both $$A$$ and $$B$$ satisfy (a) and (b).

Satisfying all three seems to be impossible though.

• If there is such an $f$, then the image of $f$ is equal to the set $P := \{A | f(A, A) = A\}$, which will be downward closed. I'm currently trying to see if this is necessarily a cover of the interval. – user44191 Oct 16 '18 at 16:40
• With respect to (c) condition (b) looks a little bit strange. Can it replaces by (b') $\lambda(f(A,B) \cap A) > 0$ and $\lambda(f(A,B) \cap B) > 0$ if $A,B \not= \emptyset$? – Dieter Kadelka Feb 25 at 9:59
• @DieterKadelka Assuming $\lambda$ refers to the length, then yes – pi66 Feb 25 at 14:07
• @pi66: And by length do you mean Lebesgue-measure $\lambda(A)$ or the span $\max A - \min A$ for $A \in U$? – Dieter Kadelka Feb 25 at 14:38
• @user44191 how do you show that $f(A, A) \subset A$ for $A$ in the image of $f$? – Jack Mar 24 at 21:32