11
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – user44191 Oct 16 '18 at 16:40
  • $\begingroup$ 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 $A,B \not= \emptyset$? $\endgroup$ – Dieter Kadelka Feb 25 at 9:59
  • $\begingroup$ @DieterKadelka Assuming $\lambda$ refers to the length, then yes $\endgroup$ – pi66 Feb 25 at 14:07
  • $\begingroup$ @pi66: And by length do you mean Lebesgue-measure $\lambda(A)$ or the span $\max A - \min A$ for $A \in U$? $\endgroup$ – Dieter Kadelka Feb 25 at 14:38
  • 1
    $\begingroup$ @user44191 how do you show that $f(A, A) \subset A$ for $A$ in the image of $f$? $\endgroup$ – Jack Mar 24 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.