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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. I posted the question on math.SE half a year ago where some remarks were made but the question remains unsolved.

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  • $\begingroup$ It seems to me that (c) is unnecessary. $\endgroup$ Dec 17, 2021 at 10:05

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Update: this is not quite a complete answer, since (d) might not hold when $A_{\epsilon/2}$ and $B_{\epsilon/2}$ are not disjoint.

Let $\epsilon < 0.0001$. For $X \in U$, let $X_t$ be the "least" subset of $X$ of length $\min(t, \lvert X \rvert)$. In other words, $X_t$ has length $t$ (or is all of $X$) and contains every element of $X$ that is less than the maximum of $X_t$. For example, $([0,1/4] \cup [1/2, 1])_{1/2} = [0, 1/4] \cup [1/2, 3/4]$.

Define $$f(A, B) := A_{\frac{\epsilon}{2}} \cup B_{\frac{\epsilon}{2}}.$$ (a): This follows from the definition of $f$.

(b): $$\lvert f(A, B) \rvert = \frac{\min(\epsilon, \lvert A \rvert) + \min(\epsilon, \lvert B \rvert)}{2} \leq \epsilon < 0.0001.$$

(c): $f(A, B) \cap A \supset A_{\frac{\epsilon}{2}}$ and therefore has positive length.

(d): The length of $f(X, B) \cap A = (X_{\frac{\epsilon}{2}} \cap A) \cup (B_{\frac{\epsilon}{2}} \cap A)$ is maximized at $X=A$ if $A_{\epsilon/2}$ and $B_{\epsilon/2}$ are disjoint).

A similarly defined $f$ also works if $X_t$ is replaced with another canonically associated subset of $X$ of length $\min(t, \lvert X \rvert)$ (e.g. the "greatest" such subset) or if $\epsilon/2$ is replaced by smaller positive values.

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    $\begingroup$ I do not think (d) always holds in this construction. Let $A=[0,\epsilon] \cup [1/2,1/2+\epsilon]$, $B=[0,\epsilon/10]\cup [1/2,1/2+\epsilon]$. Then taking $X=A$ leads to the intersection of length $\epsilon/10$ (as $A_{\epsilon/2}=[0,\epsilon/2]$), while taking $X=A\cap B$ leads to the intersection $F(X,B)\cap A$ of length $\epsilon/2$. $\endgroup$ Mar 22, 2021 at 7:17
  • $\begingroup$ That example seems fine to me: $f(A, B) = [0, \epsilon/2] \cup ([0, \epsilon/10] \cup [1/2, 1/2 + 2\epsilon/5])$, so $f(A, B) \cap A = [0, \epsilon/2] \cup [1/2, 1/2 + 2\epsilon/5]$ has length $9\epsilon/10$. It seems like you are missing the $B_{\epsilon/2}$ in $f(A, B)$? $\endgroup$ Mar 22, 2021 at 7:27
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    $\begingroup$ But in fact a slight tweak does yield a simple counterexample for (d): $A=[0, \epsilon] \cup [1/2, 1/2 + \epsilon]$, $B=[0, \epsilon]$, and $X=[1/2, 1/2 + \epsilon]$. I wonder if there is a simple modification to avoid this (which occurs only when $A_{\epsilon/2}$ and $B_{\epsilon/2}$ are not disjoint). $\endgroup$ Mar 22, 2021 at 8:12
  • $\begingroup$ Define $g(A,B)=(A\cap B)_\epsilon$, and define $f(A,B)=g(A,B)\cup(A\setminus B)_{\frac{\epsilon-|g(A,B)|}{2}}\cup(B\setminus A)_{\frac{\epsilon-|g(A,B)|}{2}}$. Then (a), (b) and (c) still hold. I don't know if this satisfies (d), but $f(A,B)$ now prioritizes the overlap of A and B over their symmetric difference. $\endgroup$
    – Tim
    Mar 22, 2021 at 10:23
  • $\begingroup$ @Tim: That seems to satisfy (d) in more cases, but here is another counterexample: $A=[0, \epsilon/2] \cup [\epsilon, 3\epsilon/2]$, $B=[0, \epsilon]$, and $X = [\epsilon, 3\epsilon/2]$. Then $f(A, B) \cap A = [0, \epsilon/2] \cup [\epsilon, 5\epsilon/4]$ but $f(X, B) \cap A = [0, \epsilon/2] \cup [\epsilon, 3\epsilon/2]$. $\endgroup$ Mar 25, 2021 at 20:59

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