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Let $X\neq \emptyset$ be a set. We say $C \subseteq {\cal P}(X)$ is a cover of $X$ if $\bigcup C = X$. For covers $C, D$ of $X$ we say that $C$ is more efficient than $D$ if $|C\setminus D| < |D \setminus C|$, and we write $C \triangleleft D$ if $C$ is more efficient than $D$.

Question. Is there a set $X$ and covers $C, D, E$ of $X$ such that $C \triangleleft D \triangleleft E \triangleleft C$?

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    $\begingroup$ Just checking: $|\cdot|$ denotes cardinality as usual? With finite covers the "more efficient than" relation would be equivalently $|C| < |D|$, so apparently you are looking for infinite covers. $\endgroup$
    – Jukka Kohonen
    Commented May 7, 2022 at 8:42
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    $\begingroup$ Never mind covers, you can't have sets $C,D,E$ with $|C\setminus D|\lt|D\setminus C|$, $|D\setminus E|\lt|E\setminus D|$, and $|E\setminus C|\lt|C\setminus E|$. This follows from the fact that (at least if the axiom of choice holds) for cardinal numbers $w,x, y,z$, if $w\lt x$ and $y\lt z$ then $w+y\lt x+z$. $\endgroup$
    – bof
    Commented May 7, 2022 at 9:45
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    $\begingroup$ $|C\setminus D|+|D\setminus E|+|E\setminus C|=|D\setminus C|+|E\setminus D|+|C\setminus E|$ $\endgroup$
    – bof
    Commented May 7, 2022 at 9:54
  • $\begingroup$ Thanks @bof - I failed to see that line of argument. If you want, you can post this as an answer (and I'll accept and upvote it). If you don't want to, I'll delete the question after a few hours. $\endgroup$
    – Dominic van der Zypen
    Commented May 7, 2022 at 13:32

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