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    $\begingroup$ This seems like a reasonable question to me and I learned something from the answer. Perhaps one of the downvoters might explain? $\endgroup$ Commented yesterday
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    $\begingroup$ Note that the bijection $\varphi$ can be constructed so that $|j\setminus\varphi(j)|=1$ for all $j\in J$. $\endgroup$ Commented yesterday
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    $\begingroup$ @bof Do you mind sharing the construction? It’s not clear to me how to do that. It certainly does not follow from the argument in my answer. $\endgroup$ Commented 20 hours ago
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    $\begingroup$ @DavidGao Call two sets equivalent if their symmetric difference is finite. On each equivalence class $C$ define a bijection $\varphi_C:C\to C$ with the desired properties. (A recursive construction of length $\omega$ similar to your argument.) Take the union. $\endgroup$ Commented 18 hours ago
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    $\begingroup$ Here is a related question. It is known that for any $n \geq 1$ and $k \leq n/2$, there is a bijection $\varphi$ from the set of $k$-subsets of $[n]=\{1,2,\ldots,n\}$ to the set of $(n-k)$-subsets of $[n]$ such that $S\subseteq \varphi(S)$ for all $S$ (look up "symmetric chain decomposition of Boolean lattice"). Is there a bijection $\varphi$ from the finite subsets of $\{1,2,\ldots\}$ to the cofinite subsets such that $S \subseteq \varphi(S)$ for all $S$? $\endgroup$ Commented 13 hours ago