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For any collection $\mathscr C$ of sets, write $\Upsilon(\mathscr C)$ for the collection of arbitrary unions in $\mathscr C$. Now, I ask the innocent question of idempotency of $\Upsilon$:

Is $\Upsilon(\Upsilon(\mathscr C)) = \Upsilon(\mathscr C)$?

"$\supseteq$" is clearly true. So, let's analyze the other direction: Let $\mathscr A\subseteq \Upsilon(\mathscr C)$. For each $A\in\mathscr A$, choose a $\mathcal C_A\subseteq\mathscr C$ such that $A = \bigcup\mathcal C_A$. Then $\bigcup\mathscr A = \bigcup_{A\in\mathscr A} \bigl(\bigcup \mathcal C_A\bigr) = \bigcup\bigl(\bigcup_{A\in\mathscr A} \mathcal C_A\bigr)\in \Upsilon(\mathscr C)$.

This usage of the full power of AC raises the following questions:

  1. Is a "better" proof known that uses milder versions of choice?
  2. Does the idempotence of $\Upsilon$ (for all $\mathscr C$'s) imply any form choice?
Atom
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  • I don't see why this is idempotent with or without the Axiom of Choice. ${{\varnothing} }$. – Asaf Karagila Jul 19 '24 at 11:32
  • @AsafKaragila I thought that it was clear that idempotence of $\Upsilon$ meant its idempotence action on all $\mathscr C$'s. Nevertheless, I have edited it in. – Atom Jul 19 '24 at 11:40
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    Can't we just choose $\mathcal C_A $ to be the set of all $B \subseteq A$? – Zoe Allen Jul 19 '24 at 11:50
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    I think it might be worth explicitly defining in your question what you mean by "collection of arbitrary unions" $\Upsilon(\mathscr C)$. If I am reading this right, you are talking about the collection of sets of the form $\bigcup_{A\in\mathscr D}A$, where $\mathscr D\subseteq\mathscr C$. – Joe Jul 19 '24 at 11:50
  • @Joe Yes, you are correct. – Atom Jul 19 '24 at 11:53
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    @ZoeAllen That solves everything! (If you post it as an answer, I will accept it.) – Atom Jul 19 '24 at 11:53
  • I don't think you understood my comment. – Asaf Karagila Jul 19 '24 at 12:26
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    @AsafKaragila If $\mathscr{C}={x}$ is any singleton, then $\Upsilon(\mathscr C)={\varnothing, x}=\Upsilon(\Upsilon(\mathscr C))$. – Alex Kruckman Jul 19 '24 at 12:59
  • @AlexKruckman: Yes, I got that, since the $\Upsilon$ operation is about unions from subsets of $\scr C$. I think this is a clarity on the definition and use of terminology rather than a mathematical issue. This was pointed out by Joe in a previous comment, but was never really addressed in an edit. – Asaf Karagila Jul 19 '24 at 13:48
  • In any case, I have closed the question as a duplicate, since essentially the same answer applies here (as remarked by Zoe Allen). – Asaf Karagila Jul 19 '24 at 13:49

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