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It's easy to prove this result with just one function, so it will also apply for a union of functions but I have no idea what to do with composition.

Edit:I think i have an solution let $H0$=$A$ ;$H(n+1)$=∪{f(x);x∈$Hn$,f∈$F$} if B=∪{$Hn$} it posses the required criteria

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closed as off-topic by mrtaurho, Leucippus, José Carlos Santos, YiFan, Aweygan Feb 13 at 16:04

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  • $\begingroup$ Why is this different when F is larger? $\endgroup$ – Asaf Karagila Feb 10 at 21:08
  • $\begingroup$ It isn't different I just had to think a little more.The fact that all these function will have the same domain (as Robert pointed out, thank you) escaped ma head. $\endgroup$ – F.Aptl Feb 11 at 9:25
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Presumably all these functions have the same domain and codomain $X$, and you're talking about sets $A \subseteq X$. Let $B$ be the intersection of all sets $S$ such that $A \subseteq S$ and $S$ is closed under $F$.

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