# If $F$ is a set of functions prove that for every set $A$ there exists a $B$ so that $A\subseteq B$, $B$ is closed for $F$ and $B$ is minimal; [closed]

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

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

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• Why is this different when F is larger? – Asaf Karagila Feb 10 at 21:08
• 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. – F.Aptl Feb 11 at 9:25

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 $$F$$.