# If $\cup \mathcal{F}=A$ then $A \in \mathcal{F}$. Prove that $A$ has exactly one element.

I'm reading through How to Prove It by Velleman and I'm having trouble with this exercise in the section about Existence and Uniqueness proofs. Here is the exercise:

Suppose $A$ is a set and for every family of sets $\mathcal{F}$, if $\cup \mathcal{F}=A$ then $A \in \mathcal{F}$. Prove that $A$ has exactly one element.

He hints that for both the existence and uniqueness parts of the proof it would be a good idea to use contradiction. I've been playing around with this proof for a while but I can't seem to make any substantial progress.

I current idea is considering some cases where for some family of sets $\mathcal{G}$, $A \in \mathcal{G}$ and $A \notin \mathcal{G}$. I thought if I could show that if $A= \varnothing$ lead to contradictions, I could at least say that there is something in $A$, and try to prove that it is unique from there. I haven't been able to make any progress with this though. Any help with this problem would be greatly appreciated!

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Hints: For $A=\varnothing$, try $\mathcal{F}=\varnothing$. Then conclude $\exists x\in A$ and consider $\mathcal{F} = \{\{x\},A\backslash\{x\}\}$.
If $\mathcal F=\emptyset$, then $\bigcup\mathcal F = \emptyset$, but $\emptyset\notin\mathcal F$, hence clearly $A\ne\emptyset$.
Assume $a\in A$. Let $\mathcal F=\{A\setminus\{a\},\{a\}\}$. Then $\bigcup F=A$ implies $A\in \mathcal F$. Since $a\in A\ne A\setminus\{a\}\not\ni a$, we conclude $A=\{a\}$.