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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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2 Answers 2

up vote 5 down vote accepted

Hints: For $A=\varnothing$, try $\mathcal{F}=\varnothing$. Then conclude $\exists x\in A$ and consider $\mathcal{F} = \{\{x\},A\backslash\{x\}\}$.

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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 $A=\{a\}$.

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