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Given a set $A$ such that for any family of sets $F$: if $\cup F = A$ then $A \in F$.

Prove that $A$ has only one element.

Please help.

Thank you all.

I tried to solve it by assuming that there are two elements in $A$. But I can't solve it.

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    Are you sure $\,F\,$ is to a family of *groups* and not of *sets*? Otherwise the cyclic group of order two is a contradiction to the claim...2012-11-28
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    @DonAntonio, you are right. I translated the word to english and use a group instead of a set.2012-11-28
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    Ok then. Take into account that when you have some doubt you can try to write down some words or sentences in your own language. There's a fair chance somebody knows that language and can help you out. Also change your post's title.2012-11-28

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Well: if $\,|A|\geq 2\,$ then the family $\,\mathcal F:=\{\,\{a\}\;\;;\;\;a\in A\}\,$ fulfills the given condition

$$\bigcup_{F\in\mathcal F}F=A\,\,\,,\,\,\text{yet}\,\,A\notin\mathcal F\,\;\;(\text{can you see why?)}$$

Thus, it must be that $\,|A|=1\,$

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    can you give me an exaple of A and F? I think I dont really know what is F.. thank you!2012-11-28
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    For example take the simplest non-trivial group, the cyclic one of order two: $\,A=\{1,x\}\,\,,\,x^2=1\,$, and take $\,\mathcal F=\{\{1\}\,,\,\{x\}\}\,$ ...:)2012-11-28
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    I know why but i dont know how to explain it.. the answer is: A doesn't belong to F because of the definition of F..2012-11-28
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    @AlonShmiel, by any chance is your mother tongue hebrew or spanish? If it is I could explain it to you in that language. WE see here an example of a group that doesn't fulfill the condition $\,\bigcup_{F\in\mathcal F}F=A\Longrightarrow A\in\mathcal F\,$, and the reason is that $\,A\,$ has more than one element...I don't know how or what else is there to explain.2012-11-28