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

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.

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... – DonAntonio Nov 28 '12 at 3:18
@DonAntonio, you are right. I translated the word to english and use a group instead of a set. – Alon Shmiel Nov 28 '12 at 10:27
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. – DonAntonio Nov 28 '12 at 11:57

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 $\,|A|=1\,$
For example take the simplest non-trivial group, the cyclic one of order two: $\,A=\{1,x\}\,\,,\,x^2=1\,$, and take $\,\mathcal F=\{\{1\}\,,\,\{x\}\}\,$ ...:) – DonAntonio Nov 28 '12 at 14:15
@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. – DonAntonio Nov 28 '12 at 16:05