Is there a mistake in this problem? I am trying to solve the following problem:
Suppose A is a set, and for every family of sets F, if ∪F = A then
A ∈ F. Prove that A has exactly one element.
However, i can come up with a counterexample. Suppose A = {1,2} and A ∈ F. Then ∪F = {1,2} = A.
Do you think the author made a mistake or am i the wrong one here ?
 A: I am not sure what the contradiction is in your example. However, if we choose $F=\{\{x\}\,\,|\,\,x\in A\}$, then $\bigcup F = A$, thus $A\in F$, thus $|A| = \{x\}$ for some $x$, thus $|A|=1$.
A: The assumption is that, for every $F$ with $\cup F=A$, $A \in F$. For illustration, fix $A=\{ 1,2 \}$ for this paragraph. You are correct that if $F=\{ \{ 1,2 \} \}$, then $\cup F=A$ and $A \in F$. But if $F=\{ \{ 1 \},\{ 2 \} \}$ then $\cup F=A$ and $A \not \in F$. That is, this $F$ shows that the hypothesis of the problem does not hold for $A=\{ 1,2 \}$. 
With this counterexample in mind, you can come up with a proof for the original statement: if $F$ is the set of all singleton subsets of $A$, then $\cup F=A$. Now if $A \in F$ then $A$ is a singleton subset of itself, that is, $A$ has one element. 
A different proof: if $F$ is the set of all singleton subsets of $A$, then $F=A$. If $A$ has more than one element, then $A$ cannot be a member of $F$ (since every member of $F$ is a singleton). Now the result follows by contraposition.
