I am going through Vellemen's How to prove it in my free time. Currently I am working on problem 6 in section 3.6: Prove that there is a unique A ∈ P(U) such that for every B ∈ P(U), A ∪ B =B.
I already guessed that the unique set sought is the empty set and there is a solution in the back of the book. But to prove uniqueness I thought whether there might be another way. So here's the outline. Suppose $A$ is not the empty set. Then there must be at least one non-trivial element in $A$, call it $x$ Since the property has to hold for all $B$ if we can find counter example then we can show that $A$ has to be empty set. So suppose $B$ is a set which doesn't contain $x$. Then $A\cup B\neq B$. Therefore $A$ has to be the empty set. Is this a valid proof for uniqueness?
How are you sure that there is a set $B$ that does not contain $x$ ?
You can actually go much faster. Suppose $A$ and $B$ both satisfy the property. Then $$B = A \cup B = B \cup A = A.$$