Help with existence & uniqueness proof involving sets. Im currently reading Velleman's How To Prove It. Exercise 6. (a) Asks to prove the following: Let $U$ be any set. Prove there is a unique $A \in \mathcal{P}(U)$ such that for every $B \in \mathcal{P}(U)$, $A \cup B = B$
Im a little puzzled with the uniqueness solution that the book proposes. It goes like this:
Proof
Let $A = \varnothing \in \mathcal{P}(U)$. Then clearly for any $B \in \mathcal{P}(U)$, $A \cup B = \varnothing \cup B = B$
To see that $A$ is unique, suppose $A' \in \mathcal{P}(U) $ & for all $B \in \mathcal{P}(U)$, $A' \cup B = B $. Then in particular taking $B = \varnothing$, we can conclude that $A' \cup \varnothing = \varnothing$. But clearly $A' \cup \varnothing = A'$ so we have,$A' = \varnothing = A$
$\square$
What's bugging me is that I don't seem to get why is it valid to set $B = \varnothing$. It seems to me that we're only showing that $A' = \varnothing$ only when $B = \varnothing$, but $B$ is supposed to be arbitrary. Why is the proof still valid?
 A: $B$ is indeed supposed to be an arbitrary element of $\mathcal{P}(U)$. That's exactly why we can take $B=\emptyset$!
We suppose $A'$ has this property. That means $A'\cup B=B$ for all $B$. If this is true, then in particular it is true for $B=\emptyset$, whence $A'\cup\emptyset=\emptyset$ and $A'=\emptyset$.
So, we did not actually need to check this property for all $B$, just the one. That doesn't make the proof any less valid.
Again, if $A'$ has this property for arbitrary $B$, then we, as the arbiters, may choose $B=\emptyset$ to test $A'$ on. Why not? We could have chosen another $B$, but we don't have to.
A: To prove existence, they let $A'$ be an arbitrary solution. This means that $A'\cup B=B$ for all $B\in\mathcal{P}(U)$. Note also that $A'$ is fixed, we are not changing it from now on.
Thus, for any choice of $B$, we will have $A'\cup B=B$. Picking $B=\emptyset$, it must hold for this $B$. Of course, it holds for other $B$'s as well, but this one is useful. We see that $B=A'\cup B = A'\cup \emptyset = A'$. Using just a single choice of $B$, we can deduce something about our fixed $A'$. But since $A'$ was arbitrary, we get that any solution has this form.
