# 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?

$$B$$ 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.
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.