Proving uniqueness in naive set theory I am doing the following exercise from Velleman's How To Prove It:
Prove that there is a unique $A \in \mathcal P (U)$ such that for every $ B\in \mathcal P (U)$, $A \cup B=A$. ($ \mathcal P (U)$ being the powerset of U)
One solution I found is as follows:
To prove the existence, try $A=U$. For any $ B\in \mathcal P (U)$, $B\subseteq U$ and so $U\cup B=U$.
To prove the uniqueness, let say there is another set $ C\in \mathcal P (U)$ s.t. $\forall B \in \mathcal P (U)(C \cup B=C)$. We can put $B=U \setminus C$ in this equation to get $C \cup (U \setminus C) =C$, therefore $U=C$ and hence $C=U=A$.
Two confusions I have:


*

*I understand the existence proof, but how could you know you should substitute $U$ for $A$? Is it just by practice and experience?

*I do not understand what's going on in the uniqueness proof at all. In the preceding pages the following existence and uniqueness form have been mentioned:
a. $\exists x(Fx \land \forall y(Fy \to y=x) $
b. $\exists x \forall y(Fy$↔$y=x)$
c. $\exists x Fx \land \forall y \forall z ((Fy \land Fz) \to y=z)  $
But none of these seem to fit into what he is trying to do here. I don't understand where is the $ C\in \mathcal P (U)$ from; it seems that he did an existential instantiation but it is not apparent to me where the existential comes from.
More importantly, why substitute $B$ (instead of $A$) for $B=U \setminus C$? And wouldn't $B=U \setminus C$ be contradictory to the conclusion? Since if we substitute $B$ with this from $(C \cup B=C)$, it becomes $x\in C \lor (x \in U \land \sim x \in C)=x \in C$. If we try to prove this by cases, the second case i.e. second disjunction would yield the negated form of the conclusion: $\sim x \in C$!
Could anyone please help me on this? Thank you so much!
 A: To answer the first question:If you think about the property that such $A\in \mathcal{P}$ should satisfy, it is:

**$(\star)$  For each $B\in\mathcal{P}(U)$, $A\cap B=B$. **

So, such $A$ is a subset of $U$, and since $A\cap B=B$ iff $B\subseteq A$, we have that such $A$ is actually the largest subset of $U$. This gives you some intuition of why $A$ should be equal to $U$.
For the second question: I don't see clearly why the intersections become unions in your argument, so I take the opportunity to change it a little bit:

Goal: Prove that if $C$ satisfy the required condition, then $C=U$.

First of all, we know that $C\in \mathcal{P}(U)$, so $C\subseteq U$. On the other hand, if we put $B=U\setminus C$, then by the condition $(\star)$ we have
$$U\setminus C=B=B\cap C=(U\setminus C) \cap C=\emptyset$$ and so, $U\subseteq C$ (because we knew already that $C\subseteq U$).
This shows that $C=U$, by double contenence.
A: To answer the first question:If you think about the property that such $A\in \mathcal{P}$ should satisfy, it is:

**$(\star)$  For each $B\in\mathcal{P}(U)$, $A\cup B=A$. **

So, such $A$ is a subset of $U$, and since $A\cup B=A$ iff $B\subseteq A$, we have that such $A$ is actually the largest subset of $U$. This gives you some intuition of why $A$ should be equal to $U$.
For the second question: I don't see clearly why the intersections become unions in your argument, so I take the opportunity to change it a little bit:

Goal: Prove that if $C$ satisfy the required condition, then $C=U$.

First of all, we know that $C\in \mathcal{P}(U)$, so $C\subseteq U$. On the other hand, if we put $B=U\setminus C$, then by the condition $(\star)$ we have
$$C=B\cup C=(U\setminus C) \cup C$$ and so, $U\setminus C\subseteq C$. This is only possible if $U\setminus C=\emptyset$, or equivalent, if $U\subseteq C$.
This shows that $C=U$, by double contenence.
