Trouble with unique existance proof
Exercise 3.7.1 of "How To Prove It" (Vellemen 2006)
Suppose F is a family of sets. Prove that there is a unique set A that has the following two properties:
(a) $\displaystyle F \subseteq \mathcal{P}(A)$
(b) $\displaystyle \forall B (F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)$
So what I want to show is:
$\displaystyle \exists !A[F \subseteq \mathcal{P}(A) \land \forall B(F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)]$
I have the existence part down:
Suppose $\displaystyle A = \cup F$. Then $\displaystyle F \subseteq \mathcal{P}(A)$.
Suppose $\displaystyle y \in F$ and $\displaystyle x \in y$ and $\displaystyle B$ such that $\displaystyle F \subseteq B$.
$\displaystyle A = \cup F$, so clearly $\displaystyle x \in A$. Since $\displaystyle F \subseteq \mathcal{P}(B)$ then it
follows that $\displaystyle x \in B$, therefore $\displaystyle A \subseteq B$. Since $\displaystyle x$, $\displaystyle y$ and
$\displaystyle B$ were arbitrary, $\displaystyle \forall B(F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)$.
Where I'm having trouble is with the uniqueness proof. I believe that in order to prove uniqueness,
I have to show that
$\displaystyle \forall C \{ [F \subseteq \mathcal{P}(C) \land \forall B(F \subseteq \mathcal{P}(B) \rightarrow C \subseteq B)] \rightarrow C = \cup F \}$
The first half of proving the equality is trivial, that is $\displaystyle F \subseteq \mathcal{P}(C)$ so clearly $\displaystyle \cup F \subseteq C$. It's the
other half that is giving me trouble, that is how do I show that $\displaystyle C \subseteq \cup F$? I can see that it is true but for some reason I can't come up with well reasoned argument as to why.
