Trouble with unique existance proof

Printable View

May 25th 2011, 07:32 PM
Ontolog

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) $F \subseteq \mathcal{P}(A)$
(b) $\forall B (F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)$

So what I want to show is:

$\exists !A[F \subseteq \mathcal{P}(A) \land \forall B(F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)]$

I have the existence part down:

Suppose $A = \cup F$ . Then $F \subseteq \mathcal{P}(A)$ .

Suppose $y \in F$ and $x \in y$ and $B$ such that $F \subseteq B$ .

$A = \cup F$ , so clearly $x \in A$ . Since $F \subseteq \mathcal{P}(B)$ then it
follows that $x \in B$ , therefore $A \subseteq B$ . Since $x$ , $y$ and
$B$ were arbitrary, $\forall B(F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)$ .

Where I'm having trouble is with the uniqueness proof. I believe that in order to prove uniqueness,
I have to show that

$\forall C \{ [F \subseteq \mathcal{P}(C) \land \forall B(F \subseteq \mathcal{P}(B) \rightarrow C \subseteq B)] \rightarrow C = \cup F \}$

The first half of proving the equality is trivial, that is $F \subseteq \mathcal{P}(C)$ so clearly $\cup F \subseteq C$ . It's the
other half that is giving me trouble, that is how do I show that $C \subseteq \cup F$ ? I can see that it is true but for some reason I can't come up with well reasoned argument as to why.
May 25th 2011, 11:17 PM
bryangoodrich

I'm tired and completely didn't think about $A = \cup F$ . I also haven't done set theory in awhile now. Good insight! You are right. The first half of the uniqueness proof is trivial since it follows from (b) and that A exists with the given properties. Given the definition you provided for uniqueness, wouldn't A be included in the $\forall B$ ? If so, does it now follow from (a) then that:

$F \subseteq P(A) \Rightarrow C\subseteq A$

Then the proof is complete, assuming everything else is correct.
May 25th 2011, 11:31 PM
Ontolog

Quote:

Originally Posted by bryangoodrich

I'm tired and completely didn't think about $A = \cup F$ . I also haven't done set theory in awhile now. Good insight! You are right. The first half of the uniqueness proof is trivial since it follows from (b) and that A exists with the given properties. Given the definition you provided for uniqueness, wouldn't A be included in the $\forall B$ ? If so, does it now follow from (a) then that:

$F \subseteq P(A) \Rightarrow C\subseteq A$

Then the proof is complete, assuming everything else is correct.

Hey that's great :) Yeah, that completes the proof. I don't know why, but sometimes if I don't get how to do a certain proof right away, I'll get stuck in this awful whirlpool of circular reasoning that I just can't seem to get out of.