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This is an exercise in How to prove it by Velleman.

Suppose $\mathcal{F}$ is a family of sets. Prove that there is a unique set $A$ that has the following two properties:

(a) $\mathcal{F} \subseteq \mathcal{P}(A)$

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

My approach so far: This set is obviously $A = \cup \mathcal{F}$. I have proven the existence part of the proof, but I am struggling with the uniqueness.

My approach so far for proving uniqueness:

Let $P(X)$ denote $\mathcal{F} \subseteq \mathcal{P}(X) \land \forall B(\mathcal{F} \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)$

My first approach was:

$\forall Y \forall Z( (P(Y) \land P(Z)) \rightarrow (Y=Z))$. I tried to somehow prove through $P(Y)$ that $Y = \cup \mathcal{F}$ and through $P(Z)$ that $Z = \cup \mathcal{F}$ and thus that $Y=Z$.

My second approach was:

Prove that $\forall X (P(X) \rightarrow X = A)$. This has also brought me nowhere so far. I tried proving the contra-positive, but got stuck in the case where I had to prove $\lnot P(X)$ when $X \nsubseteq A$.

I would be really grateful for all hints!

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Based on your remark that you allready proved existence I preassume that you have proved that $A:=\cup\mathcal F$ satisfies the conditions (a) and b).

Let it be that the set $A'$ also satisfies these conditions. So:

(a') $\mathcal{F}\subseteq\wp\left(A'\right)$

(b') $\forall B\left[\mathcal{F}\subseteq\wp\left(B\right)\implies A'\subseteq B\right]$

Then $\mathcal F\subseteq\wp(A')$ allowing the conclusion that $A\subseteq A'$. This as a consequence of (a') and (b).

Also $\mathcal F\subseteq\wp(A)$ allowing the conclusion that $A'\subseteq A$. This as a consequence of (a) and (b').

So $A=A'$ wich proves the uniqueness.

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HINT: Let $A=\bigcup\mathscr{F}$, and suppose that $B$ also has the desired properties. Then on the one hand (b) applied to $A$ tells you that $A\subseteq B$, and on the other hand (b) applied to ... ?

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  • $\begingroup$ I would say that $\cup \mathcal{F} \subseteq B$ . But how does that help with uniqueness? $\endgroup$ – geomquestion Oct 4 '15 at 15:09
  • $\begingroup$ @geomquestion: My apologies; I’m not quite awake yet and misread the question. I’ve fixed the answer to apply to the actual question. $\endgroup$ – Brian M. Scott Oct 4 '15 at 15:11

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