Suppose $F$ is a family of sets that has the property that for every $G \subseteq F$, $\cup G \in F$. Prove that there is a unique set $A$ such that $A \in F$ and $\forall B \in F (B \subseteq A)$.

Proof of existence:

Let $X$ be any set and let $F = \mathcal{P}(X)$. Then for every $G \subseteq F$, $\cup G \in F$. Let $A \in F$ such that $A=X$ Let $B \in F$ be arbitrary. Then $B \subseteq A$. This concludes the proof of existence.

I am pretty sure that this is correct. However, the uniqueness part is bothering me.

Proof of uniqueness:

Let $C \in F$ such that $\forall B \in F (B \subseteq C)$ (first assumption) and let $D \in F$ such that $\forall B \in F (B \subseteq D)$ (second assumption). From the first assumption it follows that $D \subseteq C$, similarly, from the second assumption it follows that $C \subseteq D$. Taken together it follows that $C=D$. This concludes the proof of uniqueness.

Is the proof correct? I am working through an intro to proving things, however, most solutions I found online are dubious at best.

All the best!

  • $\begingroup$ Hint: There is no need to introduce a set $X$ and put $F = P(X)$. For existence, take $F \subseteq F$. For uniqueness, a standard argument is to assume that there are two such $B$. What can we say when two sets are both subsets of one another? :) $\endgroup$
    – JJH
    May 27 '19 at 20:02
  • $\begingroup$ Thanks.I guess that since $F \subseteq F$ it follows that $\cup F \in F$. We then let $A= \cup F$. Suppose then that $B \in F$, It follows that $B \subseteq A$. However, the uniqueness part does not seem to be in need of correction or am I wrong? $\endgroup$
    – Maximilian
    May 27 '19 at 20:07
  • $\begingroup$ @JJH Can confirm my thoughts? $\endgroup$
    – Maximilian
    May 27 '19 at 20:28
  • 1
    $\begingroup$ F is given. You cannot set it to P(X). $\endgroup$ May 27 '19 at 21:01
  • 1
    $\begingroup$ Uniqueness is ok. The first part is still wrong, you cannot set F to a Power set. A = union F will suffice. $\endgroup$ May 27 '19 at 21:22

$A:=\bigcup F$ is itself a valid set and as $F \subseteq F$ by assumption it's a member of $F$ (being the union of a subset), so $A \in F$

If $B \in F$ then $B \subseteq \bigcup F$ is pretty obvious by definition of the union, so $\forall B \in F: B \subseteq A$ is also clear.

Suppose $A'$ also obeys these properties. Then for all $B \in F$ we know (by the second property) that $B \subseteq A'$ so $A=\bigcup F \subseteq A'$. We also $A' \in F$ so $A' \subseteq \bigcup F = A$ so $A'=A$.


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