# Existence and uniqueness of a set in a family of sets.

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!

• 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? :)
– JJH
May 27 '19 at 20:02
• 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? May 27 '19 at 20:07
• @JJH Can confirm my thoughts? May 27 '19 at 20:28
• F is given. You cannot set it to P(X). May 27 '19 at 21:01
• Uniqueness is ok. The first part is still wrong, you cannot set F to a Power set. A = union F will suffice. 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 $$A'=A$$.