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# Every set has a minimal superset

I credit Mikey and Atal with this proof.

For any set $A$ which is a subset of a (lets say much larger) set $K$, a minimal superset of $A$ is a set $B\subset K$ such that $A\subset B$, and if $A \subset C$, then $B\subset C$ for any superset $C$ of $A$.

I claim that any set $A$ has a minimal superset. We see this by using Zorn’s Lemma on the elements of $\mathcal{P}(K)$ which contain $A$ (I denote this collection of sets as $K_A$). The ordering will be the reverse of the typical: if $I, J \in K_A$ such that $I\subset J$, then we say $I > J$. If $S_1 < S_2 < \dots$ is a chain in $K_A$, then this chain is bounded above by $\bigcap_{i=1}^\infty S_i$. Thus, by Zorn’s Lemma, there is a maximal element $B$.

I claim this $B$ is a minimal subset. We see that $B$ must contain $A$ by definition, and further $B$ contains no other elements of $K_A$. Suppose that $C$ is in $K_A$. We have that $A\subset C$ and $A \subset C\cap B$, as both $C$ and $B$ are in $K_A$. Further, $C\cap B \subset B$. Because $C\cap B$ contains $A$ and is a subset of $B$, we are forced to conclude, by maximality of $B$, that $B = C\cap B$, meaning that $B \subset C$. Thus, $B$ is a minimal superset of $A$.

Of course, this whole proof can be greatly simplified by noting that $A$ itself will fulfill the property of being a minimal superset of $A$, so this is a proof by immediate example. Hey, the definition doesn’t forbid it!