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!

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