Let $(\mathfrak{A};\leq)$ be a poset. Let $A$ be its subset.
I will denote
- $\operatorname{Up} A = \{ x\in\mathfrak{A} \mid \forall a\in A: x \geq a \}$;
- $\operatorname{Down} A = \{ x\in\mathfrak{A} \mid \forall a\in A: x \leq a \}$.
I remind that $\sup A = \min\operatorname{Up} A$.
Question: Does it make sense to repeat these operations more than once, like: $$\max\operatorname{Down}\operatorname{Up} A$$ or further $$\min\operatorname{Up}\operatorname{Down}\operatorname{Up} A,$$ etc.?
Attempt to generalize suprema and infima
The upper set of A, written [small up arrow] A or up A = { x | some a in A with a <= x }.
The over or above set of A, the set of upper bounds of A, above A = { x | A <= x } = { x | for all a in A, a <= x }.
down A and below A are the order dual concepts.
Above and below can be used for an algebraic approach to inf and sup. sup A = min above A, for example.
It's an awkward method as compared to the usual point method. Accordingly I've little use for it. As for compounding above and below, the following may be of interest.
below A = below up A; above A = above down S A subset down A subset below above A A subset up A subset above below A
below A = below above below A; above A = above below above A
below below A = {bottom} if bottom exists, otherwise it's empty. above above A = {top} if top exists, otherwise it's empty.
Let S be the set that is (partially) ordered. below above above A = above below below A = S
below empty set = above empty set = S below S = {bottom} if bottom exists, otherwise it's empty. above S = {top} if top exists, otherwise it's empty.
Proofs, if needed, upon request. What would you suggest for generalized supremum?