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What is the name of this property when

$ (\bigotimes_{a \in A} a ) \in A $

that is the operator selects a member of A. Examples are min, max and median operators on ordered sets. It seems to be similar to idempotency but it is not.

SKV
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2 Answers2

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I'd call it closure, or an operation closed on a particular set.

Scott Caldwell
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    Hrm, I guess in a sense it is, but it's really much stronger than that, right? It's closed on every nonempty (finite maybe) subset. – rschwieb Jul 16 '14 at 18:41
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You could say that the operator is a choice function from a subset of $\mathcal{P}(A)$ into $A$.

In fact, this is pretty much exactly what you're describing ("having sets as inputs and the output for each set is a member of that set").

The only details to consider are that the function might not be defined on the empty set, or might not be defined on infinitely many inputs. Considering that, the domain of your function might only be from $\mathcal{P}(A)\setminus \{\emptyset\}$ (the nonempty subsets of $A$) or else the set $\mathcal{P}_{fin}(A)\setminus \{\emptyset\}$ ( nonempty finite subsets of $A$).

Each "rule" like min/max/median gives you the method of choosing an element out of the set.

rschwieb
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  • I think closure more sounds like a "property". – SKV Jul 16 '14 at 18:24
  • @Siamak What "closure" do you mean? – rschwieb Jul 16 '14 at 18:29
  • the one in the first answer. The domain is closed wrt the operator. – SKV Jul 16 '14 at 18:31
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    Dear @Siamak : I just discovered that Scott removed the link I'm talking about. The solution is not bad, but I don't think it is very accurate. What you're describing is an operation that's closed on every nonempty subset (maybe only finite subsets) and that's a lot stronger than just plain old closure. And closure isn't really even a property of the operation, it's a property of the operation and a set upon which the operation is closed. The fact the operation is a choice function really seems to be a lot closer to what you're describing. Regards – rschwieb Jul 16 '14 at 18:43
  • I see what you mean now. – SKV Jul 16 '14 at 21:29