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Is the membership function not equivalent to what I describe above on the query side? My difference is that I am making explicit that the set needs to be popula
by cgio 4y ago
Is the membership function not equivalent to what I describe above on the query side? My difference is that I am making explicit that the set needs to be populated. The uttering of a set of “indescribable “ objects does not a set give you. That membership function works both inwards and outwards.
- recursive 4y agoThis concept of "population" doesn't really seem to be consistent with my understanding of a set. Set A is the set of all widgets that have a frozz. The existence of Set A doesn't imply that there exists a widget that has a frozz. Perhaps the existence of such a widget is an open question. But the set is perfectly well defined.
- lisper 4y agoRight. The set of all integers which are both even and odd is a well-defined set. It just happens to be empty. The set of all counter-examples to the Goldbach conjecture is a well-defined set. We don't know if it's empty or not.
- cgio 4y agoIf I don’t know what frozz is, is the set well defined? Maybe in form, but not in essences. Lisper’s example in sibling comment is indeed well formed, but I know that if I get a counter-example to the conjecture it will pass the population interface and become part of the set. Therefore, I don’t see how we differ comparing simple membership with population and query. My feeling is we disagree over an ex nihilo vs procedural emergence of sets.