Monday, June 24, 2013

Basic Set Theory

Times out and cannot be enacted with 1 vote / Skju

Adminned at 26 Jun 2013 17:00:45 UTC

Add the following as Axioms:

* For all sets A and B:
**  A + B = C, such that for every element x in A and y in B, x and y are elements of set C
**  A & B = C, such that for every element x in A that is in B, x is is an element of set C
**  A | B = C, such that for every element x in A, but x is not in B, and every element y in B, but y is not in A, x and y are elements of set C
**  A - B = C, such that for every element x in A that is not in B, x is an element of set C
* There exists a set with no elements - the null (or empty) set - {}

introducing the fundamental set definitions. Intended to help create formulas related to the set of Atoms. For example: A = the set of Atoms a such that Has(a,“Essence”,“KY”) is true. If {A} - A, then give each atom b in set (A &{A}) 2 points. Or something of the sort.

Comments

Tavros:

24-06-2013 21:17:44 UTC

Mm, some nitpicks. The axiom defining + does not clearly state that if B has no elements, then the elements of A are nevertheless elements of C (since there is no pair (x, y) such that x and y are “an element x in A and y in B”). None of these axioms state that anything is *not* an element of C; we could consistently assume that no matter what A and B are, the sets (A + B), (A & B), (A | B), and (A - B) are all the universal set.

nqeron:

24-06-2013 21:37:02 UTC

“for every element x in A ... x and y are elements of set C” suggests that all elements of A are in C, regardless of the elements in B.

As for “not C” - perhaps, but it’s moot since it’s not provable. One can only *prove* the existence of certain elements in C, which is the important part. In principle, you’re right, though - it should be provable that the elements mentioned are the only ones.
This really matters most with subtractive logic (A|B and A-B). I would just argue that it only refers to items provably in A or B.

RaichuKFM: she/her

26-06-2013 18:51:57 UTC

imperial