A group that's a collection of sets

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SUMMARY

The discussion focuses on the mathematical operation defined for a set of subsets P of a set S, specifically the operation A*B = [(S-A) ∩ B] ∪ [A ∩ (S-B)]. Participants are tasked with proving that this operation is commutative, identifying the identity element, and finding the inverse of a subset A. Key insights include the use of unions and intersections, which are commutative operations, to demonstrate A*B = B*A. The identity element I is sought such that A*I = I*A = A for all A in P.

PREREQUISITES
  • Understanding of set theory, including subsets and set operations.
  • Familiarity with the concepts of unions and intersections in mathematics.
  • Knowledge of commutative properties of operations.
  • Basic skills in mathematical proofs and logic.
NEXT STEPS
  • Study the properties of set operations in detail, focusing on unions and intersections.
  • Learn about identity elements in algebraic structures.
  • Explore the concept of inverses in set operations.
  • Review examples of commutative operations in various mathematical contexts.
USEFUL FOR

Mathematics students, educators, and anyone interested in advanced set theory and algebraic structures will benefit from this discussion.

halvizo1031
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Homework Statement


Let S be a set of things and let P be the set of subsets of S. For A,B in P, define
A*B=[(S-A)intersection B] union [A intersection (S-B)]
I'm suppose to show that (p,*) is commutative, find the identity, and given that A is a subset of S, find the inverse of A. How do i even start this?



Homework Equations





The Attempt at a Solution


I need help starting it.
 
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Show that A*B=B*A.

Find I \in S such that A*I=I*A=A , \quad\forall A \in P.

Find B \in S such that A*B=I.
 
Donaldos said:
Show that A*B=B*A.

Find I \in S such that A*I=I*A=A , \quad\forall A \in P.

Find B \in S such that A*B=I.


I'm sorry but i couldn't understand what you wrote. And by that i mean that i can't read it. Could you rewrite it please?
 
then try the follwing multiplications: A with the empty set & A with its complement
 

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