Set theory proof

  • #1
189
4

Homework Statement


Assume {B, C, D} is a partition of the universal set U, A is a subset of U and A is not a subset of B complement, A is not a subset of C complement, A is not a subset of D complement. Prove that {A ∩ B, A ∩ C, A ∩ D} is a partition of A.

Homework Equations




The Attempt at a Solution


I know that this is right intuitively. I know how explain it with words, but I don't know how mathematically.

A is not a subset of B complement, A is not a subset of C complement, A is not a subset of D complement implies that A has to be distributed among the three subsets. There is nothing left of A because B,C and D is a partition of the universal set. Therefore, the union of the pieces in which A overlaps with B,C,D is A. This pieces are going to be disjoint. Mathematically, I can prove that

(AnB)n(AnC)n(AnD)=An(BnCnD)= An empty set =empty set
 

Answers and Replies

  • #2
Ssnow
Gold Member
532
156
You must prove that the intersection of every two pairs of your new family ##\{A\cap B, A\cap C, A\cap D\}## is empty (using the fact that ##\{B,C,D\}## is a partition of ##U## and that in fact ##A## is not contained in the complement of every set of the partition (that is the same that ##A## has intersection noempty with every set in the partition...)) and that the union of all is ##A##. You will use the distributive law for intersection and union ...
 

Related Threads on Set theory proof

  • Last Post
Replies
7
Views
775
Replies
5
Views
2K
  • Last Post
Replies
5
Views
771
  • Last Post
Replies
4
Views
9K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
4
Views
941
Replies
3
Views
5K
Top