A description/proof of that complement of product of subsets is union of products of whole sets 1 of which is replaced with complement of subset
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set.
Target Context
- The reader will have a description and a proof of the proposition that the complement of the product of any subsets is the union of the products of the whole sets, 1 of which is replaced with the complement of a subset for each product.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any sets, \(S_1, S_2, . . ., S_n\), and any subsets, \(S'_i \subseteq S_i\), \((S_1 \times S_2 \times . . . \times S_n) \setminus (S'_1 \times S'_2 \times . . . \times S'_n) = ((S_1 \setminus S'_1) \times S_2 \times . . . \times S_n) \cup (S_1 \times (S_2 \setminus S'_2) \times . . . \times S_n) \cup . . . \cup (S_1 \times S_2 \times . . . \times (S_n \setminus S'_n))\).
2: Proof
For any \(p = (p^1, p^2, . . ., p^n) \in (S_1 \times S_2 \times . . . \times S_n) \setminus (S'_1 \times S'_2 \times . . . \times S'_n)\), \(p \notin S'_1 \times S'_2 \times . . . \times S'_n\), \(p^i \notin S'_i\) for an \(i\), \(p^i \in S_i \setminus S'_i\) for an \(i\), \(p \in S_1 \times . . . \times (S_i \setminus S'_i) \times . . . \times S_n\) for an \(i\), \(p \in ((S_1 \setminus S'_1) \times S_2 \times . . . \times S_n) \cup (S_1 \times (S_2 \setminus S'_2) \times . . . \times S_n) \cup . . . \cup (S_1 \times S_2 \times . . . \times (S_n \setminus S'_n))\)\); for any \(p \in ((S_1 \setminus S'_1) \times S_2 \times . . . \times S_n) \cup (S_1 \times (S_2 \setminus S'_2) \times . . . \times S_n) \cup . . . \cup (S_1 \times S_2 \times . . . \times (S_n \setminus S'_n))\), \(p \in S_1 \times . . . \times (S_i \setminus S'_i) \times . . . \times S_n\) for an \(i\), \(p^i \in S_i \setminus S'_i\) for an \(i\), \(p^i \notin S'_i\) for an \(i\), \(p \notin S'_1 \times S'_2 \times . . . \times S'_n\), \(p \in (S_1 \times S_2 \times . . . \times S_n) \setminus (S'_1 \times S'_2 \times . . . \times S'_n)\)).
