A description/proof of that product of any complements is product of whole sets minus union of products of whole sets 1 of which is replaced with subset for each product
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 product of any complements is the product of the whole sets minus the union of the products of the whole sets, 1 of which is replaced with 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, \(S'_i \subseteq S_i\), \((S_1 \setminus S'_1) \times (S_2 \setminus S'_2) \times . . . \times (S_n \setminus S'_n) = S_1 \times S_2 \times . . . \times S_n \setminus ((S'_1 \times S_2 \times . . . \times S_n) \cup (S_1 \times S'_2 \times . . . \times S_n) \cup . . . \cup (S_1 \times S_2 \times . . . \times S'_n))\).
2: Proof
For any \(p = (p_1, p_2, . . ., p_n) \in (S_1 \setminus S'_1) \times (S_2 \setminus S'_2) \times . . . \times (S_n \setminus S'_n)\), \(p_i \in S_i \setminus S'_i\), \(p_i \notin S'_i\), \((p_1, p_2, . . ., p_n) \notin S_1 \times S_2 \times . . . \times S'_i \times ... \times S_n\), so, \(p \in S_1 \times S_2 \times . . . \times S_n \setminus ((S'_1 \times S_2 \times . . . \times S_n) \cup (S_1 \times S'_2 \times . . . \times S_n) \cup . . . \cup (S_1 \times S_2 \times . . . \times S'_n))\); for any \(p \in S_1 \times S_2 \times . . . \times S_n \setminus ((S'_1 \times S_2 \times . . . \times S_n) \cup (S_1 \times S'_2 \times . . . \times S_n) \cup . . . \cup (S_1 \times S_2 \times . . . \times S'_n))\), \(p \notin (S'_1 \times S_2 \times . . . \times S_n) \cup (S_1 \times S'_2 \times . . . \times S_n) \cup . . . \cup (S_1 \times S_2 \times . . . \times S'_n)\), \(p \notin S_1 \times S_2 \times . . . \times S'_i \times . . . \times S_n\) for each \(i\), \(p_i \notin S'_i\) for each \(i\), \(p_i \in S_i \setminus S'_i\) for each \(i\), so, \(p \in (S_1 \setminus S'_1) \times (S_2 \setminus S'_2) \times . . . \times (S_n \setminus S'_n)\).
