description/proof of that product of complements of subsets 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 for any product set, the product of the complements of any subsets 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S'_j \in \{\text{ the sets }\} \vert j \in J\}\):
\(\times_{j \in J} S'_j\): \(= \text{ the product set }\)
\(\{S_j \subseteq S'_j \vert j \in J\}\):
//
Statements:
\(\times_{j \in J} (S'_j \setminus S_j) = (\times_{j \in J} S'_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j})\) where \(S^`_{j', j'} = S_{j'}\) and \(S^`_{j', j} = S'_j\) for each \(j \in J \setminus \{j'\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(\times_{j \in J} (S'_j \setminus S_j) \subseteq (\times_{j \in J} S'_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j})\); Step 2: see that \((\times_{j \in J} S'_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j}) \subseteq \times_{j \in J} (S'_j \setminus S_j)\); Step 3: conclude the proposition.
Step 1:
Let \(p \in \times_{j \in J} (S'_j \setminus S_j)\) be any.
\(p^j \in S'_j \setminus S_j\) for each \(j \in J\).
\(p^j \notin S_j\) for each \(j \in J\).
For each \(j' \in J\), \(p^{j'} \notin S_{j'} = S^`_{j', j'}\), so, \(p \notin \times_{j \in J} S^`_{j', j}\) for each \(j'\).
So, \(p \notin \cup_{j' \in J} \times_{j \in J} S^`_{j', j}\).
So, \(p \in (\times_{j \in J} S'_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j})\).
So, \(\times_{j \in J} (S'_j \setminus S_j) \subseteq (\times_{j \in J} S'_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j})\).
Step 2:
Let \(p \in (\times_{j \in J} S'_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j})\) be any.
\(p \notin \cup_{j' \in J} \times_{j \in J} S^`_{j', j}\).
\(p \notin \times_{j \in J} S^`_{j', j}\) for each \(j' \in J\).
\(p^{j'} \notin S^`_{j', j'} = S_{j'}\) for each \(j' \in J\), because for each \(j \in J \setminus \{j'\}\), \(p^j \in S'_j = S^`_{j', j}\).
\(p^{j'} \in S'_{j'} \setminus S_{j'}\) for each \(j' \in J\).
So, \(p \in \times_{j \in J} (S'_j \setminus S_j)\).
So, \((\times_{j \in J} S'_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j}) \subseteq \times_{j \in J} (S'_j \setminus S_j)\).
Step 3:
\(\times_{j \in J} (S'_j \setminus S_j) = (\times_{j \in J} S'_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j})\).
