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