description/proof of that for product set, subsets, and element of subproduct, cross section of intersection of subsets is intersection of cross sections of subsets
Topics
About: set
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any product set, any subsets, and any element of any subproduct, the cross section of the intersection of the subsets by the element is the intersection of the cross sections of the subsets by the element.
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:
\(J'\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_{j'} \in \{\text{ the sets }\} \vert j' \in J'\}\):
\(\times_{j' \in J'} S_{j'}\): \(= \text{ the product set }\)
\(K\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{Q_k \subseteq \times_{j' \in J'} S_{j'} \vert k \in K\}\):
\(J\): \(\subset J'\), such that \(J \neq \emptyset\)
\(\times_{j \in J} s_j\): \(\in \times_{j \in J} S_j\)
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Statements:
\((\cap_{k \in K} Q_k)_{[\times_{j \in J} s_j]} = \cap_{k \in K} ({Q_k}_{[\times_{j \in J} s_j]})\)
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2: Proof
Whole Strategy: Step 1: see that \((\cap_{k \in K} Q_k)_{[\times_{j \in J} s_j]} \subseteq \cap_{k \in K} ({Q_k}_{[\times_{j \in J} s_j]})\); Step 2: see that \(\cap_{k \in K} ({Q_k}_{[\times_{j \in J} s_j]}) \subseteq (\cap_{k \in K} Q_k)_{[\times_{j \in J} s_j]}\); Step 3: conclude the proposition.
Step 1:
Let \(\times_{l \in J' \setminus J} s_l \in (\cap_{k \in K} Q_k)_{[\times_{j \in J} s_j]}\) be any.
\(\times_{j' \in J'} s_{j'} \in \cap_{k \in K} Q_k\).
For each \(k \in K\), \(\times_{j' \in J'} s_{j'} \in Q_k\).
So, for each \(k \in K\), \(\times_{l \in J' \setminus J} s_l \in {Q_k}_{[\times_{j \in J} s_j]}\).
So, \(\times_{l \in J' \setminus J} s_l \in \cap_{k \in K} ({Q_k}_{[\times_{j \in J} s_j]})\).
So, \((\cap_{k \in K} Q_k)_{[\times_{j \in J} s_j]} \subseteq \cap_{k \in K} ({Q_k}_{[\times_{j \in J} s_j]})\).
Step 2:
Let \(\times_{l \in J' \setminus J} s_l \in \cap_{k \in K} ({Q_k}_{[\times_{j \in J} s_j]})\) be any.
For each \(k \in K\), \(\times_{l \in J' \setminus J} s_l \in {Q_k}_{[\times_{j \in J} s_j]}\).
For each \(k \in K\), \(\times_{j' \in J'} s_{j'} \in Q_k\).
So, \(\times_{j' \in J'} s_{j'} \in \cap_{k \in K} Q_k\).
So, \(\times_{l \in J' \setminus J} s_l \in (\cap_{k \in K} Q_k)_{[\times_{j \in J} s_j]}\).
So, \(\cap_{k \in K} ({Q_k}_{[\times_{j \in J} s_j]}) \subseteq (\cap_{k \in K} Q_k)_{[\times_{j \in J} s_j]}\).
Step 3:
So, \((\cap_{k \in K} Q_k)_{[\times_{j \in J} s_j]} = \cap_{k \in K} ({Q_k}_{[\times_{j \in J} s_j]})\).
