description/proof of that for product set, \(2\) subsets, and element of subproduct, cross section of 1st subset minus 2nd subset is cross section of 1st subset minus cross section of 2nd subset
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 \(2\) subsets, and any element of any subproduct, the cross section of the 1st subset minus the 2nd subset by the element is the cross section of the 1st subset by the element minus the cross section of the 2nd subset 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:
\(J'\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_{j'} \in \{\text{ the sets }\} \vert j' \in J'\}\):
\(\times_{j' \in J'} S_{j'}\): \(= \text{ the product set }\)
\(Q_1\): \(\subseteq \times_{j' \in J'} S_{j'}\)
\(Q_2\): \(\subseteq \times_{j' \in J'} S_{j'}\)
\(J\): \(\subset J'\), such that \(J \neq \emptyset\)
\(\times_{j \in J} s_j\): \(\in \times_{j \in J} S_j\)
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Statements:
\((Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]} = (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\)
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2: Proof
Whole Strategy: Step 1: see that \((Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]} \subseteq (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\); Step 2: see that \((Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]} \subseteq (Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]}\); Step 3: conclude the proposition.
Step 1:
Let \(\times_{l \in J' \setminus J} s_l \in (Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]}\) be any.
\(\times_{j' \in J'} s_{j'} \in Q_1 \setminus Q_2\).
\(\times_{j' \in J'} s_{j'} \in Q_1\) and \(\times_{j' \in J'} s_{j'} \notin Q_2\).
So, \(\times_{l \in J' \setminus J} s_l \in (Q_1)_{[\times_{j \in J} s_j]}\) and \(\times_{l \in J' \setminus J} s_l \notin (Q_2)_{[\times_{j \in J} s_j]}\).
So, \(\times_{l \in J' \setminus J} s_l \in (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\).
So, \((Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]} \subseteq (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\).
Step 2:
Let \(\times_{l \in J' \setminus J} s_l \in (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\) be any.
\(\times_{l \in J' \setminus J} s_l \in (Q_1)_{[\times_{j \in J} s_j]}\) and \(\times_{l \in J' \setminus J} s_l \notin (Q_2)_{[\times_{j \in J} s_j]}\).
\(\times_{j' \in J'} s_{j'} \in Q_1\) and \(\times_{j' \in J'} s_{j'} \notin Q_2\).
So, \(\times_{j' \in J'} s_{j'} \in Q_1 \setminus Q_2\).
So, \(\times_{l \in J' \setminus J} s_l \in (Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]}\).
So, \((Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]} \subseteq (Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]}\).
Step 3:
So, \((Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]} = (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\).
