description/proof of that for product set, subset, and element of subproduct, cross section of subset is contained in projection of subset
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of cross section of subset of product set by element of subproduct set.
- The reader knows a definition of projection from product set onto subproduct set.
Target Context
- The reader will have a description and a proof of the proposition that for any product set, any subset, and any element of any subproduct, the cross section of the subset by the element is contained in the projection of the subset onto the complemental subproduct.
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:
\(J'\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_{j'} \in \{\text{ the sets }\} \vert j' \in J'\}\):
\(\times_{j' \in J'} S_{j'}\): \(= \text{ the product set }\)
\(Q\): \(\subseteq \times_{j' \in J'} S_{j'}\)
\(J\): \(\subset J'\), such that \(J \neq \emptyset\)
\(\times_{j \in J} S_j\): \(= \text{ the product set }\)
\(\times_{j \in J} s_j\): \(\in \times_{j \in J} S_j\)
\(\pi^{J' \setminus J}\): \(: \times_{j' \in J'} S_{j'} \to \times_{l \in J' \setminus J} S_l\), \(= \text{ the projection }\)
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Statements:
\(Q_{[\times_{j \in J} s_j]} \subseteq \pi^{J' \setminus J} (Q)\)
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2: Proof
Whole Strategy: Step 1: see that for each \(\times_{l \in J' \setminus J} s_l \in Q_{[\times_{j \in J} s_j]}\), \(\times_{l \in J' \setminus J} s_l \in \pi^{J' \setminus J} (Q)\).
Step 1:
Let \(\times_{l \in J' \setminus J} s_l \in Q_{[\times_{j \in J} s_j]}\) be any.
\(\times_{j' \in J'} s_{j'} \in Q\).
\(\pi^{J' \setminus J} (\times_{j' \in J'} s_{j'}) = \times_{l \in J' \setminus J} s_l\), which means that \(\times_{l \in J' \setminus J} s_l \in \pi^{J' \setminus J} (Q)\).
So, \(Q_{[\times_{j \in J} s_j]} \subseteq \pi^{J' \setminus J} (Q)\).
