description/proof of that for product set and subsets, projection of union of subsets is union of projections of subsets
Topics
About: set
The table of contents of this article
Starting Context
- 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 and any subsets, the projection of the union of the subsets is the union of the projections of the subsets.
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:
\(J'\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S'_{j'} \in \{\text{ the sets }\} \vert j' \in J'\}\):
\(\times_{j' \in J'} S'_{j'}\): \(= \text{ the product set }\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq \times_{j' \in J'} S'_{j'} \vert j \in J\}\):
\(l'\): \(\in J'\)
\(\pi^{l'}\): \(: \times_{j' \in J'} S'_{j'} \to S'_{l'}, f \mapsto f (l')\), \(= \text{ the projection }\)
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Statements:
\(\pi^{l'} (\cup_{j \in J} S_j) = \cup_{j \in J} \pi^{l'} (S_j)\)
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2: Note
Compare with the proposition that for any product set and any subsets, the projection of the intersection of the subsets is contained in but not necessarily is the intersection of the projections of the subsets.
3: Proof
Whole Strategy: Step 1: see that \(\pi^{l'} (\cup_{j \in J} S_j) \subseteq \cup_{j \in J} \pi^{l'} (S_j)\); Step 2: see that \(\cup_{j \in J} \pi^{l'} (S_j) \subseteq \pi^{l'} (\cup_{j \in J} S_j)\); Step 3: conclude the proposition.
Step 1:
Let \(s'_{l'} \in \pi^{l'} (\cup_{j \in J} S_j)\) be any.
There is an \(f \in \cup_{j \in J} S_j\) such that \(\pi^{l'} (f) = f (l') = s'_{l'}\).
\(f \in S_j\) for a \(j \in J\).
\(s'_{l'} = \pi^{l'} (f)\) implies that \(s'_{l'} \in \pi^{l'} (S_j)\).
So, \(s'_{l'} \in \cup_{j \in J} \pi^{l'} (S_j)\).
So, \(\pi^{l'} (\cup_{j \in J} S_j) \subseteq \cup_{j \in J} \pi^{l'} (S_j)\).
Step 2:
Let \(s'_{l'} \in \cup_{j \in J} \pi^{l'} (S_j)\) be any.
\(s'_{l'} \in \pi^{l'} (S_j)\) for a \(j \in J\).
There is an \(f \in S_j\) such that \(\pi^{l'} (f) = f (l') = s'_{l'}\).
\(f \in \cup_{j \in J} S_j\).
\(s'_{l'} = \pi^{l'} (f)\) implies that \(s'_{l'} \in \pi^{l'} (\cup_{j \in J} S_j)\).
So, \(\cup_{j \in J} \pi^{l'} (S_j) \subseteq \pi^{l'} (\cup_{j \in J} S_j)\).
Step 3:
So, \(\pi^{l'} (\cup_{j \in J} S_j) = \cup_{j \in J} \pi^{l'} (S_j)\).
