description/proof of that for product set, intersection of projection-preimages of subsets of constituent sets is product of intersections 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 of any possibly uncountable number of constituent sets and any subsets of the constituent sets, the intersection of the projection-preimages of the subsets is the product of the intersections 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:
\(J_1\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(J_2\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S'_j \in \{\text{ the sets }\}\vert j \in J_1\}\):
\(S'\): \(= \times_{j \in J_1} S'_j\)
\(\{S_l \in \{\text{ the sets }\} \vert l \in J_2 \land \exists j \in J_1 (S_l \subseteq S'_j)\}\)
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Statements:
\(\cap_{l \in J_2} {\pi_{j (l)}}^{-1} (S_l) = \times_{j \in J_1} \cap_{\{l \in J_2 \vert S_l \subseteq S'_j\}} S_l\) where "\(j (l)\)" means that \(j\) is uniquely determined by \(l\) by \(S_l \subseteq S'_j\) and \(\cap_{\emptyset} S_l = S'_j\)
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2: Proof
Whole Strategy: Step 1: see that \(\cap_{l \in J_2} {\pi_{j (l)}}^{-1} (S_l) \subseteq \times_{j \in J_1} \cap_{\{l \in J_2 \vert S_l \subseteq S'_j\}} S_l\); Step 2: see that \(\times_{j \in J_1} \cap_{\{l \in J_2 \vert S_l \subseteq S'_j\}} S_l \subseteq \cap_{l \in J_2} {\pi_{j (l)}}^{-1} (S_l)\).
Step 1:
Let \(p \in \cap_{l \in J_2} {\pi_{j (l)}}^{-1} (S_l)\) be any.
\(p \in {\pi_{j (l)}}^{-1} (S_l)\) for each \(l \in J_2\), so, \(p_{j (l)} := \pi_{j (l)} (p) \in S_l\) for each \(l \in J_2\).
Let \(j \in J_1\) be any such that \(\{l \in J_2 \vert S_l \subseteq S'_j\} \neq \emptyset\).
\(p_j \in \cap_{\{l \in J_2 \vert S_l \subseteq S'_j\}} S_l\).
Let \(j \in J_1\) be any such that \(\{l \in J_2 \vert S_l \subseteq S'_j\} = \emptyset\).
\(p_j \in S'_j = \cap_{\emptyset} S_l = \cap_{\{l \in J_2 \vert S_l \subseteq S'_j\}} S_l\).
So, \(p \in \times_{j \in J_1} \cap_{\{l \in J_2 \vert S_l \subseteq S'_j\}} S_l\).
Step 2:
Let \(p \in \times_{j \in J_1} \cap_{\{l \in J_2 \vert S_l \subseteq S'_j\}} S_l\) be any.
\(p_j \in \cap_{\{l \in J_2 \vert S_l \subseteq S'_j\}} S_l\).
\(p_{j'} \in \pi_{j'} ({\pi_{j (l)}}^{-1} (S_l))\) for each \(l\)?
When \(j' = j (l)\), \(\pi_{j'} ({\pi_{j (l)}}^{-1} (S_l)) = S_l\), by the proposition that for any map between any sets, the composition of the map after the preimage of any subset of the codomain is identical if the map is surjective with respect to the argument subset, but as \(p_{j'} \in \cap_{\{l' \in J_2 \vert S_{l'} \subseteq S'_{j'}\}} S_{l'}\) and \(l\) is such an \(l'\), \(p_{j'} \in S_l\), so, \(p_{j'} \in \pi_{j'} ({\pi_{j (l)}}^{-1} (S_l))\).
When \(j' \neq j (l)\), \(\pi_{j'} ({\pi_{j (l)}}^{-1} (S_l)) = S'_l\), but as \(p_{j'} \in S'_l\), \(p_{j'} \in \pi_{j'} ({\pi_{j (l)}}^{-1} (S_l))\).
So, yes, and \(p \in {\pi_{j (l)}}^{-1} (S_l)\) for each \(l\), and so, \(p \in \cap_{l \in J_2} {\pi_{j (l)}}^{-1} (S_l)\).
