description/proof of that for maps from same domain and map from same domain into product set with image as product of images, preimage of product of subsets is intersection of preimages of subsets
Topics
About: set
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description 1
- 2: Proof 1
- 3: Structured Description 2
- 4: Proof 2
Starting Context
- The reader knows a definition of map.
- The reader knows a definition of product set.
Target Context
- The reader will have a description and a proof of the proposition that for any possibly uncountable number of maps from any same domain and the map from the same domain into the product set of the codomains with image as the product of the images, the preimage of the product of any subsets of the codomains is the intersection of the preimages 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 1
Here is the rules of Structured Description.
Entities:
\(S\): \(\in \{\text{ the sets }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S'_j \in \{\text{ the sets }\} \vert j \in J\}\):
\(\{f_j: S \to S'_j \vert j \in J\}\):
\(S'\): \(= \times_{j \in J} S'_j\)
\(f\): \(: S \to S', s \mapsto (: J \to \cup_{j \in J} S_j, j \mapsto f_j (s))\)
\(\{S_j \subseteq S'_j\} \vert j \in J\}\):
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Statements:
\(f^{-1} (\times_{j \in J} S_j) = \cap_{j \in J} {f_j}^{-1} (S_j)\)
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2: Proof 1
Whole Strategy: Step 1: see that \(f^{-1} (\times_{j \in J} S_j) \subseteq \cap_{j \in J} {f_j}^{-1} (S_j)\); Step 2: see that \(\cap_{j \in J} {f_j}^{-1} (S_j) \subseteq f^{-1} (\times_{j \in J} S_j)\).
Step 1:
Let \(p \in f^{-1} (\times_{j \in J} S_j)\) be any.
\(f (p) = (:J \to \cup_{j \in J} S_j, j \mapsto f_j (p)) \in \times_{j \in J} S_j\).
\(f_j (p) \in S_j\) for each \(j\), so, \(p \in {f_j}^{-1} (S_j)\) for each \(j\), so, \(p \in \cap_{j \in J} {f_j}^{-1} (S_j)\).
Step 2:
Let \(p \in \cap_{j \in J} {f_j}^{-1} (S_j)\) be any.
\(p \in {f_j}^{-1} (S_j)\) for each \(j\).
\(f_j (p) \in S_j\) for each \(j\). \(f (p) = (: J \to \cup_{j \in J} S_j, j \mapsto f_j (p)) \in \times_{j \in J} S_j\), so, \(p \in f^{-1} (\times_{j \in J} S_j)\).
3: Structured Description 2
Here is the rules of Structured Description.
Entities:
\(S\): \(\in \{\text{ the sets }\}\)
\(\{S'_1, ..., S'_n\}\): \(S'_j \in \{\text{ the sets }\}\)
\(\{f_1, ..., f_n\}\): \(f_j: S \to S'_j\)
\(S'\): \(= S'_1 \times ... \times S'_n\)
\(f\): \(: S \to S', s \mapsto (f_1 (s), ..., f_n (s))\)
\(\{S_1, ..., S_n\}\): \(S_j \subseteq S'_j\)
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Statements:
\(f^{-1} (S_1 \times ... \times S_n) = {f_1}^{-1} (S_1) \cap ... \cap {f_n}^{-1} (S_n)\)
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4: Proof 2
Whole Strategy: Step 1: see that \(f^{-1} (S_1 \times ... \times S_n) \subseteq {f_1}^{-1} (S_1) \cap ... \cap {f_n}^{-1} (S_n)\); Step 2: see that \({f_1}^{-1} (S_1) \cap ... \cap {f_n}^{-1} (S_n) \subseteq f^{-1} (S_1 \times ... \times S_n)\).
Step 1:
Let \(p \in f^{-1} (S_1 \times ... \times S_n)\) be any.
\(f (p) = (f_1 (p), ..., f_n (p)) \in S_1 \times ... \times S_n\).
\(f_j (p) \in S_j\) for each \(j\), so, \(p \in {f_j}^{-1} (S_j)\) for each \(j\), so, \(p \in {f_1}^{-1} (S_1) \cap ... \cap {f_n}^{-1} (S_n)\).
Step 2:
Let \(p \in {f_1}^{-1} (S_1) \cap ... \cap {f_n}^{-1} (S_n)\) be any.
\(p \in {f_j}^{-1} (S_j)\) for each \(j\). \(f_j (p) \in S_j\) for each \(j\). \(f (p) = (f_1 (p), f_2 (p), ..., f_n (p)) \in S_1 \times ... \times S_n\), so, \(p \in f^{-1} (S_1 \times ... \times S_n)\).
