description/proof of that for map between sets, image of intersection of intersection of preimages of codomain subsets and domain subset is intersection of intersection of codomain subsets and image of domain subset
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of map.
- The reader admits the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.
Target Context
- The reader will have a description and a proof of the proposition that for any map between any sets, the image of the intersection of the intersection of the preimages of any codomain subsets and any domain subset is the intersection of the intersection of the codomain subsets and the image of the domain subset.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(: S_1 \to S_2\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_{2, j} \subseteq S_2 \vert j \in J\}\):
\(S^`_1\): \(\subseteq S_1\)
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Statements:
\(f (\cap_{j \in J} f^{-1} (S_{2, j}) \cap S^`_1) = \cap_{j \in J} S_{2, j} \cap f (S^`_1)\)
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2: Proof
Whole Strategy: Step 1: see that \(f (\cap_{j \in J} f^{-1} (S_{2, j}) \cap S^`_1) = f (f^{-1} (S^`_2) \cap S^`_1)\) where \(S^`_2 := \cap_{j \in J} S_{2, j}\); Step 2: see that \(f (f^{-1} (S^`_2) \cap S^`_1) = S^`_2 \cap f (S^`_1)\).
Step 1:
\(\cap_{j \in J} f^{-1} (S_{2, j}) = f^{-1} (\cap_{j \in J} S_{2, j})\), by the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.
Let \(S^`_2 := \cap_{j \in J} S_{2, j}\).
\(f (\cap_{j \in J} f^{-1} (S_{2, j}) \cap S^`_1) = f (f^{-1} (\cap_{j \in J} S_{2, j}) \cap S^`_1) = f (f^{-1} (S^`_2) \cap S^`_1)\).
Step 2:
Let us see that \(f (f^{-1} (S^`_2) \cap S^`_1) = S^`_2 \cap f (S^`_1)\).
Let \(s \in f (f^{-1} (S^`_2) \cap S^`_1)\) be any.
There is an \(s' \in f^{-1} (S^`_2) \cap S^`_1\) such that \(f (s') = s\).
As \(s' \in f^{-1} (S^`_2)\), \(s = f (s') \in S^`_2\).
As \(s' \in S^`_1\), \(s = f (s') \in f (S^`_1)\).
So, \(s \in S^`_2 \cap f (S^`_1)\).
Let \(s \in S^`_2 \cap f (S^`_1)\) be any.
There is an \(s' \in S^`_1\) such that \(f (s') = s\).
As \(s = f (s') \in S^`_2\), \(s' \in f^{-1} (S^`_2)\).
So, \(s' \in f^{-1} (S^`_2) \cap S^`_1\), and \(s = f (s') \in f (f^{-1} (S^`_2) \cap S^`_1)\).
So, \(f (f^{-1} (S^`_2) \cap S^`_1) = S^`_2 \cap f (S^`_1)\).
