description/proof of that for map between sets, image of intersection of preimages of codomain subsets is intersection of images of preimages of codomain subsets
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 preimages of any codomain subsets is the intersection of the images of the preimages of the codomain 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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(: S_1 \to S_2\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_{2, j} \subseteq S_2 \vert j \in J\}\):
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Statements:
\(f (\cap_{j \in J} f^{-1} (S_{2, j})) = \cap_{j \in J} f (f^{-1} (S_{2, j}))\)
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2: Proof
Whole Strategy: Step 1: see that \(f (\cap_{j \in J} f^{-1} (S_{2, j})) \subseteq \cap_{j \in J} f (f^{-1} (S_{2, j}))\); Step 2: see that \(\cap_{j \in J} f (f^{-1} (S_{2, j})) \subseteq f (\cap_{j \in J} f^{-1} (S_{2, j}))\); Step 3: conclude the proposition.
Step 1:
Let \(s \in f (\cap_{j \in J} f^{-1} (S_{2, j}))\) be any.
There is an \(s' \in \cap_{j \in J} f^{-1} (S_{2, j})\) such that \(f (s') = s\).
\(s' \in f^{-1} (S_{2, j})\) for each \(j \in J\).
\(s = f (s') \in f (f^{-1} (S_{2, j}))\) for each \(j\).
So, \(s \in \cap_{j \in J} f (f^{-1} (S_{2, j}))\).
So, \(f (\cap_{j \in J} f^{-1} (S_{2, j})) \subseteq \cap_{j \in J} f (f^{-1} (S_{2, j}))\).
Step 2:
Let \(s \in \cap_{j \in J} f (f^{-1} (S_{2, j}))\) be any.
\(s \in f (f^{-1} (S_{2, j}))\) for each \(j \in J\).
Let \(l \in J\) be any fixed one.
As \(s \in f (f^{-1} (S_{2, l}))\), there is an \(s' \in f^{-1} (S_{2, l})\) such that \(f (s') = s\).
Let \(m \in J\) be any.
\(s = f (s') \in f (f^{-1} (S_{2, m}))\).
So, \(s' \in f^{-1} (f (f^{-1} (S_{2, m})))\).
But \(f (f^{-1} (S_{2, m})) \subseteq S_{2, m}\).
So, \(f^{-1} (f (f^{-1} (S_{2, m}))) \subseteq f^{-1} (S_{2, m})\).
\(s' \in f^{-1} (f (f^{-1} (S_{2, m}))) \subseteq f^{-1} (S_{2, m})\).
So, \(s' \in \cap_{j \in J} f^{-1} (S_{2, j})\).
So, \(s = f (s') \in f (\cap_{j \in J} f^{-1} (S_{2, j}))\).
So, \(\cap_{j \in J} f (f^{-1} (S_{2, j})) \subseteq f (\cap_{j \in J} f^{-1} (S_{2, j}))\).
Step 3:
So, \(f (\cap_{j \in J} f^{-1} (S_{2, j})) = \cap_{j \in J} f (f^{-1} (S_{2, j}))\).
