description/proof of that map preimage of subset minus subset is preimage of 1st subset minus preimage of 2nd subset
Topics
About: set
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
- 4: Note
Starting Context
- The reader knows a definition of map.
Target Context
- The reader will have a description and a proof of the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd 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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(S_1 \to S_2\)
\(S_{2, 1}\): \(\subseteq S_2\)
\(S_{2, 2}\): \(\subseteq S_2\)
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Statements:
\(f^{-1} (S_{2, 1} \setminus S_{2, 2}) = f^{-1} (S_{2, 1}) \setminus f^{-1} (S_{2, 2})\)
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2: Natural Language Description
For any sets, \(S_1, S_2\), any map, \(f: S_1 \to S_2\), and any subsets, \(S_{2, 1}, S_{2, 2} \subseteq S_2\), \(f^{-1} (S_{2, 1} \setminus S_{2, 2}) = f^{-1} (S_{2, 1}) \setminus f^{-1} (S_{2, 2})\).
3: Proof
Whole Strategy: Step 1: see that \(f^{-1} (S_{2, 1} \setminus S_{2, 2}) \subseteq f^{-1} (S_{2, 1}) \setminus f^{-1} (S_{2, 2})\); Step 2: see that \(f^{-1} (S_{2, 1}) \setminus f^{-1} (S_{2, 2}) \subseteq f^{-1} (S_{2, 1} \setminus S_{2, 2})\).
Step 1:
For each point, \(p \in f^{-1} (S_{2, 1} \setminus S_{2, 2})\), \(f (p) \in S_{2, 1} \setminus S_{2, 2}\), \(f (p) \in S_{2, 1}\) and \(f (p) \notin S_{2, 2}\), \(p \in f^{-1} (S_{2, 1})\) and \(p \notin f^{-1} (S_{2, 2})\), and so, \(p \in f^{-1} (S_{2, 1}) \setminus f^{-1} (S_{2, 2})\).
Step 2:
For each point, \(p \in f^{-1} (S_{2, 1}) \setminus f^{-1} (S_{2, 2})\), \(p \in f^{-1} (S_{2, 1})\) and \(p \notin f^{-1} (S_{2, 2})\), \(f (p) \in S_{2, 1}\) and \(f (p) \notin S_{2, 2}\), \(f (p) \in S_{2, 1} \setminus S_{2, 2}\), and so, \(p \in f^{-1} (S_{2, 1} \setminus S_{2, 2})\).
4: Note
Compare this with the proposition on images.
