description/proof of that for map, image of subset minus subset contains image of 1st subset minus image 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 image of any subset minus any subset contains the image of the 1st subset minus the image 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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(S_1 \to S_2\)
\(S_{1, 1}\): \(\subseteq S_1\)
\(S_{1, 2}\): \(\subseteq S_1\)
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Statements:
\(f (S_{1, 1}) \setminus f (S_{1, 2}) \subseteq f (S_{1, 1} \setminus S_{1, 2})\)
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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, \(S_{1, 1}, S_{1, 2} \subseteq S_1\), \(f (S_{1, 1}) \setminus f (S_{1, 2}) \subseteq f (S_{1, 1} \setminus S_{1, 2})\).
3: Proof
Whole Strategy: Step 1: for each \(p \in f (S_{1, 1}) \setminus f (S_{1, 2})\), take a \(p' \in S_{1, 1} \setminus S_{1, 2}\) such that \(p = f (p')\), and conclude the proposition.
Step 1:
For each \(p \in f (S_{1, 1}) \setminus f (S_{1, 2})\), there is a \(p' \in S_{1, 1} \setminus S_{1, 2}\) such that \(p = f (p')\), because while \(p\) is realized as \(p = f (p')\) for a \(p' \in S_{1, 1}\), if \(p' \in S_{1, 2}\), \(f (p') \in f (S_{1, 2})\), which could not be \(p\) because \(p \notin f (S_{1, 2})\), so, \(p\) is realized by a \(p' \in S_{1, 1} \setminus S_{1, 2}\).
So, \(p = f (p') \in f (S_{1, 1} \setminus S_{1, 2})\).
4: Note
The equality holds when \(f\) is injective, by the proposition that for any injective map, the image of any subset minus any subset is the image of the 1st subset minus the image of the 2nd subset.
