description/proof of that for injective map, image of subset minus subset is 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
Starting Context
Target Context
- The reader will have a description and a proof of 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.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(S_1 \to S_2\), \(\in \{\text{ the injections }\}\)
\(S_{1, 1}\): \(\subseteq S_1\)
\(S_{1, 2}\): \(\subseteq S_1\)
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Statements:
\(f (S_{1, 1} \setminus S_{1, 2}) = f (S_{1, 1}) \setminus f (S_{1, 2})\)
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2: Natural Language Description
For any sets, \(S_1, S_2\), any injection, \(f: S_1 \to S_2\), and any subsets, \(S_{1, 1}, S_{1, 2} \subseteq S_1\), \(f (S_{1, 1} \setminus S_{1, 2}) = f (S_{1, 1}) \setminus f (S_{1, 2})\).
3: Proof
Whole Strategy: Step 1: see that \(f (S_{1, 1}) \setminus f (S_{1, 2}) \subseteq f (S_{1, 1} \setminus S_{1, 2})\); Step 2: see that \(f (S_{1, 1} \setminus S_{1, 2}) \subseteq f (S_{1, 1}) \setminus f (S_{1, 2})\).
Step 1:
\(f (S_{1, 1}) \setminus f (S_{1, 2}) \subseteq f (S_{1, 1} \setminus S_{1, 2})\), by 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.
Step 2:
Let us see that \(f (S_{1, 1} \setminus S_{1, 2}) \subseteq f (S_{1, 1}) \setminus f (S_{1, 2})\).
For each \(p \in f (S_{1, 1} \setminus S_{1, 2})\), there is a \(p' \in S_{1, 1} \setminus S_{1, 2}\) such that \(p = f (p')\). \(f (p') \in f (S_{1, 1})\). \(p' \notin S_{1, 2}\). \(f (p') \notin f (S_{1, 2})\), because if \(f (p') \in f (S_{1, 2})\), \(f (p') = f (p'')\) for a \(p'' \in S_{1, 2}\), but \(p' \neq p''\) because \(p' \notin S_{1, 2}\) and \(p'' \in S_{1, 2}\), a contradiction against \(f\)'s being injective. So, \(p = f (p') \in f (S_{1, 1}) \setminus f (S_{1, 2})\).
