description/proof of that for map and subset of codomain, complement of preimage of subset is preimage of complement of subset
Topics
About: set
The table of contents of this article
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 between any sets and any subset of the codomain, the complement of the preimage of the subset is the preimage of the complement of the 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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(: S_1 \to S_2\)
\(S\): \(\subseteq S_2\)
//
Statements:
\(S_1 \setminus f^{-1} (S) = f^{-1} (S_2 \setminus S)\)
//
2: Proof
Whole Strategy: Step 1: see that for each \(s \in S_1 \setminus f^{-1} (S)\), \(s \in f^{-1} (S_2 \setminus S)\), and for each \(s \in f^{-1} (S_2 \setminus S)\), \(s \in S_1 \setminus f^{-1} (S)\).
Step 1:
Let \(s \in S_1 \setminus f^{-1} (S)\) be any.
\(s \notin f^{-1} (S)\), \(f (s) \notin S\), \(f (s) \in S_2 \setminus S\), so, \(s \in f^{-1} (S_2 \setminus S)\).
So, \(S_1 \setminus f^{-1} (S) \subseteq f^{-1} (S_2 \setminus S)\).
Let \(s \in f^{-1} (S_2 \setminus S)\) be any.
\(f (s) \in S_2 \setminus S\), \(f (s) \notin S\), \(s \notin f^{-1} (S)\), so, \(s \in S_1 \setminus f^{-1} (S)\).
So, \(f^{-1} (S_2 \setminus S) \subseteq S_1 \setminus f^{-1} (S)\).
So, \(S_1 \setminus f^{-1} (S) = f^{-1} (S_2 \setminus S)\).
