description/proof of that for map, subset of domain, and subset of codomain, image of subset is contained in subset and image of complement of subset is contained in complement of subset, iff preimage of subset is subset and preimage of complement of subset is complement of 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
- 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, any subset of the domain, and any subset of the codomain, the image of the subset is contained in the subset and the image of the complement of the subset is contained in the complement of the subset, if and only if the preimage of the subset is the subset and the preimage of the complement of the subset is 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:
\(S'_1\): \(\in \{\text{ the sets }\}\)
\(S'_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(S'_1 \to S'_2\)
\(S_1\): \(\subseteq S'_1\)
\(S_2\): \(\subseteq S'_2\)
//
Statements:
(
\(f (S_1) \subseteq S_2\)
\(\land\)
\(f (S'_1 \setminus S_1) \subseteq S'_2 \setminus S_2\)
)
\(\iff\)
(
\(S_1 = f^{-1} (S_2)\)
\(\land\)
\(S'_1 \setminus S_1 = f^{-1} (S'_2 \setminus S_2)\)
).
//
2: Natural Language Description
For any sets, \(S'_1, S'_2\), any map, \(f: S'_1 \to S'_2\), and any subsets, \(S_1 \subseteq S'_1, S_2 \subseteq S'_2\), \(f (S_1) \subseteq S_2\) and \(f (S'_1 \setminus S_1) \subseteq S'_2 \setminus S_2\), if and only if \(S_1 = f^{-1} (S_2)\) and \(S'_1 \setminus S_1 = f^{-1} (S'_2 \setminus S_2)\).
3: Proof
Whole Strategy: Step 1: suppose that \(f (S_1) \subseteq S_2\) and \(f (S'_1 \setminus S_1) \subseteq S'_2 \setminus S_2\), and see that \(S_1 = f^{-1} (S_2)\) and \(S'_1 \setminus S_1 = f^{-1} (S'_2 \setminus S_2)\); Step 2: suppose that \(S_1 = f^{-1} (S_2)\) and \(S'_1 \setminus S_1 = f^{-1} (S'_2 \setminus S_2)\), and see that \(f (S_1) \subseteq S_2\) and \(f (S'_1 \setminus S_1) \subseteq S'_2 \setminus S_2\).
Step 1:
Let us suppose that \(f (S_1) \subseteq S_2\) and \(f (S'_1 \setminus S_1) \subseteq S'_2 \setminus S_2\).
\(S_1 \subseteq f^{-1} (S_2)\) is immediate.
Let us prove that \(f^{-1} (S_2) \subseteq S_1\). For any \(p \in f^{-1} (S_2)\), \(f (p) \in S_2\), \(f (p) \notin S'_2 \setminus S_2\), \(f (p) \notin f (S'_1 \setminus S_1)\), \(p \notin S'_1 \setminus S_1\), and so, \(p \in S_1\) as \(p \in S'_1\).
\(S'_1 \setminus S_1 \subseteq f^{-1} (S'_2 \setminus S_2)\) is immediate.
Let us prove that \(f^{-1} (S'_2 \setminus S_2) \subseteq S'_1 \setminus S_1\). For any \(p \in f^{-1} (S'_2 \setminus S_2)\), \(f (p) \in S'_2 \setminus S_2\), \(f (p) \notin S_2\) as \(f (p) \in S'_2\), \(p \notin S_1\), and so, \(p \in S'_1 \setminus S_1\) as \(p \in S'_1\).
Step 2:
Let us suppose that \(S_1 = f^{-1} (S_2)\) and \(S'_1 \setminus S_1 = f^{-1} (S'_2 \setminus S_2)\).
\(f (S_1) \subseteq S_2\) and \(f (S'_1 \setminus S_1) \subseteq S'_2 \setminus S_2\) are immediate.
