description/proof of that for map into Euclidean set and characteristic function, preimage of subset under map multiplied by characteristic function is this
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of map.
- The reader knows a definition of characteristic function over subset of set.
- The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
Target Context
- The reader will have a description and a proof of the proposition that for any map into any Euclidean set and any characteristic function, the preimage of any subset under the map multiplied by the characteristic function is this.
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:
\(S'\): \(\in \{\text{ the sets }\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean set }\)
\(f\): \(: S' \to \mathbb{R}^d\)
\(S\): \(\subseteq S'\)
\(\chi_S\): \(= \text{ the characteristic function over } S\)
\(S''\): \(\subseteq \mathbb{R}^d\)
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Statements:
when \(0 \notin S''\), \((\chi_S f)^{-1} (S'') = S \cap f^{-1} (S'')\)
\(\land\)
when \(0 \in S''\), \((\chi_S f)^{-1} (S'') = (S \cap f^{-1} (S'' \setminus \{0\})) \cup ((S' \setminus S) \cup f^{-1} (0))\)
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2: Proof
Whole Strategy: Step 1: suppose \(0 \notin S''\), and see that \((\chi_S f)^{-1} (S'') = S \cap f^{-1} (S'')\); Step 2: suppose \(0 \in S''\), and see that \((\chi_S f)^{-1} (S'') = (S \cap f^{-1} (S'' \setminus \{0\})) \cup ((S' \setminus S) \cup f^{-1} (0))\).
Step 1:
Let us suppose that \(0 \notin S''\).
Let us see that \((\chi_S f)^{-1} (S'') = S \cap f^{-1} (S'')\).
Let \(s \in (\chi_S f)^{-1} (S'')\) be any.
\((\chi_S f) (s) = \chi_S (s) f (s) \in S''\).
As \((\chi_S f) (s) \neq 0\), \(\chi_S (s) \neq 0\), so, \(s \in S\).
\(\chi_S (s) f (s) = f (s) \in S''\), so, \(s \in f^{-1} (S'')\).
So, \(s \in S \cap f^{-1} (S'')\).
Let \(s \in S \cap f^{-1} (S'')\) be any.
\((\chi_S f) (s) = f (s) \in S''\), so, \(s \in (\chi_S f)^{-1} (S'')\).
So, \((\chi_S f)^{-1} (S'') = S \cap f^{-1} (S'')\).
Step 2:
Let us suppose that \(0 \in S''\).
\(S'' = (S'' \setminus \{0\}) \cup \{0\}\).
\((\chi_S f)^{-1} (S'') = (\chi_S f)^{-1} ((S'' \setminus \{0\}) \cup \{0\}) = (\chi_S f)^{-1} (S'' \setminus \{0\}) \cup (\chi_S f)^{-1} (\{0\})\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
\((\chi_S f)^{-1} (S'' \setminus \{0\}) = S \cap f^{-1} (S'' \setminus \{0\})\), by Step 1.
Let us see that \((\chi_S f)^{-1} (\{0\}) = (S' \setminus S) \cup f^{-1} (0)\).
Let \(s \in (\chi_S f)^{-1} (\{0\})\) be any.
\(\chi_S f (s) = \chi_S (s) f (s) = 0\), so, \(\chi_S (s) = 0\) or \(f (s) = 0\), so, \(s \in S' \setminus S\) or \(s \in f^{-1} (0)\), so, \(s \in (S' \setminus S) \cup f^{-1} (0)\).
Let \(s \in (S' \setminus S) \cup f^{-1} (0)\) be any.
\((\chi_S f) (s) = \chi_S (s) f (s)\), but while \(s \in S' \setminus S\) or \(s \in f^{-1} (0)\), \(s \in S' \setminus S\) implies that \(\chi_S (s) = 0\) and \(s \in f^{-1} (0)\) implies that \(f (s) = 0\), so, \(\chi_S (s) f (s) = 0\) anyway, so, \(s \in (\chi_S f)^{-1} (\{0\})\).
So, \((\chi_S f)^{-1} (\{0\}) = (S' \setminus S) \cup f^{-1} (0)\).
So, \((\chi_S f)^{-1} (S'') = (S \cap f^{-1} (S'' \setminus \{0\})) \cup ((S' \setminus S) \cup f^{-1} (0))\).
