description/proof of that map preimage of empty subset is empty set
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of map preimage of subset of codomain.
Target Context
- The reader will have a description and a proof of the proposition that for any map, the map preimage of the empty subset is the empty set.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(: S_1 \to S_2\)
//
Statements:
\(f^{-1} (\emptyset) = \emptyset\)
//
2: Proof
Whole Strategy: Step 1: suppose that there was a \(p \in f^{-1} (\emptyset)\), and find a contradiction.
Step 1:
Let us suppose there was a \(p \in f^{-1} (\emptyset)\).
\(f (p) \in \emptyset\), by the definition of map preimage of subset of codomain.
That is a contradiction against that \(f (p)\) was a point on \(S_2\).
So, \(f^{-1} (\emptyset)\) has no element, which means that \(f^{-1} (\emptyset) = \emptyset\).
