A description/proof of that map image of subset is contained in subset iff subset is contained in preimage of subset
Topics
About: set
About: map
The table of contents of this article
Starting Context
- The reader knows a definition of set.
- 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 sets, the image of any subset is contained in any subset if and only if the former subset is contained in the preimage of the latter 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: Description
For any sets, \(S_1, S_2\), any map, \(f: S_1 \rightarrow S_2\), and any subsets, \(S_3 \in S_1, S_4 \in S_2\), \(f (S_3) \subseteq S_4\) if and only if \(S_3 \subseteq f^{-1} (S_4)\).
2: Proof
Suppose that \(S_3 \subseteq f^{-1} (S_4)\). \(f (S_3) \subseteq f (f^{-1} (S_4)) \subseteq S_4\), by the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.
Suppose that \(f (S_3) \subseteq S_4\). For any \(p \in S_3\), \(f (p) \in f (S_3) \subseteq S_4\), so, \(p \in f^{-1} (S_4)\) by the proposition that for any map between sets, the image of any point is on any subset if and only if the point is on the preimage of the subset, so, \(S_3 \subseteq f^{-1} (S_4)\).
