A description/proof of that map preimage of union of sets is union of map preimages of sets
Topics
About: set
About: map
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, the map preimage of any union of sets is the union of the map preimages of the sets.
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, \(S_1\) and \(S_2\), any map, \(f: S_1 \rightarrow S_2\), and any possibly uncountable number of subsets of \(S_2\), \(S_{2_i} \subseteq S_2\), the map preimage of the union of the subsets, \(f^{-1} (\cup_i S_{2_i})\), is the union of the map preimages of the subsets, \(\cup_i f^{-1} (S_{2_i})\), which is \(f^{-1} (\cup_i S_{2_i}) = \cup_i f^{-1} (S_{2_i})\).
2: Proof
For any element, \(p \in f^{-1} (\cup_i S_{2_i})\), \(f (p) \in \cup_{i} S_{2_i})\), so, \(f (p) \in S_{2_i}\) for an i, so, \(p \in f^{-1} (S_{2_i})\), so, \(p \in \cup_i f^{-1} (S_{2_i})\). For any element, \(p \in \cup_i f^{-1} (S_{2_i})\), \(p \in f^{-1} (S_{2_i})\) for an i, so, \(f (p) \in S_{2_i}\), so, \(f (p) \in \cup_i S_{2_i}\), so, \(p \in f^{-1} (\cup_i S_{2_i})\).
3: Note
It is important to be aware of that there is no such thing as a "limit element" in \(\cup_i S_{2_i}\), which does not belong to any \(S_{2_i}\) but to which a sequence of elements infinitely nears.
