A description/proof of that map image of union of sets is union of map images 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 image of any union of sets is the union of the map images 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 \(S_1\), \(S_{1_i} \subseteq S_1\), the map image of the union of the subsets, \(f (\cup_i S_{1_i})\), is the union of the map images of the subsets, \(\cup_i f (S_{1_i})\), which is \(f (\cup_i S_{1_i}) = \cup_i f (S_{1_i})\).
2: Proof
For any element, \(p \in f (\cup_i S_{1_i})\), \(p \in f (S_{1_i})\) for an i, because as there is an element in \(\cup_i S_{1_i}\) that maps to p, the element has to be in an \(S_{1_i}\), which means that \(p \in \cup_i f (S_{1_i})\). For any element, \(p \in \cup_i f (S_{1_i})\), \(p \in f (S_{1_i})\) for an i, which means that \(p \in f (\cup_i S_{1_i})\), because as there is an element in \(S_{1_i}\) that maps to p, the element is also in \(\cup_i S_{1_i}\).
3: Note
It is important to be aware of that there is no such thing as a "limit element" in \(\cup_i S_{1_i}\), which does not belong to any \(S_{1_i}\) but to which a sequence of elements infinitely nears.
