A description/proof of that map image of intersection of sets is contained in intersection of map images of sets
Topics
About: set
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 between sets, the map image of any intersection of subsets is contained in the intersection of the map images of the subsets.
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 \(S_1\), \(S_{1_\alpha} \subseteq S_1\), the map image of the intersection of the subsets, \(f (\cap_\alpha S_{1_\alpha})\), is contained in the intersection of the map images of the subsets, \(\cap_\alpha f (S_{1_\alpha})\), which is, \(f (\cap_\alpha S_{1_\alpha}) \subseteq \cap_\alpha f (S_{1_\alpha})\).
2: Proof
For any \(p \in f (\cap_\alpha S_{1_\alpha})\), there is a \(p' \in \cap_\alpha S_{1_\alpha}\) such that \(p = f (p')\), which means that \(p' \in S_{1_\alpha}\) for every \(\alpha\). So, \(p \in f (S_{1_\alpha})\) for every \(\alpha\). So, \(p \in \cap_\alpha f (S_{1_\alpha})\).
3: Note
\(f (\cap_\alpha S_{1_\alpha}) = \cap_\alpha f (S_{1_\alpha})\) does not necessarily hold as is proved in another proposition.
\(f (\cup_i S_{1_i}) = \cup_i f (S_{1_i})\) always holds as is proved in another proposition.
