A description/proof of that injective map image of intersection of sets is intersection 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 injective map, the map image of the intersection of any sets is the intersection 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 injective 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 the intersection of the map images of the subsets, \(\cap_\alpha f (S_{1_\alpha})\), which is, \(f (\cap_\alpha S_{1_\alpha}) = \cap_\alpha f (S_{1_\alpha})\).
2: Proof
For any element, \(p \in f (\cap_\alpha S_{1_\alpha})\), there is an element, \(p' \in \cap_\alpha S_{1_\alpha}\) such that \(p = f (p')\), which means that for each \(\alpha\), \(p' \in S_{1_\alpha}\). So, \(p \in f (S_{1_\alpha})\) for each \(\alpha\), so, \(p \in \cap_\alpha f (S_{1_\alpha})\).
For any element, \(p \in \cap_\alpha f (S_{1_\alpha})\), \(p \in f (S_{1_\alpha})\) for each \(\alpha\), so, there is a \(p'_\alpha \in S_{1_\alpha}\) for each \(\alpha\) such that \(p = f (p'_\alpha)\), but as \(f\) is injective, all the \(p'_\alpha\)s are the same \(p' \in S_1\), so, \(p' \in \cap_\alpha S_{1_\alpha}\), so, \(p \in f (\cap_\alpha S_{1_\alpha})\).
3: Note
The map has to be injective for this proposition. Otherwise, see another proposition.
