A description/proof of that map image of intersection of sets is not necessarily 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 a map, the map image of an intersection of sets is not necessarily 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 some sets, \(S_1\) and \(S_2\), a map, \(f: S_1 \rightarrow S_2\), and a possibly uncountable number of subsets of \(S_1\), \(S_{1_i} \subseteq S_1\), the map image of the intersection of the subsets, \(f (\cap_i S_{1_i})\), is not necessarily the intersection of the map images of the subsets, \(\cap_i f (S_{1_i})\), which is, not necessarily \(f (\cap_i S_{1_i}) = \cap_i f (S_{1_i})\).
2: Proof
It suffices to show a counter-example. If each element in \(S_1\) maps to the same element, \(p \in S_2\), \(S_{1_1}\) is a non-empty proper subset, \(S_{1_1} \subset S_1\), and \(S_{1_2}\) is the complement of \(S_{1_1}\) with respect to \(S_1\), \(S_{1_2} = S_1 \setminus S_{1_1}\), \(\cap_i f (S_{1_i}) = {p}\), but \(\cap_i S_{1_i} = \emptyset\), so \(f (\cap_i S_{1_i}) = \emptyset\).
3: Note
Although \(f (\cup_i S_{1_i}) = \cup_i f (S_{1_i})\) always holds, the intersection counterpart does not necessarily holds.
