A description/proof of that continuous map preimage of closed set is closed set
Topics
About: topological space
About: map
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of open set.
- The reader knows a definition of closed set.
- The reader knows a definition of continuous map.
- The reader admits the proposition that the preimage of the range minus any range subset of any map is the domain minus the preimage of the subset.
Target Context
- The reader will have a description and a proof of the proposition that the preimage of any closed set of any continuous map is a closed set.
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 topological spaces, \(M_1\) and \(M_2\), any continuous map, \(f: M_1 \rightarrow M_2\), and any closed subset of the range, \(C \subseteq M_2\), the preimage of the closed set, \(f^{-1} (C)\), is a closed set.
2: Proof
By the proposition that the preimage of the range minus any range subset of any map is the domain minus the preimage of the subset, \(f^{-1} (C) = f^{-1} (M_2 \setminus U) = M_1 \setminus f^{-1} (U)\) where \(U = M_2 \setminus C\), an open set. As f is continuous, \(f^{-1} (U)\) is open, so, \(M_1 \setminus f^{-1} (U)\) is closed, which means that \(f^{-1} (C)\) is closed.
