A description/proof of that preimage of non-zero determinants of matrix of continuous functions is open
Topics
About: topological space
About: map
The table of contents of this article
Starting Context
- 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.
- The reader admits the proposition that that the preimage of any closed set of any continuous map is a closed set.
Target Context
- The reader will have a description and a proof of the proposition that the preimage of the non-zero determinants of any matrix of any continuous functions is open.
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 matrix of any continuous functions, \([M_{ij}]\), where \(M_{ij}: U \rightarrow \mathbb{R}\) are continuous where U is an open set, \(U \subseteq \mathbb{R}^d\), and the determinant map of the matrix, \(f := det [M_{ij}]: U \rightarrow \mathbb{R}\), the preimage of the non-zero values of the map, \(f^{-1} (\{\forall p \in \mathbb{R}| p \neq 0\})\), is open on U.
2: Proof
As taking the determinant of the matrix is continuous with respect to the matrix components, f is continuous as a compound of continuous maps. 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, \(f^{-1} ({\mathbb{R} \setminus {0}}) = U \setminus f^{-1} ({0})\), and by the the proposition that that the preimage of any closed set of any continuous map is a closed set, \(f^{-1} ({0})\) is closed, so, \(U \setminus f^{-1} ({0})\) is open, so, \(f^{-1} ({\mathbb{R} \setminus {0}})\) is open.
