285: Preimage of Non-Zero Determinants of Matrix of Continuous Functions Is Open

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A description/proof of that preimage of non-zero determinants of matrix of continuous functions is open

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Topics

About: topological space
About: map

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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1: Description

For any matrix of any continuous functions, $[M_{ij}]$, where $M_{ij}:U\to \mathbb{R}$ are continuous where U is an open set, $U\subseteq {\mathbb{R}}^d$, and the determinant map of the matrix, $f:=det[M_{ij}]:U\to \mathbb{R}$, the preimage of the non-zero values of the map, $f^{-1}(\{\mathrm{\forall }p\in \mathbb{R}|p\ne 0\})$, is open on U.

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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.

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References

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