284: Continuous Map Preimage of Closed Set Is Closed Set

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A description/proof of that continuous map preimage of closed set is closed set

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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 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.

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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.

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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 topological spaces, $M_1$ and $M_2$, any continuous map, $f:M_1\to M_2$, and any closed subset of the range, $C\subseteq M_2$, the preimage of the closed set, $f^{-1}(C)$, is a closed set.

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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.

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References

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