description/proof of that for continuous map from closed subset of topological space and closed subset and open subset containing closed subset of codomain, there is open subset of domain space that contains preimage of closed subset s.t. complement of open subset contains preimage of complement of open subset
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of closed subset of topological space.
- The reader knows a definition of topological subspace.
- The reader knows a definition of continuous, topological spaces map.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map from any closed subset of any topological space and any closed subset and any open subset containing the closed subset of the codomain, there is an open subset of the domain space that contains the preimage of the closed subset such that the complement of the open subset contains the preimage of the complement of the open subset.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(C_1\): \(\in \{\text{ the closed subsets of } T_1\}\), with the subspace topology
\(f\): \(: C_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(C_2\): \(\in \{\text{ the closed subsets of } T_2\}\)
\(U_2\): \(\in \{\text{ the open subsets of } T_2\}\), such that \(C_2 \subseteq U_2\)
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Statements:
\(\exists U_1 \in \{\text{ the open subsets of } T_1\} \text{ such that } f^{-1} (U_2) = U_1 \cap C_1 (f^{-1} (C_2) \subseteq U_1 \land f^{-1} (T_2 \setminus U_2) \subseteq T_1 \setminus U_1)\)
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2: Note
How does that matter? \(f^{-1} (C_2) \subseteq T_1\) is a closed subset, and \(f^{-1} (C_2) \subseteq U_1 \land f^{-1} (T_2 \setminus U_2) \subseteq T_1 \setminus U_1\) implies that when \(T_1\) is normal, the proposition that for any normal topological space, any closed subset, and any open subset that contains the closed subset, there is a continuous map into any closed Interval that maps the closed subset to any boundary of the closed interval and the complement of the open subset to the other boundary (Urysohn's lemma) applies with a continuous \(g: T_1 \to [r_1, r_2]\) such that \(g (f^{-1} (C_2)) = \{r_1\} \land g (f^{-1} (T_2 \setminus U_2)) = \{r_2\}\) or \(g (f^{-1} (C_2)) = \{r_2\} \land g (f^{-1} (T_2 \setminus U_2)) = \{r_1\}\).
3: Proof
Whole Strategy: Step 1: see that there is an open \(U_1 \subseteq T_1\) such that \(f^{-1} (U_2) = U_1 \cap C_1\); Step 2: see that \(f^{-1} (C_2) \subseteq U_1\); Step 3: see that \(f^{-1} (T_2 \setminus U_2) \subseteq T_1 \setminus U_1\).
Step 1:
\(f^{-1} (U_2) \subseteq C_1\) is open, because \(f\) is continuous.
\(f^{-1} (U_2) = U_1 \cap C_1\), where \(U_1 \subseteq T_1\) is an open subset, by the definition of subspace topology.
Step 2:
As \(C_2 \subseteq U_2\), \(f^{-1} (C_2) \subseteq f^{-1} (U_2) = U_1 \cap C_1 \subseteq U_1\).
Step 3:
Let us see that \(f^{-1} (T_2 \setminus U_2) \subseteq T_1 \setminus U_1\).
Let \(c_1 \in f^{-1} (T_2 \setminus U_2)\) be any.
\(f (c_1) \in T_2 \setminus U_2\).
\(c_1 \notin U_1\), because if \(c_1 \in U_1\), \(c_1 \in U_1 \cap C_1 = f^{-1} (U_2)\), so, \(f (c_1) \in U_2\), a contradiction against \(f (c_1) \in T_2 \setminus U_2\).
So, \(c_1 \in T_1 \setminus U_1\).
So, \(f^{-1} (T_2 \setminus U_2) \subseteq T_1 \setminus U_1\).
