description/proof of that closure of continuous map preimage of subset equals preimage of closure of subset if map is open especially if map is surjective and open subset of domain is preimage of open subset of codomain
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of closure of subset of topological space.
- The reader knows a definition of continuous map.
- The reader admits the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of the closure of the subset.
- The reader admits the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.
- The reader admits the proposition that any map between any topological spaces is open if but not only if the map is surjective and any open subset of the domain is the preimage of an open subset of the codomain.
- The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain equals the preimage of the closure of the subset if the map is open especially if the map is surjective and any open subset of the domain is the preimage of an open subset of the codomain.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(S\): \(\subseteq T_2\)
//
Statements:
(
\(f \in \{\text{ the open maps }\}\)
\(\implies\)
\(\overline{f^{-1} (S)} = f^{-1} (\overline{S})\)
)
\(\land\)
(
(
\(f \in \{\text{ the surjections }\}\)
\(\land\)
\(\forall U_1 \in \{\text{ the open subsets of } T_1\} (\exists U_2 \in \{\text{ the open subsets of } T_2\} (U_1 = f^{-1} (U_2)))\)
)
\(\implies\)
\(\overline{f^{-1} (S)} = f^{-1} (\overline{S})\)
)
//
2: Proof
Whole Strategy: Step 1: see that \(\overline{f^{-1} (S)} \subseteq f^{-1} (\overline{S})\); Step 2: suppose that \(f\) is open, and see that \(f^{-1} (\overline{S}) \subseteq \overline{f^{-1} (S)}\); Step 3: suppose that \(f\) is surjective and any open subset of the domain is the preimage of an open subset of the codomain, and see that \(f\) is open.
Step 1:
\(\overline{f^{-1} (S)} \subseteq f^{-1} (\overline{S})\), by the proposition that for any continuous map between topological spaces, the closure of the map preimage of any subset is contained in but not necessarily equal to the preimage of the closure of the subset.
So, it is about \(f^{-1} (\overline{S}) \subseteq \overline{f^{-1} (S)}\).
Step 2:
Let us suppose that \(f\) is open.
For any \(t \in f^{-1} (\overline{S})\), \(t \in \overline{f^{-1} (S)}\)?
For each open neighborhood of \(t\), \(U_t \subseteq T_1\), \(U_t \cap f^{-1} (S) \neq \emptyset\)?
As \(f (t) \in f (U_t)\) and \(f (U_t)\) is open on \(T_2\), \(f (U_t) \subseteq T_2\) is an open neighborhood of \(f (t)\).
As \(f (t) \in \overline{S}\), \(f (U_t) \cap S \neq \emptyset\). There is a point, \(t' \in f (U_t) \cap S\). There is a point, \(t'' \in U_t\), such that \(t' = f (t'')\). \(f (t'') \in S\), so, \(t'' \in f^{-1} (S)\). So, \(t'' \in U_t \cap f^{-1} (S)\), so, \(U_t \cap f^{-1} (S) \neq \emptyset\).
So, \(t \in \overline{f^{-1} (S)}\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Step 3:
Let us suppose that \(f\) is surjective and any open subset of the domain is the preimage of an open subset of the codomain.
\(f\) is open, by the proposition that any map between any topological spaces is open if but not only if the map is surjective and any open subset of the domain is the preimage of an open subset of the codomain.
So, the conclusion follows by Step 2.
