description/proof of that proper continuous map between locally compact Hausdorff topological spaces is closed
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of locally compact topological space.
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of proper map.
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of closed map.
- The reader knows a definition of \(1\)-point compactification of locally compact Hausdorff topological space.
- The reader admits the proposition that for any locally compact Hausdorff topological space, the topology of the \(1\)-point compactification is the only topology that makes the \(1\)-point-augmented set compact Hausdorff with the original space as the subspace.
- The reader admits the proposition that for any map and its any extension that maps the extended area outside the original codomain, the extension preimage of any subset of the extension codomain is the union of the original map preimage of the intersection of the subset and the original codomain and the extension preimage of the subset minus the original codomain.
- The reader admits the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd subset.
- The reader admits the proposition that the preimage of the whole codomain of any map is the whole domain.
- The reader admits the proposition that any compact subset of any Hausdorff topological space is closed.
- The reader admits the proposition that for any set, the union of any subsets minus any subset is the union of (each of the former subsets minus the latter subset) s.
- The reader admits the proposition that for any map and its any extension that maps the extended area outside the original codomain, the original map image of any subset of the original domain is the intersection of the extension image of the union of the subset and the extended area and the original codomain.
- The reader admits the proposition that for any set, any subset minus the union of any subsets is the intersection of the 1st subset minus the 2nd chunk of subsets.
- The reader admits the proposition that the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd set.
- The reader admits the proposition that any closed subset of any compact topological space is compact.
- The reader admits the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain.
- The reader admits the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
Target Context
- The reader will have a description and a proof of the proposition that any proper continuous map between any locally compact Hausdorff topological spaces is closed.
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:
\(T_1\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\)
\(T_2\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the proper maps }\} \cap \{\text{ the continuous maps }\}\)
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Statements:
\(f \in \{\text{ the closed maps }\}\)
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2: Proof
Whole Strategy: Step 1: take the \(1\)-point compactifications, \({T_1}^+\) and \({T_2}^+\), and the continuous extension of \(f\), \(f': {T_1}^+ \to {T_2}^+\); Step 2: see that for each closed \(C_1 \subseteq T_1\), \(f (C_1) = f' (C_1 \cup \{\infty\}) \cap T_2\) is closed on \(T_2\).
Step 1:
Let us take the \(1\)-point compactification of \(T_1\), \({T_1}^+\), and the \(1\)-point compactification of \(T_2\), \({T_2}^+\).
\({T_1}^+\) and \({T_2}^+\) are some compact Hausdorff topological spaces such that \(T_1\) and \(T_2\) are their topological subspaces, by the proposition that for any locally compact Hausdorff topological space, the topology of the \(1\)-point compactification is the only topology that makes the \(1\)-point-augmented set compact Hausdorff with the original space as the subspace.
Let us define \(f': {T_1}^+ \to {T_2}^+\) as the extension of \(f\) such that \(f' (\infty) = \infty\).
Let us see that \(f'\) is continuous.
Let \(U_2 \subseteq {T_2}^+\) be any open subset.
\(U_2\) is an open subset of \(T_2\) or \(U_2 = {T_2}^+ \setminus K\) where \(K \subseteq T_2\) is a compact subset.
When \(U_2\) is an open subset of \(T_2\), \(f'^{-1} (U_2) = f^{-1} (U_2)\), by the proposition that for any map and its any extension that maps the extended area outside the original codomain, the extension preimage of any subset of the extension codomain is the union of the original map preimage of the intersection of the subset and the original codomain and the extension preimage of the subset minus the original codomain, which is open on \(T_1\), because \(f\) is continuous, so, is open on \({T_1}^+\).
When \(U_2 = {T_2}^+ \setminus K\), \(f'^{-1} (U_2) = f'^{-1} ({T_2}^+ \setminus K) = f'^{-1} ({T_2}^+) \setminus f'^{-1} (K)\), by the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd subset, \(= {T_1}^+ \setminus f^{-1} (K)\), by the proposition that the preimage of the whole codomain of any map is the whole domain and the proposition that for any map and its any extension that maps the extended area outside the original codomain, the extension preimage of any subset of the extension codomain is the union of the original map preimage of the intersection of the subset and the original codomain and the extension preimage of the subset minus the original codomain, but \(f^{-1} (K) \subseteq T_1\) is compact, because \(f\) is proper, so, \({T_1}^+ \setminus f^{-1} (K)\) is open on \({T_1}^+\).
So, \(f'\) is continuous.
Step 2:
Let \(C_1 \subseteq T_1\) be any closed subset.
\(f (C_1) = f' (C_1 \cup \{\infty\}) \cap T_2\), by the proposition that for any map and its any extension that maps the extended area outside the original codomain, the original map image of any subset of the original domain is the intersection of the extension image of the union of the subset and the extended area and the original codomain.
But \(C_1 \cup \{\infty\} \subseteq {T_1}^+\) is closed, because \({T_1}^+ \setminus (C_1 \cup \{\infty\}) = ({T_1}^+ \setminus C_1) \cap ({T_1}^+ \setminus \{\infty\})\), by the proposition that for any set, any subset minus the union of any subsets is the intersection of the 1st subset minus the 2nd chunk of subsets, \(= ({T_1}^+ \setminus C_1) \cap T_1 = ({T_1}^+ \cap T_1) \setminus (C_1 \cap T_1)\), by the proposition that the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd set, \(= T_1 \setminus C_1\), which is open on \(T_1\), so, is open on \({T_1}^+\).
So, \(C_1 \cup \{\infty\}\) is compact on \({T_1}^+\), by the proposition that any closed subset of any compact topological space is compact.
\(f' (C_1 \cup \{\infty\}) \subseteq {T_2}^+\) is compact, by the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain, so, is closed on \({T_2}^+\), by the proposition that any compact subset of any Hausdorff topological space is closed.
So, \(f (C_1) = f' (C_1 \cup \{\infty\}) \cap T_2\) is closed on \(T_2\), by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
So, \(f\) is closed.
