description/proof of \(1\)-point compactification of 2nd-countable locally compact Hausdorff topological space is 2nd-countable and metrizable
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of 2nd-countable topological space.
- 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 metrizable topological space.
- The reader admits the proposition that for any Hausdorff topological space, if and only if each point has a compact neighborhood, the space is locally compact.
- The reader admits the proposition that for any topological space, the union of any finite compact subsets is compact.
- The reader admits the proposition that any compact subset of any Hausdorff topological space is closed.
- The reader admits the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
- 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.
- The reader admits the proposition that any closed subset of any compact topological space is compact.
- The reader admits the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
- The reader admits the proposition that for any countable set, the set of the finite subsets is countable.
- The reader admits the proposition that any subset of any countable set is countable.
- The reader admits the proposition that any finite composition of surjections is a surjection, if the codomains of the constituent surjections equal the domains of the succeeding surjections.
- The reader admits the proposition that for any infinite set, if there is a surjection from the natural numbers set onto the set, there is a bijection from the natural numbers set onto the set.
- The reader admits the proposition that any countable union of countable sets is countable.
- The reader admits the proposition that for any topological space and any point on any subspace, the intersection of any neighborhood of the point on the base space and the subspace is a neighborhood on the subspace.
- 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 any compact Hausdorff topological space is normal.
- The reader admits the proposition that any normal topological space is completely regular.
- The reader admits the proposition that any 2nd-countable completely regular topological space is metrizable (Urysohn metrization theorem).
Target Context
- The reader will have a description and a proof of the proposition that the \(1\)-point compactification of any 2nd-countable locally compact Hausdorff topological space is 2nd-countable and metrizable.
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:
\(T\): \(\in \{\text{ the 2nd-countable locally compact Hausdorff topological spaces }\}\)
\(T^+\): \(= \text{ the 1-point compactification of } T\)
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Statements:
\(T^+\): \(\in \{\text{ the 2nd-countable topological spaces }\} \cap \{\text{ the metrizable topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: take any countable basis for \(T\), \(B\), and for each compact \(K \subseteq T\), take \(U_K := T \setminus \cup_{j \in J_K} \overline{U_{k_j}}\) where \(J_K\) is a finite index set and \(U_{k_j} \in B\) such that \(\overline{U_{k_j}}\) is compact and \(K \subseteq \cup_{j \in J_K} U_{k_j}\); Step 2: see that \(B^+ := B \cup \{U_K \cup \{\infty\} \vert K \subseteq T\}\) is a countable basis for \(T^+\); Step 3: see that \(T^+\) is metrizable.
Step 1:
Let \(B\) be any countable basis for \(T\).
Let \(K \subseteq T\) be any compact subset.
Let \(k \in K\) be any.
There is a compact neighborhood of \(k\), \(K_k \subseteq T\), by the proposition that for any Hausdorff topological space, if and only if each point has a compact neighborhood, the space is locally compact.
There is an open neighborhood of \(k\), \(U_k \in B\), such that \(U_k \subseteq K_k\), by the definition of basis for topological space.
\(K_k\) is closed on \(T\), by the proposition that any compact subset of any Hausdorff topological space is closed, and \(\overline{U_k} \subseteq K_k\), where \(\overline{U_k}\) is the closure with respect to \(T\), because the closure is the intersection of all the closed subsets that contain \(U_k\), of which, \(K_k\) is one.
\(K_k\) is a compact subspace of \(T\), by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace, and \(\overline{U_k}\) is a closed subset of \(K_k\), 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.
\(\overline{U_k}\) is a compact subset of \(K_k\), by the proposition that any closed subset of any compact topological space is compact, and is a compact subset of \(T\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
\(K \subseteq \cup_{k \in K} U_k\).
As \(K\) is compact, there is a finite subcover, \(\{U_{k_j} \vert j \in J_K\}\).
Let us define \(U_K := T \setminus \cup_{j \in J_K} \overline{U_{k_j}}\).
\(U_K \cup \{\infty\} \subseteq T^+\) is open, because \(U_K \cup \{\infty\} = T^+ \setminus \cup_{j \in J_K} \overline{U_{k_j}}\) and \(\cup_{j \in J_K} \overline{U_{k_j}} \subseteq T\) is compact, by the proposition that for any topological space, the union of any finite compact subsets is compact.
\(U_K \cup \{\infty\} \subseteq T^+ \setminus K\), because for each \(p \in U_K \cup \{\infty\}\), when \(p \in U_K = T \setminus \cup_{j \in J_K} \overline{U_{k_j}}\), \(p \notin \cup_{j \in J_K} \overline{U_{k_j}}\), but as \(K \subseteq \cup_{j \in J_K} U_{k_j} \subseteq \cup_{j \in J_K} \overline{U_{k_j}}\), \(p \notin K\), so, \(p \in T^+ \setminus K\), and when \(p = \infty\), \(p \in T^+ \setminus K\).
Step 2:
Let us take \(B^+ := B \cup \{U_K \cup \{\infty\} \vert K \in \{\text{ the compact subsets of } T\}\}\).
Each \(\{U_{k_j} \vert j \in J_K\}\) is a finite subset of \(B\), and \(\{\{U_{k_j} \vert j \in J_K\} \vert K \in \{\text{ the compact subsets of } T\}\} \subseteq \{\text{ the finite subsets of } B\}\): \(\{\{U_{k_j} \vert j \in J_K\} \vert K \in \{\text{ the compact subsets of } T\}\}\) does not necessarily have the number of the compact subsets of \(T\) elements, because any set has no duplication in the elements and \(\{U_{k_j} \vert j \in J_K\} = \{U_{k_j} \vert j \in J_{K'}\}\) for some \(K \neq K'\) and they are a single element.
\(\{\{U_{k_j} \vert j \in J_K\} \vert K \in \{\text{ the compact subsets of } T\}\}\) is countable, by the proposition that for any countable set, the set of the finite subsets is countable and the proposition that any subset of any countable set is countable, so, there is a bijection, \(f: \mathbb{N} \to \{\{U_{k_j} \vert j \in J_K\} \vert K \in \{\text{ the compact subsets of } T\}\}\).
There is the surjection, \(g: \{\{U_{k_j} \vert j \in J_K\} \vert K \in \{\text{ the compact subsets of } T\}\} \to \{(T \setminus \cup_{j \in J_K} \overline{U_{k_j}}) \cup \{\infty\} \vert K \in \{\text{ the compact subsets of } T\}\} = \{U_K \cup \{\infty\} \vert K \in \{\text{ the compact subsets of } T\}\}\).
So, \(g \circ f: \mathbb{N} \to \{U_K \cup \{\infty\} \vert K \in \{\text{ the compact subsets of } T\}\}\) is a surjection, by the proposition that any finite composition of surjections is a surjection, if the codomains of the constituent surjections equal the domains of the succeeding surjections, and there is a bijection, \(f': \mathbb{N} \to \{U_K \cup \{\infty\} \vert K \in \{\text{ the compact subsets of } T\}\}\), by the proposition that for any infinite set, if there is a surjection from the natural numbers set onto the set, there is a bijection from the natural numbers set onto the set.
So, \(\{U_K \cup \{\infty\} \vert K \in \{\text{ the compact subsets of } T\}\}\) is countable.
\(B^+\) is countable, by the proposition that any countable union of countable sets is countable.
Let us see that \(B^+\) is a basis for \(T^+\).
For each \(b^+ \in B^+\), \(b^+ \in B\) or \(b^+ \in \{U_K \cup \{\infty\} \vert K \in \{\text{ the compact subsets of } T\}\}\), but when \(b^+ \in B\), \(b^+\) is open on \(T\), so, is open on \(T^+\), and when \(b^+ \in \{U_K \cup \{\infty\} \vert K \in \{\text{ the compact subsets of } T\}\}\), \(b^+\) is open on \(T^+\), as has been seen above.
Let \(t^+ \in T^+\) be any.
Let \(N^+_{t^+} \subseteq T^+\) be any neighborhood of \(t^+\).
\(t^+ \in T\) or \(t^+ = \infty\).
When \(t^+ \in T\), \(N^+_{t^+} \cap T \subseteq T\) is a neighborhood of \(t^+\) on \(T\), by the proposition that for any topological space and any point on any subspace, the intersection of any neighborhood of the point on the base space and the subspace is a neighborhood on the subspace, and there is a \(U_{t^+} \in B \subseteq B^+\), such that \(t^+ \in U_{t^+} \subseteq N^+_{t^+} \cap T \subseteq N^+_{t^+}\).
When \(t^+ = \infty\), there is an open neighborhood of \(t^+\), \(U^+_{t^+} \subseteq T^+\), such that \(U^+_{t^+} \subseteq N^+_{t^+}\), but \(U^+_{t^+} = T^+ \setminus K\) for a compact \(K \subseteq T\), and \(\infty \in U_K \cup \{\infty\} \subseteq T^+ \setminus K\), as has been seen above, so, \(\infty \in U_K \cup \{\infty\} \subseteq N^+_{t^+}\) where \(U_K \cup \{\infty\} \in B^+\).
So, \(B^+\) is a basis for \(T^+\).
So, \(B^+\) is a countable basis for \(T^+\).
Step 3:
\(T^+\) is compact Hausdorff, 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, \(T^+\) is normal, by the proposition that any compact Hausdorff topological space is normal, \(T^+\) is completely regular, by the proposition that any normal topological space is completely regular, and \(T^+\) is metrizable, by the proposition that any 2nd-countable completely regular topological space is metrizable (Urysohn metrization theorem) and Step 2.
