description/proof of that for locally compact Hausdorff topological space, topology of \(1\)-point compactification is only topology that makes \(1\)-point-augmented set compact Hausdorff with original space as subspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of \(1\)-point compactification of locally compact Hausdorff topological space.
- The reader knows a definition of compact topological space.
- 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 compact subset of any Hausdorff topological space is closed.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
- 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 any closed subset of any compact topological space is compact.
- The reader admits the proposition that for any topological space, any compact subset of the space that is contained in any subspace is compact on the subspace.
- The reader admits the proposition that for any Hausdorff topological space, any 1 point subset is closed.
- The reader admits the proposition that any open set on any open topological subspace is open on the base space.
- The reader admits the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
Target Context
- The reader will have a description and a proof of 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.
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:
\(T\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\), with topology, \(O\)
\(T^+\): \(= \text{ the 1-point compactification of } T\) with the topology, \(O^+\)
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Statements:
\(\{\text{ the topologies that make } T^+ \text{ compact Hausdorff with } T \text{ as the subspace }\} = \{O^+\}\)
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2: Proof
Whole Strategy: Step 1: suppose that \(T^+\) has \(O^+\); Step 2: see that \(T\) is the subspace of \(T^+\); Step 3: for any open cover of \(T^+\), \(\{U_j \vert j \in J\}\), find a finite subcover; Step 4: see that \(T^+\) is Hausdorff; Step 5: suppose that \({O^+}'\) makes \(T^+\) compact Hausdorff with \(T\) as the subspace, and see that \(O^+ = {O^+}'\).
Step 1:
Let us suppose that \(T^+\) has \(O^+\).
Step 2:
Let us see that \(T\) is the topological subspace of \(T^+\).
Let \(U \in O\) be any.
\(U \in O^+\), and \(U = U \cap T\).
Let \(U' \in O^+\) be any.
\(U' \in O\) or \(U' = T^+ \setminus K\) where \(K \subseteq T\) is compact.
When \(U' \in O\), \(U' \cap T = U' \in O\).
When \(U' = T^+ \setminus K\), \(U' \cap T = (T^+ \setminus K) \cap T = (T^+ \cap T) \setminus (K \cap T)\), 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, \(= T \setminus K\), but \(K \subseteq T\) is closed, by the proposition that any compact subset of any Hausdorff topological space is closed, so, \(T \setminus K\) is open on \(T\).
So, any subset of \(T\) is open if and only if it is the intersection of an open subset of \(T^+\) and \(T\), so, \(T\) is the subspace of \(T^+\).
Step 3:
Let \(\{U_j \vert j \in J\}\) where \(J\) is any possibly uncountable index set be any open cover of \(T^+\).
There is a \(U_l\) such that \(\infty \in U_l\).
\(U_l = T^+ \setminus K\) where \(K \subseteq T\) is a compact subset.
\(T^+ \subseteq U_l \cup K\).
\(K \subseteq \cup_{j \in J \setminus \{l\}} U_j\), because for each \(k \in K\), as \(k \in T^+\), \(k \in U_j\) for a \(j \in J\), but \(k \notin U_l\), so, \(k \in U_j\) for a \(j \in J \setminus \{l\}\).
\(K \subseteq (\cup_{j \in J \setminus \{l\}} U_j) \cap T = \cup_{j \in J \setminus \{l\}} (U_j \cap T)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, but \(U_j \cap T\) is open on \(T\), because \(T\) is a topological subspace of \(T^+\), so, \(\{U_j \cap T \vert j \in J \setminus \{l\}\}\) is an open cover of \(K\).
As \(K \subseteq T\) is compact, there is a finite subcover, \(\{U_j \cap T \vert j \in J^`\}\), where \(J^` \subseteq J \setminus \{l\}\) is a finite index set.
\(K \subseteq \cup_{j \in J^`} (U_j \cap T) \subseteq \cup_{j \in J^`} U_j\).
So, \(T^+ \subseteq U_l \cup K \subseteq U_l \cup (\cup_{j \in J^`} U_j) = \cup_{j \in (\{l\} \cup J^`)} U_j\).
As \(\{U_j \vert j \in (\{l\} \cup J^`)\}\) is a finite subcover of \(\{U_j \vert j \in J\}\), \(T^+\) is compact.
Step 4:
Let \(p_1, p_2 \in T^+\) be any such that \(p_1 \neq p_2\).
\(p_1, p_2 \in T\) or \(p_1 \in T\) and \(p_2 = \infty\), without loss of generality.
When \(p_1, p_2 \in T\), there are an open neighborhood of \(p_1\), \(U_{p_1} \subseteq T\), and an open neighborhood of \(p_2\), \(U_{p_2} \subseteq T\), such that \(U_{p_1} \cap U_{p_2} = \emptyset\), because \(T\) is Hausdorff, but \(U_{p_1}\) and \(U_{p_2}\) are open on \(T^+\).
When \(p_1 \in T\) and \(p_2 = \infty\), there is a compact neighborhood of \(p_1\), \(K_{p_1} \subseteq T\), 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 \(p_1\), \(U_{p_1} \subseteq T\), such that \(U_{p_1} \subseteq K_{p_1}\), because \(K_{p_1}\) is a neighborhood, and \(T^+ \setminus K_{p_1}\) is an open neighborhood of \(p_2\), and \(U_{p_1} \cap (T^+ \setminus K_{p_1}) = \emptyset\).
So, \(T^+\) is Hausdorff.
Step 5:
Let us suppose that a topology, \({O^+}'\), makes \(T^+\) compact Hausdorff with \(T\) as the topological subspace.
Let \(U \in O^+\) be any.
\(U \in O\) or \(U = T^+ \setminus K\) where \(K \subseteq T\) is compact.
When \(U \in O\), \(T \subseteq T^+\) is open with respect to \({O^+}'\), because \(T = T^+ \setminus \{\infty\}\) and \(\{\infty\} \subseteq T^+\) is closed, by the proposition that for any Hausdorff topological space, any 1 point subset is closed, and \(U = U \cap T \in {O^+}'\), by the proposition that any open set on any open topological subspace is open on the base space.
When \(U = T^+ \setminus K\), \(K\) is compact on \(T^+\) with respect to \({O^+}'\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space, so, \(K\) is closed on \(T^+\) with respect to \({O^+}'\), by the proposition that any compact subset of any Hausdorff topological space is closed, so, \(U \in {O^+}'\).
So, \(O^+ \subseteq {O^+}'\).
Let \(U' \in {O^+}'\) be any.
\(U' \subseteq T\) or not.
When \(U' \subseteq T\), \(U' = U' \cap T \in O\), because \(T\) is the subspace, so, \(U' \in O^+\).
Otherwise, \(S := T^+ \setminus U' \subseteq T^+\) is closed and is compact with respect to \({O^+}'\), by the proposition that any closed subset of any compact topological space is compact, and \(T^+ \setminus U' \subseteq T\) is compact on \(T\), by the proposition that for any topological space, any compact subset of the space that is contained in any subspace is compact on the subspace, so, \(U' = T^+ \setminus S \in O^+\).
So, \({O^+}' \subseteq O^+\).
So, \(O^+ = {O^+}'\).
