definition of \(1\)-point compactification of locally compact Hausdorff topological space
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 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 any open set minus any closed set is open.
- The reader admits the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets.
- 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 closed subset of any compact topological space is compact.
- 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 for any topological space, any compact subset of any subspace is compact on the base space.
Target Context
- The reader will have a definition of \(1\)-point compactification of locally compact Hausdorff topological space.
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, \(O\)
\(*T^+\): \(= T \cup \{\infty\}\) with topology, \(O^+ := O \cup \{T^+ \setminus K \vert K \in \{\text{ the compact subsets of } T\}\}\)
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Conditions:
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\(T^+\) is, in fact, a compact Hausdorff topological space with \(T\) as the subspace, 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.
2: Note
The added point, \(\infty\), does not have any special meaning of being infinitely large or something, but it is just any point not contained in \(T\), so, any point, \(p \notin T\), can be used instead of \(\infty\).
Let us see that \(O^+\) is indeed a topology.
\(\emptyset \in O^+\), because \(\emptyset \in O\).
\(T^+ = T^+ \setminus \emptyset \in O^+\), because \(\emptyset \subseteq T\) is compact.
Let \(U_1, U_2 \in O^+\) be any.
\(U_1, U_2 \in O\), \(U_1 \in O\) and \(U_2 = T^+ \setminus K\), or \(U_1 = T^+ \setminus K_1\) and \(U_2 = T^+ \setminus K_2\), without loss of generality.
When \(U_1, U_2 \in O\), \(U_1 \cap U_2 \in O\), so, \(U_1 \cap U_2 \in O^+\).
When \(U_1 \in O\) and \(U_2 = T^+ \setminus K\), \(U_1 \cap U_2 = U_1 \cap (T^+ \setminus K) = (U_1 \cap T^+) \setminus (U_1 \cap K)\), 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, \(= U_1 \setminus K\), but \(K \subseteq T\) is closed, by the proposition that any compact subset of any Hausdorff topological space is closed, so, \(U_1 \setminus K \subseteq T\) is open, by the proposition that any open set minus any closed set is open, so, \(U_1 \cap U_2 \in O\), so, \(U_1 \cap U_2 \in O^+\).
When \(U_1 = T^+ \setminus K_1\) and \(U_2 = T^+ \setminus K_2\), \(U_1 \cap U_2 = (T^+ \setminus K_1) \cap (T^+ \setminus K_2) = T^+ \setminus (K_1 \cup K_2)\), by the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets, but \(K_1 \cup K_2 \subseteq T\) is compact, by the proposition that for any topological space, the union of any finite compact subsets is compact, so, \(U_1 \cap U_2 \in O^+\).
Let \(\{U_j \in O^+ \vert j \in J\}\) where \(J\) is any possibly uncountable index set be any.
\(\{U_j \in O^+ \vert j \in J\} \subseteq O\) or not.
When \(\{U_j \in O^+ \vert j \in J\} \subseteq O\), \(\cup_{j \in J} U_j \in O\), so, \(\cup_{j \in J} U_j \in O^+\).
Otherwise, there is at least \(1\), \(U_l\) such that \(U_l \notin O\), which means that \(U_l = T^+ \setminus K\), \(S := T^+ \setminus (\cup_{j \in J} U_j) = \cap_{j \in J} (T^+ \setminus U_j)\), by the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets, \(= (T^+ \setminus U_l) \cap (\cap_{j \in J \setminus \{l\}} (T^+ \setminus U_j)) = K \cap (\cap_{j \in J \setminus \{l\}} (T^+ \setminus U_j)) = K \cap (\cap_{j \in J \setminus \{l\}} (T \setminus U_j))\), but \(T \setminus U_j\) is closed on \(T\) when \(U_j \in O\) and is closed on \(T\) when \(U_j = T^+ \setminus K_j\), because \(T \setminus U_j = T \setminus (T^+ \setminus K_j) = K_j\), closed on \(T\), by the proposition that any compact subset of any Hausdorff topological space is closed, so, \(\cap_{j \in J \setminus \{l\}} (T \setminus U_j)\) is closed on \(T\), \(K \cap (\cap_{j \in J \setminus \{l\}} (T \setminus U_j))\) is closed on \(K\) and is compact on \(K\), by the proposition that any closed subset of any compact topological space is compact: \(K\) is a compact subspace, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace, so, \(S\) is compact on \(T\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space, and \(\cup_{j \in J} U_j = T^+ \setminus S\), so, \(\cup_{j \in J} U_j \in O^+\).
So, \(O^+\) is a topology.
