description/proof of that for Hausdorff topological space, iff each point has compact neighborhood, space is locally compact
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Hausdorff topological space.
- The reader admits the proposition that any compact subset of any Hausdorff topological space is closed.
- 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 the compactness of any topological subset as a subset equals the compactness as a subspace.
- 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 any subspace of any Hausdorff topological space is Hausdorff.
- The reader admits the proposition that any compact Hausdorff topological space is normal.
- The reader admits the proposition that any normal topological space is regular.
- The reader admits the proposition that any topological space is regular if and only if for each point of the space, the 1-point subset is closed and the set of the closed neighborhoods of the point is a neighborhoods basis at the point.
- The reader admits the proposition that any closed set on any closed topological subspace is closed on the base space.
Target Context
- The reader will have a description and a proof of the proposition that for any Hausdorff topological space, if and only if each point has a compact neighborhood, the space is locally compact.
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:
\(T\): \(\in \{\text{ the Hausdorff topological spaces }\}\)
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Statements:
\(\forall t \in T (\exists K_t \subseteq T \in \{\text{ the compact neighborhoods of } t\})\)
\(\iff\)
\(T \in \{\text{ the locally compact topological spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that each point has a compact neighborhood; Step 2: take any neighborhood of \(t\), \(N_t\), any open neighborhoods of \(t\), \(U_{t, 1} \subseteq K_t\) and \(U_{t, 2} \subseteq N_t\); Step 3: see that \(\overline{U_{t, 1} \cap U_{t, 2}}\) is a compact Hausdorff space and is regular; Step 4: take a closed neighborhood of \(t\) on \(\overline{U_{t, 1} \cap U_{t, 2}}\) such that \(C_t \subseteq U_{t, 1} \cap U_{t, 2}\), and see that \(C_t\) is a compact neighborhood of \(t\) on \(T\) such that \(C_t \subseteq N_t\); Step 5: suppose that \(T\) is locally compact; Step 6: see that each point has a compact neighborhood.
Step 1:
Let us suppose that \(\forall t \in T (\exists K_t \subseteq T \in \{\text{ the compact neighborhoods of } t\})\).
Step 2:
There is an open neighborhood of \(t\), \(U_{t, 1} \subseteq T\), such that \(U_{t, 1} \subseteq K_t\).
Let \(N_t \subseteq T\) be any neighborhood of \(t\).
There is an open neighborhood of \(t\), \(U_{t, 2} \subseteq T\), such that \(U_{t, 2} \subseteq N_t\).
\(U_{t, 1} \cap U_{t, 2} \subseteq T\) is an open neighborhood of \(t\).
Step 3:
\(K_t\) is closed on \(T\), by the proposition that any compact subset of any Hausdorff topological space is closed.
\(\overline{U_{t, 1} \cap U_{t, 2}} \subseteq K_t\), because \(U_{t, 1} \cap U_{t, 2} \subseteq U_{t, 1} \subseteq K_t\) and the closure is the intersection of all the closed subsets that contain \(U_{t, 1} \cap U_{t, 2}\), and \(\overline{U_{t, 1} \cap U_{t, 2}}\) is a closed subset of \(K_t\), 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: \(\overline{U_{t, 1} \cap U_{t, 2}} = \overline{U_{t, 1} \cap U_{t, 2}} \cap K_t\).
\(K_t\) 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, \(\overline{U_{t, 1} \cap U_{t, 2}}\) is a compact subset of \(K_t\), by the proposition that any closed subset of any compact topological space is compact, 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, and 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.
\(\overline{U_{t, 1} \cap U_{t, 2}} \subseteq T\) is Hausdorff, by the proposition that any subspace of any Hausdorff topological space is Hausdorff.
\(\overline{U_{t, 1} \cap U_{t, 2}}\) is normal and is regular, by the proposition that any compact Hausdorff topological space is normal and the proposition that any normal topological space is regular.
Step 4:
On \(\overline{U_{t, 1} \cap U_{t, 2}}\), the set of the closed neighborhoods of \(t\) is a neighborhoods basis at \(t\), by the proposition that any topological space is regular if and only if for each point of the space, the 1-point subset is closed and the set of the closed neighborhoods of the point is a neighborhoods basis at the point, so, there is a closed neighborhood of \(t\), \(C_t \subseteq \overline{U_{t, 1} \cap U_{t, 2}}\), such that \(C_t \subseteq U_{t, 1} \cap U_{t, 2}\): \(U_{t, 1} \cap U_{t, 2} \subseteq \overline{U_{t, 1} \cap U_{t, 2}}\) is an open neighborhood of \(t\).
\(C_t\) is closed on \(T\), because \(\overline{U_{t, 1} \cap U_{t, 2}}\) is closed on \(T\), and the proposition that any closed set on any closed topological subspace is closed on the base space applies.
\(C_t \subseteq U_{t, 1} \cap U_{t, 2} \subseteq U_{t, 1} \subseteq K_t\), and \(C_t\) is closed on \(K_t\), because \(C_t = C_t \cap K_t\), 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, and is compact on \(K_t\), by the proposition that any closed subset of any compact topological space is compact, and is compact on \(T\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
\(C_t\) is a neighborhood of \(t\) on \(T\), because there is an open neighborhood of \(t\), \(U_t \subseteq \overline{U_{t, 1} \cap U_{t, 2}}\), such that \(U_t \subseteq C_t\), \(U_t = U'_t \cap \overline{U_{t, 1} \cap U_{t, 2}}\) where \(U'_t \subseteq T\) is an open neighborhood of \(t\) on \(T\), \(= U'_t \cap \overline{U_{t, 1} \cap U_{t, 2}} \cap U_{t, 1} \cap U_{t, 2}\), because \(U_t \subseteq C_t \subseteq U_{t, 1} \cap U_{t, 2}\), \(= U'_t \cap U_{t, 1} \cap U_{t, 2}\), which is an open neighborhood of \(t\) on \(T\), because \(U_{t, 1} \cap U_{t, 2}\) is open on \(T\).
So, \(C_t \subseteq T\) is a compact neighborhood of \(t\) on \(T\).
\(C_t \subseteq U_{t, 1} \cap U_{t, 2} \subseteq U_{t, 2} \subseteq N_t\).
So, \(T\) is locally compact.
Step 5:
Let us suppose that \(T\) is locally compact.
Step 6:
For each \(t\), \(T\) is a neighborhood of \(t\), and there is a compact neighborhood of \(t\) contained in \(T\), so, \(t\) has a compact neighborhood.
