Topology is Locally Compact iff Ordered Set of Topology is Continuous

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $L = \struct {\tau, \preceq}$ be the ordered set where $\preceq$ is the subset relation.


Then

$(1): \quad T$ is locally compact implies $L$ is continuous
$(2): \quad T$ is $T_3$ space and $L$ is continuous implies $T$ is locally compact


Proof

Condition $(1)$

Let $T$ be locally compact.

By Topology forms Complete Lattice:

$L$ is complete lattice.

Thus by Way Below Closure is Directed in Bounded Below Join Semilattice:

$\forall x \in \tau: x^\ll$ is directed

Thus by definition:

$L$ is up-complete.

Let $x \in \tau$.

We will prove that

$x \subseteq \map \sup {x^\ll}$

Let $a \in x$.

By definition:

$x$ is open set in $T$

By definition of locally compact:

there exists local basis $\BB$ of $a$ such that all elements are compact.

By definition of local basis:

$\exists y \in \BB: y \subseteq x$

Then

$y$ is compact.

By Way Below if Between is Compact Set in Ordered Set of Topology:

$y \ll x$

By definition of way below closure:

$y \in x^\ll$

By Set is Subset of Union/Set of Sets:

$\ds y \subseteq \bigcup \paren {x^\ll}$

By definition of subset:

$\ds a \in \bigcup \paren {x^\ll}$

Thus by proof of Topology forms Complete Lattice:

$a \in \sup \paren {x^\ll}$

$\Box$


According to definition of set equality it remains to prove that:

$\map \sup {x^\ll} \subseteq x$

Let $a \in \map \sup {x^\ll}$.

By proof of Topology forms Complete Lattice:

$\ds a \in \bigcup \paren {x^\ll}$

By definition of union:

$\exists y \in x^\ll: a \in y$

By definition of way below closure:

$y \ll x$

By Way Below implies Preceding:

$y \preceq x$

Then

$y \subseteq x$

Thus by definition of subset:

$a \in x$

By definition:

$L$ satisfies the axiom of approximation.

Thus by definition

$L$ is continuous.

$\Box$


Condition $(2)$

Let $T$ be a $T_3$ space.

Let $L$ be continuous.

Let $x \in S, A \in \tau$ such that:

$x \in A$

By definition of continuous:

$L$ satisfies the axiom of approximation.
\(\ds A\) \(=\) \(\ds \map \sup {A^\ll}\) Axiom of Approximation
\(\ds \) \(=\) \(\ds \bigcup \paren {A^\ll}\) Topology forms Complete Lattice

By definition of union:

$\exists y \in A^\ll: x \in y$

By definition of way below closure:

$y \ll A$

By definition of Definition:T3 Space/Definition 2:

$\exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq y$

By definition of interior:

$x \in N_x^\circ$

where $N_x^\circ$ denotes the interior of $N_x$.


We will prove that $N_x$ is topologically compact.

Let $\FF$ be a set of open subsets of $S$ such that:

$\FF$ is a cover of $N_x$

Define a set of open subsets of $S$:

$\FF' := \FF \cup \set {\relcomp S {N_x} }$
\(\ds \bigcup \FF'\) \(=\) \(\ds \paren {\bigcup \FF} \cup \bigcup \set {\relcomp S {N_x} }\) Set Union is Self-Distributive/Sets of Sets
\(\ds \) \(=\) \(\ds \paren {\bigcup \FF} \cup \relcomp S {N_x}\) Union of Singleton
\(\ds \) \(\supseteq\) \(\ds N_x \cup \relcomp S {N_x}\) {Definition of Cover of Set
\(\ds \) \(=\) \(\ds S\) Definition of Relative Complement
\(\ds \) \(\supseteq\) \(\ds A\)

Then

$\ds A \subseteq \bigcup \FF'$

By Way Below in Ordered Set of Topology:

there exists a finite subset $\GG'$ of $\FF'$: $\ds y \subseteq \bigcup \GG'$

Define a finite subset of $\FF$:

$\GG := \GG' \setminus \set {\relcomp S {N_x} }$

By definition of union:

$\ds N_x \subseteq \bigcup \GG$

Thus by definition:

$N_x$ has finite subcover of $\FF$

Thus by definition:

$N_x$ is compact.

$\Box$


Thus by Locally Compact iff Open Neighborhood contains Compact Set;

$T$ is locally compact.

$\blacksquare$


Sources

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