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:
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
- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices
- Mizar article WAYBEL_3:41
- Mizar article WAYBEL_3:42