Way Below Compact is Topological Compact

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Theorem

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

Let $L = \struct {\tau, \preceq}$ be an ordered set where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$

Let $x \in \tau$.


Then

$x$ is compact in $L$ (it means: $x \ll x$)

if and only if

$T_x$ is compact (topologically)

where $T_x = \struct {x, \tau_x}$ denotes the topological subspace of $x$.


Proof

Sufficient Condition

Let

$x \ll x$

Let $F \subseteq \tau_x$ be a open cover of $x$.

By definition of cover:

$x \subseteq \bigcup F$

By definition of topological space:

$\forall y \in \tau: x \cap y \in \tau$

By definition of subset:

$\tau_x \subseteq \tau$

By Subset Relation is Transitive:

$F \subseteq \tau$

By Way Below in Ordered Set of Topology:

there exists a finite subset $G$ of $F$: $\ds x \subseteq \bigcup G$

By definition:

$G$ is cover of $x$

Thus by definition:

$x$ has finite subcover.

Thus by definition:

$T_x$ is compact.

$\Box$

Necessary Condition

Let

$T_x$ is compact.

Let $F$ be a set of open subsets of $S$ such that

$\ds x \subseteq \bigcup F$

Define $Y := \set {x \cap y: y \in F}$

By definition of subset:

$Y \subseteq \tau_x$

By definition of union:

$\ds x \subseteq \bigcup Y$

By definition

$Y$ is open cover of $x$.

By definition of compact:

$x$ has finite subcover $G$.

By Axiom of Choice:

$\exists g: G \to F: \forall y \in G: \map g y \cap x = y$

By Cardinality of Image of Set not greater than Cardinality of Set:

$\card {\map {g^\to} G} \le \card G$

where $\card G$ denotes the cardinality of $G$.

Thus

$\map {g^\to} G$ is finite.

Thus by definitions of subset and image of set:

$\map {g^\to} G \subseteq F$

By Intersection is Subset:

$\forall y \in G: y \subseteq \map g y$

By Set Union Preserves Subsets/Families of Sets:

$\ds \bigcup G = \bigcup_{y \mathop \in G}y \subseteq \bigcup_{y \mathop \in G} \map g y = \bigcup \map {g^\to} G$

By definition of cover:

$\ds x \subseteq \bigcup G$

Thus by Subset Relation is Transitive:

$\ds x \subseteq \bigcup \map {g^\to} G$

Thus by Way Below in Ordered Set of Topology;

$x \ll x$

$\blacksquare$


Sources

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