Characterization of Even Cover
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $\UU$ be a cover of $S$.
Let $\tau_{S \times S}$ denote the product topology on the cartesian product $S \times S$.
Then $\UU$ is an even cover of $T$ if and only if there exists an open neighborhood $V$ of the diagonal $\Delta_S$ of $S \times S$ in the product space $\struct {S \times S, \tau_{S \times S} }$:
- $\set{\map V x : x \in S}$ is a refinement of $\UU$
where:
Proof
Necessary Condition
Let $\UU$ be an even cover of $T$.
By definition of even cover, there exists a neighborhood $N$ of the diagonal $\Delta_S$ of $S \times S$ in the product space $\struct {S \times S, \tau_{S \times S} }$:
- $\set{\map N x : x \in S}$ is a refinement of $\UU$
By definition of a neighborhood, there exists an open set $V$ of $\struct {S \times S, \tau_{S \times S} }$:
- $\Delta_S \subseteq V \subseteq N$
From Set is Open iff Neighborhood of all its Points and Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
- $V$ is an open neighborhood of the $\Delta_S$
It remains to show that $\set{\map V x : x \in S}$ is a refinement of $\UU$.
Let $x \in S$.
We have:
- $\tuple{x, x} \in \Delta_S \subseteq V$
By definition of image:
- $x \in \map V x$
From Corollary to Image under Subset of Relation is Subset of Image under Relation:
- $\map V x \subseteq \map N x$
By definition of refinement:
- $\exists U \in \UU : \map N x \subseteq U$
From Subset Relation is Transitive:
- $\map V x \subseteq U$
Since $x$ was arbitrary, we have:
- $\forall x \in S : x \in \map V x$
and
- $\forall x \in S : \exists U \in \UU : \map V x \subseteq U$
It follows that $\set{\map V x : x \in S}$ is a cover of $S$ and a refinement of $\UU$ by definition.
$\Box$
Sufficient Condition
Let $V$ be an open neighborhood of the diagonal $\Delta_S$ of $S \times S$ in the product space $\struct {S \times S, \tau_{S \times S} }$:
- $\set{\map V x : x \in S}$ is a refinement of $\UU$
By definition of open neighborhood:
- $V$ is a neighborhood of $\Delta_S$ in $\struct {S \times S, \tau_{S \times S}}$
It follows that $\UU$ is an even cover of $T$ by definition.
$\blacksquare$