Neighborhood of Diagonal induces Open Cover

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

Let $X \times X$ denote the cartesian product of $X$ with itself.

Let $\tau_{X \times X}$ denote the product topology on $X \times X$.

Let $T \times T$ denote the product space $\struct {X \times X, \tau_{X \times X} }$.

Let $U$ be a neighborhood of the diagonal $\Delta_X$ of $X \times X$ in $T \times T$.

Let $\VV = \set{V \in \tau : V \times V \subseteq U}$

Then:
 * $\VV$ is an open cover of $T$

Proof
By definition of product topology:
 * $\BB = \set {V_1 \times V_2: V_1, V_2 \in \tau}$ is a basis on $T \times T$

Let $x \in X$.

From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
 * $U$ is a neighborhood of the point $\tuple{x,x}$

From Characterization of Neighborhood by Basis:
 * $\exists V_1, V_2 \in \tau : \tuple{x, x} \in V_1 \times V_2 : V_1 \times V_2 \subseteq U$

By definition of Cartesian product:
 * $x \in V_1$ and $x \in V_2$

Let $V = V_1 \cap V_2$.

By definition of set intersection:
 * $x \in V$

By :
 * $V \in \tau$

From Intersection is Subset:
 * $V \subseteq V_1$ and $V \subseteq V_2$

From Cartesian Product of Subsets:
 * $V \times V \subseteq V_1 \times V2$

From Subset Relation is Transitive:
 * $V \times V \subseteq U$

Hence:
 * $\exists V \in \VV : x \in V$

Since $x$ was arbitrary, it follows that:
 * $\forall x \in X : \exists V \in \VV : x \in V$

By definition, $\VV$ is an open cover of $T$.