Inverse of Neighborhood of Diagonal Point is Neighborhood

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

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

Let $x \in X$.

Let $R$ be a neighborhood of $\tuple{x,x}$ in the product space $\struct{X \times X, \tau_{X \times X}}$.

Let $R^{-1}$ denote the inverse relation of $R$ where $R$ is viewed as a relation on $X \times X$.

Then:
 * $R^{-1}$ is a neighborhood of $\tuple{x,x}$ in $\struct{X \times X, \tau_{X \times X}}$.

Proof
By definition of neighborhood:
 * $\tuple{x, x} \in R$

By definition of product topology:
 * $\BB = \set {V_1 \times V_2: V_1, V_2 \in \tau}$ is a basis for $\tau_{X \times X}$

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

By definition of symmetric relation:
 * $\tuple{x, x} \in V_2 \times V_1 : V_2 \times V_1 \subseteq R^{-1}$

By definition of product topology:
 * $V_2 \times V_1 \in \tau_{X \times X}$

Hence $R^{-1}$ is a neighborhood $\tuple{x, x}$ by definition.