Inverse of Neighborhood of Diagonal Point is Neighborhood
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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.
$\blacksquare$