Inverse of Neighborhood of Diagonal Point is Neighborhood

From ProofWiki
Jump to navigation Jump to search



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.

$\blacksquare$