Image of Point under Neighborhood of Diagonal is Neighborhood of Point

Theorem
Let $T = \struct{X, \tau}$ be a $T_3$ Space.

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

Let $V$ be a neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.

Then:
 * $\forall x \in X : \map V x$ is a neighborhood of $x$ in $T$

Proof
Let $x \in X$.

By definition of diagonal:
 * $\tuple{x, x} \in \Delta_X$

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

By definition of the product topology:
 * $\BB = \set {U_1 \times U_2: U_1, U_2 \in \tau}$ is a basis for $\tau_{X \times X}$

From User:Leigh.Samphier/Topology/Characterization of Neighborhood by Basis:
 * $\exists U_1, U_2 \in \tau : \tuple{x, x} \in U_1 \times U_2$ and $U_1 \times U_2 \subseteq V$

By definition of Cartesian product:
 * $x \in U_1, x \in U_2$

We have:

Hence $\map V x$ is a neighborhood of $x$ in $T$ by definition.

The result follows.