Image of Subset under Neighborhood of Diagonal is Neighborhood of Subset

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 $V$ 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:
 * $\forall A \subseteq X : V \sqbrk A$ is a neighborhood of $A$ in $T$

Proof
Let $A \subseteq X$.

From Image of Subset under Relation equals Union of Images of Elements:
 * $V \sqbrk A = \ds \bigcup_{x \in A} \map V x$

From Subset of Union:
 * $\forall x \in A : \map V x \subseteq V \sqbrk A$

From Image of Point under Neighborhood of Diagonal is Neighborhood of Point:
 * $\forall x \in A : \map V x$ is a neighborhood of $x$ in $T$

From Superset of Neighborhood in Topological Space is Neighborhood:
 * $\forall x \in A : \map V A$ is a neighborhood of $x$ in $T$

From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
 * $\map V A$ is a neighborhood of $A$ in $T$