Image of Subset under Open Neighborhood of Diagonal is Open 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 $U$ be an open neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.

Then:
 * $\forall A \subseteq X : U \sqbrk A$ is an open neighborhood of $A$ in $T$

Proof
Let $A \subseteq X$.

From Image of Subset under Neighborhood of Diagonal is Neighborhood of Subset:
 * $U \sqbrk A$ is a neighborhood of $A$ in $T$

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

From User:Leigh.Samphier/Topology/Image of Point under Open Neighborhood of Diagonal is Open Neighborhood of Point:
 * $\forall x \in A : \map U x$ is an open neighborhood of $x$ in $T$

By :
 * $U \sqbrk A \in \tau$

Since $A$ was arbitrary:
 * $\forall A \subseteq X : U \sqbrk A$ is an open neighborhood of $A$ in $T$