Image of Subset under Neighborhood of Diagonal is Neighborhood of Subset

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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 $V$ 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 $T$

$\blacksquare$