Image of Subset under Open Neighborhood of Diagonal is Open 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 $U$ 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 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 Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:
- $U \sqbrk A \in \tau$
Since $A$ was arbitrary:
- $\forall A \subseteq X : U \sqbrk A$ is an open neighborhood of $A$ in $T$
$\blacksquare$