Neighborhood Space is Topological Space

Theorem

Let $\struct {S, \NN}$ be a neighborhood space.

Let $\tau = \set {N: N \in \NN}$ be the set of all open sets of $\struct {S, \NN}$.

Then $\struct {S, \tau}$ forms a topological space.

Proof

Each of the open set axioms is examined in turn:

Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets

From Union of Open Sets of Neighborhood Space is Open, it follows that {{Open-set-axiom|1} is fulfilled.

$\Box$

Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets

From Intersection of two Open Sets of Neighborhood Space is Open, it follows that Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets is fulfilled.

$\Box$

Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology

From Whole Space is Open in Neighborhood Space, it follows that Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology is fulfilled.

$\Box$

All the open set axioms are fulfilled, and the result follows.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces