Empty Set is Open in Neighborhood Space
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Theorem
Let $\struct {S, \NN}$ be a neighborhood space.
Then the empty set $\O$ is an open set of $\struct {S, \NN}$.
Proof
Suppose $\O$ were not an open set of $\struct {S, \NN}$.
Then $\exists x \in \O$ such that $\O$ is not a neighborhood of $\O$.
By definition of empty set, such an $x$ does not exist.
Hence the result.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Lemma $3.6$