Whole Space is Open in Neighborhood Space

Theorem

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

Then $S$ itself is an open set of $\struct {S, \NN}$.

Proof

Let $x \in S$.

Then by neighborhood space axiom $\text N 1$ there exists a neighborhood $N$ of $x$.

As $N \subseteq S$ it follows from neighborhood space axiom $\text N 3$ that $S$ is a neighborhood of $x$.

As this holds for all $x \in S$ it follows that $S$ is an open set of $\struct {S, \NN}$.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Lemma $3.6$