Niemytzki Plane is Topology

Theorem
Niemytzki plane is a topological space.

Proof
By definition $T = \left({S, \tau}\right)$ is the Niemytzki plane if

According to Topology Defined by Neighborhood System it should be proved
 * $(N0): \quad \forall \left({x, y}\right) \in S: \mathcal B\left({x, y}\right)$ is non-empty set of subsets of $S$
 * $(N1): \quad \forall \left({x, y}\right) \in S, U \in \mathcal B\left({x, y}\right): \left({x, y}\right) \in U$
 * $(N2): \quad \forall \left({x, z}\right) \in S, U \in \mathcal B\left({x, z}\right), \left({y, s}\right) \in U:\exists V \in \mathcal B\left({y, s}\right): V \subseteq U$
 * $(N3): \quad \forall \left({x, y}\right) \in S, U_1, U_2 \in \mathcal B\left({x, y}\right): \exists U \in \mathcal B\left({x, y}\right): U \subseteq U_1 \cap U_2$