By definition $T = \struct {S, \tau}$ is the Niemytzki plane {{iff}}: {{begin-eqn}} {{eqn | n = 1       | l = S       | r = \set {\tuple {x, y}: y \ge 0} }} {{eqn | n = 2       | l = \map \BB {x, y}       | r = \set {\map {B_r} {x, y} \cap S: r > 0}       | c = if $x, y \in \R, y > 0$ }} {{eqn | n = 3       | l = \map \BB {x, 0}       | r = \set {\map {B_r} {x, r} \cup \set {\tuple {x, 0} }: r > 0}       | c = if $x \in \R$ }} {{eqn | n = 4       | l = \tau       | r = \set {\bigcup \GG: \GG \subseteq \bigcup_{\tuple {x, y} \in S} \map \BB {x, y} } }} {{end-eqn}} where $\map {B_r} {x, y}$ denotes the open $r$-ball of $\tuple {x, y}$ in the real number plane $\R^2$ with the usual (Euclidean) metric. According to Topology Defined by Neighborhood System it should be proved: :$(\text N 0): \quad \forall \tuple {x, y} \in S: \map \BB {x, y}$ is non-empty set of subsets of $S$ :$(\text N 1): \quad \forall \tuple {x, y} \in S, U \in \map \BB {x, y}: \tuple {x, y} \in U$ :$(\text N 2): \quad \forall \tuple {x, z} \in S, U \in \map \BB {x, z}, \tuple {y, s} \in U: \exists V \in \map \BB {y, s}: V \subseteq U$ :$(\text N 3): \quad \forall \tuple {x, y} \in S, U_1, U_2 \in \map \BB {x, y}: \exists U \in \map \BB {x, y}: U \subseteq U_1 \cap U_2$ Ad $(\text N 0)$: Let $\tuple {x, y} \in S$. In both cases: $y > 0$ and $y = 0$, by $(2)$ and $(3)$: :$\map \BB {x, y}$ is non-empty Let $U \in \map \BB {x, y}$. In a case when $y > 0$: :$\exists r > 0: U = \map {B_r} {x, y} \cap S$ By Intersection is Subset: :$U$ is a subset of $S$. In a case when $y = 0$: :$\exists r > 0: U = \map {B_r} {x, r} \cup \set {\tuple {x, 0} }$ By definition of subset: :$\map {B_r} {x, r} \subseteq S$ and $\set {\tuple {x, 0} } \subseteq S$ By Union is Smallest Superset: :$U$ is a subset of $S$. Ad $(\text N 1)$: Let $\tuple {x, y} \in S$. Let $U \in \map \BB {x, y}$. In a case when $y > 0$ by $(2)$: :$\exists r > 0: U = \map {B_r} {x, y} \cap S$ By metric spce axioms: :$\map d {\tuple {x, y}, \tuple {x, y} } = 0$ where $d$ denotes the distance function of $\R^2$ Euclidean space. By definition of open ball: :$\tuple {x, y} \in \map {B_r} {x, y}$ Thus by definition of intersection: :$\tuple {x, y} \in U$ In a case when $y = 0$ by $(3)$: :$\exists r > 0: U = \map {B_r} {x, r} \cup \set {\tuple {x, 0} }$ By definition of singleton: :$\tuple {x, 0} \in \set {\tuple {x, 0} }$ Thus by definition of union: :$\tuple {x, 0} \in U$ Thus in both cases: :$\tuple {x, 0} \in U$ Ad $(\text N 2)$: Let $\tuple {x, 0} \in S, U \in \map \BB {x, y}, \tuple {z, s} \in U$. In a case when $y > 0$ by $(2)$: :$\exists r > 0: U = \map {B_r} {x, y} \cap S$ By definition of intersection: :$\tuple {z, s} \in \map {B_r} {x, y}$ By definition of open ball: :$\map d {\tuple {x, y}, \tuple {z, s} } < r$ In a subcase when $s > 0$ define :$r' := r - \map d {\tuple {x, y}, \tuple {z, s} }$ and :$V := \map {B_{r'} } {z, s} \cap S$ By $(2)$: :$V \in \map \BB {z, s}$ By proof of Open Ball of Point Inside Open Ball: :$\map {B_{r'} } {z, s} \subseteq \map {B_r} {x, y}$ Thus by Set Intersection Preserves Subsets/Corollary: :$V \subseteq U$ In a subcase when $s = 0$ define :$r'' := \dfrac {r - \map d {\tuple {x, y}, \tuple {z, s} } } 2$ and :$V := \map {B_{r''} } {z, r''} \cup \set {\tuple {z, 0} }$ By $(3)$: :$V \in \map \BB {z, s}$ {{begin-eqn}} {{eqn | l = \map d {\tuple {x, y}, \tuple {z, r''} }       | o = \le       | r = \map d {\tuple {x, y}, \tuple {z, 0} } + \map d {\tuple {z, 0}, \tuple {z, r''} }       | c = {{Defof|Metric Space Axioms}} }} {{eqn | r = \map d {\tuple {x, y}, \tuple {z, 0} } + r''       | c = {{Defof|Real Number Plane with Euclidean Metric}} }} {{eqn | r = r - r''}} {{end-eqn}} By Open Ball is Subset of Open Ball: :$\map {B_{r''} } {z, r''} \subseteq \map {B_r} {x, y}$ and :$\map {B_{r''} } {z, r''} \subseteq S$ By Intersection is Largest Subset: :$\map {B_{r''} } {z, r''} \subseteq U$ By definition of subset: :$\set {\tuple {z, 0} } \subseteq U$ Thus by Union is Smallest Superset: :$V \subseteq U$ In a case when $y = 0$ by $(3)$: :$\exists r > 0: U = \map {B_r} {x, r} \cup \set {\tuple {x, 0} }$ By definition of onion and singleton: :$\tuple {z, s} \in \map {B_r} {x, r}$ or $\tuple {z, s} = \tuple {x, 0}$ In a subcase when $\tuple {z, s} = \tuple {x, 0}$ define: :$V := U$ By $(3)$: :$V \in \map \BB {z, s}$ Thus by Set is Subset of Itself: :$V \subseteq U$ In a subcase when $\tuple {z, s} \in \map {B_r} {x, r}$, define: :$r' := r - \map d {\tuple {x, r}, \tuple {z, s} }$ and :$V := \map {B_{r'} } {z, s} \cap S$ By definition of open ball: :$\map d {\tuple {x, r}, \tuple {z, s} } < r$ Then $r' > 0$ By $(2)$: :$V \in \map \BB {z, s}$ By Open Ball of Point Inside Open Ball: :$\map {B_{r'} } {z, s} \subseteq \map {B_{r'} } {x, r}$ By Intersection is Subset: :$V = \map {B_{r'} } {z, s} \cap S \subseteq \map {B_{r'} } {z, s}$ By Set is Subset of Union: :$\map {B_{r'} } {x, r} \subseteq U$ Thus by Subset Relation is Transitive: $V \subseteq U$ Ad $(\text N 3)$: Let $\tuple {x, y} \in S, U_1, U_2 \in \map \BB {x, y}$. In a case when $y > 0$ by $(2)$: :$\exists r_1 > 0: U_1 = \map {B_{r_1} } {x, y} \cap S$ and :$\exists r_2 > 0: U_2 = \map {B_{r_2} } {x, y} \cap S$ By Trichotomy Law for Real Numbers: :$r_1 \le r_2$ or $r_1 \ge r_2$ {{WLOG}}, suppose  :$r_1 \le r_2$ {{begin-eqn}} {{eqn | l = r_2 - r_1       | o = \ge       | r = 0 }} {{eqn | r = \map d {\tuple {x, y}, \tuple {x, y} }       | c = metric space axioms }} {{end-eqn}} Then by Open Ball is Subset of Open Ball: :$\map {B_{r_1} } {x, y} \subseteq \map {B_{r_2} } {x, y}$ By Set Intersection Preserves Subsets/Corollary: :$U_1 \subseteq U_2$ By Intersection with Subset is Subset: :$U_1 = U_1 \cap U_2$ Thus by Set is Subset of Itself: :$U_1 \subseteq U_1 \cap U_2$ In a case when $y = 0$ by $(3)$: :$\exists r_1 > 0: U_1 = \map {B_{r_1} } {x, r_1} \cup \set {\tuple {x, 0} }$ and :$\exists r_2 > 0: U_2 = \map {B_{r_2} } {x, r_2} \cup \set {\tuple {x, 0} }$ By Trichotomy Law for Real Numbers: :$r_1 \le r_2$ or $r_1 \ge r_2$ {{WLOG}}, suppose  :$r_1 \le r_2$ {{begin-eqn}} {{eqn | l = r_2 - r_1       | r = \map d {\tuple {x, r_2}, \tuple {x, r_1} }       | c = {{Defof|Real Number Plane with Euclidean Metric}} }} {{end-eqn}} Then by Open Ball is Subset of Open Ball: :$\map {B_{r_1} } {x, r_1} \subseteq \map {B_{r_2} } {x, r_2}$ By Set Union Preserves Subsets/Corollary: :$U_1 \subseteq U_2$ By Intersection with Subset is Subset: :$U_1 = U_1 \cap U_2$ Thus by Set is Subset of Itself: :$U_1 \subseteq U_1 \cap U_2$ Thus by $(\text N 0)$-$(\text N 3)$, $(4)$ and Topology Defined by Neighborhood System: :$T = \struct {S, \tau}$ is a topological space. {{qed}}