Niemytzki Plane is Topology

== Theorem ==
Niemytzki plane is a topological space.
== Proof ==
By definition $T = \struct {S, \tau}$ is the Niemytzki plane :
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 the metric space 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 $(3)$:
:$V \in \map \BB {z, s}$
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 $r_1 \ge r_2$
, suppose 
:$r_1 \le r_2$
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 $r_1 \ge r_2$
, suppose 
:$r_1 \le r_2$
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.
== Sources ==
* 
* 
Category:Niemytzki Plane