# Niemytzki Plane is Topology

## Proof

By definition $T = \struct {S, \tau}$ is the Niemytzki plane if and only if:

 $\text {(1)}: \quad$ $\displaystyle S$ $=$ $\displaystyle \set {\tuple {x, y}: y \ge 0}$ $\text {(2)}: \quad$ $\displaystyle \map \BB {x, y}$ $=$ $\displaystyle \set {\map {B_r} {x, y} \cap S: r > 0}$ if $x, y \in \R, y > 0$ $\text {(3)}: \quad$ $\displaystyle \map \BB {x, 0}$ $=$ $\displaystyle \set {\map {B_r} {x, r} \cup \set {\tuple {x, 0} }: r > 0}$ if $x \in \R$ $\text {(4)}: \quad$ $\displaystyle \tau$ $=$ $\displaystyle \set {\bigcup \GG: \GG \subseteq \bigcup_{\tuple {x, y} \in S} \map \BB {x, y} }$

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 $y > 0$:

$\exists r > 0: U = \map {B_r} {x, y} \cap S$
$U$ 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 $\set {\tuple {x, 0} } \subseteq S$
$U$ 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 $(2)$:

$\exists r > 0: U = \map {B_r} {x, y} \cap S$
$\map d {\tuple {x, y}, \tuple {x, y} } = 0$

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:

$\map {B_{r'} } {z, s} \subseteq \map {B_r} {x, y}$
$V \subseteq U$

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 $(3)$:

$V \in \map \BB {z, s}$
 $\displaystyle \map d {\tuple {x, y}, \tuple {z, r''} }$ $\le$ $\displaystyle \map d {\tuple {x, y}, \tuple {z, 0} } + \map d {\tuple {z, 0}, \tuple {z, r''} }$ Definition of Metric Space Axioms $\displaystyle$ $=$ $\displaystyle \map d {\tuple {x, y}, \tuple {z, 0} } + r''$ Definition of Real Number Plane with Euclidean Metric $\displaystyle$ $=$ $\displaystyle r - r''$
$\map {B_{r''} } {z, r''} \subseteq \map {B_r} {x, y}$

and

$\map {B_{r''} } {z, r''} \subseteq S$
$\map {B_{r''} } {z, r''} \subseteq U$

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 $(2)$:

$V \in \map \BB {z, s}$
$\map {B_{r'} } {z, s} \subseteq \map {B_{r'} } {x, r}$
$V = \map {B_{r'} } {z, s} \cap S \subseteq \map {B_{r'} } {z, s}$
$\map {B_{r'} } {x, r} \subseteq U$

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

$\exists r_2 > 0: U_2 = \map {B_{r_2} } {x, y} \cap S$
$r_1 \le r_2$ or $r_1 \ge r_2$

Without loss of generality, suppose

$r_1 \le r_2$
 $\displaystyle r_2 - r_1$ $\ge$ $\displaystyle 0$ $\displaystyle$ $=$ $\displaystyle \map d {\tuple {x, y}, \tuple {x, y} }$ metric space axioms

Then by Open Ball is Subset of Open Ball:

$\map {B_{r_1} } {x, y} \subseteq \map {B_{r_2} } {x, y}$
$U_1 \subseteq U_2$
$U_1 = U_1 \cap U_2$

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

$\exists r_2 > 0: U_2 = \map {B_{r_2} } {x, r_2} \cup \set {\tuple {x, 0} }$
$r_1 \le r_2$ or $r_1 \ge r_2$

Without loss of generality, suppose

$r_1 \le r_2$
 $\displaystyle r_2 - r_1$ $=$ $\displaystyle \map d {\tuple {x, r_2}, \tuple {x, r_1} }$ Definition of Real Number Plane with Euclidean Metric

Then by Open Ball is Subset of Open Ball:

$\map {B_{r_1} } {x, r_1} \subseteq \map {B_{r_2} } {x, r_2}$
$U_1 \subseteq U_2$
$U_1 = U_1 \cap U_2$

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.

$\blacksquare$