# Niemytzki Plane is Topology

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## Proof

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

 $\displaystyle (1): \ \$ $\displaystyle S$ $=$ $\displaystyle \left\{ {\left({x, y}\right): y \geq 0}\right\}$ $\displaystyle (2): \ \$ $\displaystyle \mathcal B \left({x, y}\right)$ $=$ $\displaystyle \left\{ {B_r \left({x, y}\right) \cap S: r > 0}\right\}$ if $x, y \in \R, y > 0$ $\displaystyle (3): \ \$ $\displaystyle \mathcal B \left({x, 0}\right)$ $=$ $\displaystyle \left\{ {B_r \left({x, r}\right) \cup \left\{ {\left({x, 0}\right)}\right\}: r > 0}\right\}$ if $x \in \R$ $\displaystyle (4): \ \$ $\displaystyle \tau$ $=$ $\displaystyle \left\{ {\bigcup \mathcal G: \mathcal G \subseteq \bigcup_{\left( {x, y}\right) \in S} \mathcal B \left( {x, y}\right)}\right\}$

where $B_r \left({x, y}\right)$ denotes the open $r$-ball of $\left({x, y}\right)$ in the $\R^2$ Euclidean space.

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 $S$
$(N1): \quad \forall \left({x, y}\right) \in S, U \in \mathcal B\left({x, y}\right): \left({x, y}\right) \in U$
$(N2): \quad \forall \left({x, z}\right) \in S, U \in \mathcal B\left({x, z}\right), \left({y, s}\right) \in U:\exists V \in \mathcal B\left({y, s}\right): V \subseteq U$
$(N3): \quad \forall \left({x, y}\right) \in S, U_1, U_2 \in \mathcal B\left({x, y}\right): \exists U \in \mathcal B\left({x, y}\right): U \subseteq U_1 \cap U_2$

Ad $(N0)$:

Let $\left({x, y}\right) \in S$.

In both cases: $y > 0$ and $y = 0$, by $(2)$ and $(3)$:

$\mathcal B \left({x, y}\right)$ is non-empty

Let $U \in \mathcal B \left({x, y}\right)$.

In a case when $y > 0$:

$\exists r > 0: U = B_r\left({x, y}\right) \cap S$
$U$ is a subset of $S$.

In a case when $y = 0$:

$\exists r > 0: U = B_r\left({x, r}\right) \cup \left\{{\left({x, 0}\right)}\right\}$

By definition of subset:

$B_r\left({x, r}\right) \subseteq S$ and $\left\{{\left({x, 0}\right)}\right\} \subseteq S$
$U$ is a subset of $S$.

Ad $(N1)$:

Let $\left({x, y}\right) \in S$.

Let $U \in \mathcal B \left({x, y}\right)$.

In a case when $y > 0$ by $(2)$:

$\exists r > 0: U = B_r\left({x, y}\right) \cap S$
$d\left({\left({x, y}\right), \left({x, y}\right)}\right) = 0$

where $d$ denotes the distance function of $\R^2$ Euclidean space.

By definition of open ball:

$\left({x, y}\right) \in B_r\left({x, y}\right)$

Thus by definition of intersection:

$\left({x, y}\right) \in U$

In a case when $y = 0$ by $(3)$:

$\exists r > 0: U = B_r\left({x, r}\right) \cup \left\{{\left({x, 0}\right)}\right\}$

By definition of singleton:

$\left({x, 0}\right) \in \left\{{\left({x, 0}\right)}\right\}$

Thus by definition of union:

$\left({x, 0}\right) \in U$

Thus in both cases:

$\left({x, 0}\right) \in U$

Ad $(N2)$:

Let $\left({x, 0}\right) \in S, U \in \mathcal B \left({x, y}\right), \left({z, s}\right) \in U$.

In a case when $y > 0$ by $(2)$:

$\exists r > 0: U = B_r\left({x, y}\right) \cap S$

By definition of intersection:

$\left({z, s}\right) \in B_r\left({x, y}\right)$

By definition of open ball:

$d \left({\left({x, y}\right), \left({z, s}\right)}\right) < r$

In a subcase when $s > 0$ define

$r' := r-d \left({\left({x, y}\right), \left({z, s}\right)}\right)$

and

$V := B_{r'}\left({z, s}\right) \cap S$

By $(2)$:

$V \in \mathcal B \left({z, s}\right)$

By proof of Open Ball of Point Inside Open Ball:

$B_{r'} \left({z, s}\right) \subseteq B_r\left({x, y}\right)$
$V \subseteq U$

In a subcase when $s = 0$ define

$r'' := \displaystyle \frac {r-d \left({\left({x, y}\right), \left({z, s}\right)}\right)} 2$

and

$V := B_{r''}\left({z, r''}\right) \cup \left\{{\left({z, 0}\right)}\right\}$

By $(3)$:

$V \in \mathcal B \left({z, s}\right)$
 $\displaystyle d\left({\left({x, y}\right), \left({z, r''}\right)}\right)$ $\leq$ $\displaystyle d\left({\left({x, y}\right), \left({z, 0}\right)}\right)+d\left({\left({z, 0}\right), \left({z, r''}\right)}\right)$ metric space axioms $\displaystyle$ $=$ $\displaystyle d\left({\left({x, y}\right), \left({z, 0}\right)}\right)+r''$ definition of $\R^2$ Euclidean metric $\displaystyle$ $=$ $\displaystyle r-r''$
$B_{r''}\left({z, r''}\right) \subseteq B_r\left({x, y}\right)$

and

$B_{r''}\left({z, r''}\right) \subseteq S$
$B_{r''}\left({z, r''}\right) \subseteq U$

By definition of subset:

$\left\{{\left({z, 0}\right)}\right\} \subseteq U$

Thus by Union is Smallest Superset:

$V \subseteq U$

In a case when $y = 0$ by $(3)$:

$\exists r > 0: U = B_r\left({x, r}\right) \cup \left\{{\left({x, 0}\right)}\right\}$

By definition of onion and singleton:

$\left({z, s}\right) \in B_r\left({x, r}\right)$ or $\left({z, s}\right) = \left({x, 0}\right)$

In a subcase when $\left({z, s}\right) = \left({x, 0}\right)$ defint

$V := U$

By $(3)$:

$V \in \mathcal B \left({z, s}\right)$

Thus by Set is Subset of Itself:

$V \subseteq U$

In a subcase when $\left({z, s}\right) \in B_r\left({x, r}\right)$ define

$r' := r-d \left({\left({x, r}\right), \left({z, s}\right)}\right)$

and

$V := B_{r'}\left({z, s}\right) \cap S$

By definition of open ball:

$d \left({\left({x, r}\right), \left({z, s}\right)}\right) < r$

Then $r' > 0$

By $(2)$:

$V \in \mathcal B \left({z, s}\right)$

By proof of Open Ball of Point Inside Open Ball:

$B_{r'}\left({z, s}\right) \subseteq B_{r'}\left({x, r}\right)$
$V = B_{r'}\left({z, s}\right) \cap S \subseteq B_{r'}\left({z, s}\right)$
$B_{r'}\left({x, r}\right) \subseteq U$

Thus by Subset Relation is Transitive: $V \subseteq U$

Ad $(N3)$:

Let $\left({x, y}\right) \in S, U_1, U_2 \in \mathcal B \left({x, y}\right)$.

In a case when $y > 0$ by $(2)$:

$\exists r_1 > 0: U_1 = B_{r_1}\left({x, y}\right) \cap S$

and

$\exists r_2 > 0: U_2 = B_{r_2}\left({x, y}\right) \cap S$
$r_1 \leq r_2$ or $r_1 \geq r_2$

WLOG: suppose

$r_1 \leq r_2$
 $\displaystyle r_2-r_1$ $\geq$ $\displaystyle 0$ $\displaystyle$ $=$ $\displaystyle d \left({\left({x, y}\right), \left({x, y}\right)}\right)$ metric space axioms

Then by Open Ball is Subset of Open Ball:

$B_{r_1}\left({x, y}\right) \subseteq B_{r_2}\left({x, y}\right)$
$U_1 \subseteq U_2$
$U_1 = U_1 \cap U_2$

Thus by Set is Subset of Itself:

$U_1 \subseteq U_1 \cap U_2$

In a case when $y = 0$ by $(3)$:

$\exists r_1 > 0: U_1 = B_{r_1}\left({x, r_1}\right) \cup \left\{{\left({x, 0}\right)}\right\}$

and

$\exists r_2 > 0: U_2 = B_{r_2}\left({x, r_2}\right) \cup \left\{{\left({x, 0}\right)}\right\}$
$r_1 \leq r_2$ or $r_1 \geq r_2$

WLOG: suppose

$r_1 \leq r_2$
 $\displaystyle r_2-r_1$ $=$ $\displaystyle d \left({\left({x, r_2}\right), \left({x, r_1}\right)}\right)$ definition of $\R^2$ Euclidean metric

Then by Open Ball is Subset of Open Ball:

$B_{r_1}\left({x, r_1}\right) \subseteq B_{r_2}\left({x, r_2}\right)$
$U_1 \subseteq U_2$
$U_1 = U_1 \cap U_2$

Thus by Set is Subset of Itself:

$U_1 \subseteq U_1 \cap U_2$

Thus by $(N0)$-$(N3)$, $(4)$ and Topology Defined by Neighborhood System:

$T = \left({S, \tau}\right)$ is a topological space.

$\blacksquare$