Right Order Topology on Strictly Positive Integers as Neighborhood Space

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Theorem

Let $\Z_{>0}$ denote the set of (strictly) positive integers.

Let $n \in \Z_{>0}$.

Let $U \subseteq \Z_{>0}$ be defined as being a neighborhood of $n$ if and only if

$\forall m \in \Z: m \ge n \implies m \in U$


Then the set $\NN$ of all $U$ for all $n \in \Z_{>0}$ forms a neighborhood space which is the same as the right order space on $\Z_{>0}$.


Proof

First it is noted that a neighborhood of $n$ is exactly an element of the right order topology on $\Z_{>0}$.

It remains to be shown that $\NN$ actually forms a neighborhood space.


Let $\NN_n$ denote the set of all neighborhood of a given $n \in \Z_{>0}$.

Checking the neighborhood space axioms in turn:


Neighborhood Space Axiom $\text N 1$

We have that $n \ge n$.

Hence the set defined as:

$U_n = \set {m \in \Z: m \ge n}$

is an element of $\NN_n$.

Thus $\NN_n$ is not empty.

$n \in U_n$.

Hence $\NN$ satisfies neighborhood space axiom $\text N 1$.

$\Box$


Neighborhood Space Axiom $\text N 2$

Let $U \in \NN_n$.

Then $U$ is of the form:

$\forall m \in \Z: m \ge n \implies m \in U$

As $n \ge n$, it follows that $n \in U$.

Hence $\NN$ satisfies neighborhood space axiom $\text N 2$.

$\Box$


Neighborhood Space Axiom $\text N 3$

Let $U \in \NN_n$.

Then:

$\forall m \in \Z: m \ge n \implies m \in U$

Let $V \subseteq \Z_{>0}$ such that $V \supseteq U$.

By definition of superset:

$x \in U \implies x \in V$

Hence:

$\forall m \in \Z: m \ge n \implies m \in V$

and it is seen that $V \in \NN_n$.

Hence $\NN$ satisfies neighborhood space axiom $\text N 3$.

$\Box$


Neighborhood Space Axiom $\text N 4$

Let $U, V \in \NN_n$.

Then by definition:

$\forall m \in \Z: m \ge n \implies m \in U$
$\forall m \in \Z: m \ge n \implies m \in V$

and so by definition of set intersection:

$\forall m \in \Z: m \ge n \implies m \in U \cap V$

Hence $\NN$ satisfies neighborhood space axiom $\text N 4$.

$\Box$


Neighborhood Space Axiom $\text N 5$

Let $U \in \NN_n$.

Then by definition:

$\forall m \in \Z: m \ge n \implies m \in U$

Let $U_n$ be the set defined as:

$U_n = \set {m \in \Z: m \ge n}$

It follows that $U_n \subseteq U$.

Let $y \in U_n$.

Then:

$y \ge n$

Let $\NN_y$ denote the set of all neighborhoods of $y$.

We have that:

$\forall m \in \Z: m \ge y \implies m \ge n$

and so:

$U_n \in \NN_y$.

Hence $\NN$ satisfies neighborhood space axiom $\text N 5$.

$\Box$


It follows from Correspondence between Neighborhood Space and Topological Space that $\NN$ is the right order space on $\Z_{>0}$.

$\blacksquare$


Sources

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