nyuuzyou/wikis huggingface dataset:
Right Inverse Mapping is Injection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $g: T \to S$ be a right inverse of $f$.
Then $g$ is an injection.
== Proof ==
By the definition of right inverse:
:$f \circ g = I_T$
where $I_T$ is the identity mapping on $T$.
By Identity Mapping is Injection, $I_T$ is an injection.
By Injection if Composite is Injection, it follows that $g$ is an injection.
== Also see ==
* Left Inverse Mapping is Surjection
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
* : Chapter $1$: Theory of Sets: $\S 8$: Composition of Functions and Diagrams: Exercise $3$
Category:Injections
Category:Right Inverse Mappings
Right Inverse for All is Left Inverse

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$ such that:
:$\forall x \in S: \exists x_R: x \circ x_R = e_R$
That is, every element of $S$ has a right inverse with respect to the right identity.
Then $x_R \circ x = e_R$, that is, $x_R$ is also a left inverse with respect to the right identity.
== Proof ==
Let $y = x_R \circ x$. Then:
So $x_R \circ x = e_R$, and $x_R$ behaves as a left inverse as well as a right inverse with respect to the right identity.
== Also see ==
* Left Inverse for All is Right Inverse
* Right Identity while exists Right Inverse for All is Identity
* Left Identity while exists Left Inverse for All is Identity
== Sources ==
* : $\S 1.4$: Lemmas $3$
Category:Semigroups
Category:Inverse Elements
Right Limit Function is Right-Continuous

== Theorem ==
Let $I$ be an open interval.
Let $f : I \to \R$ be a real function such that:
:for each $x \in I$, the right limit:
::$\ds \lim_{y \mathop \to x^+} \map f y$ 
:exists.
Define $f_\leftarrow : I \to \R$ by:
:$\ds \map {f_\leftarrow} x = \lim_{y \mathop \to x^+} \map f y$
for each $x \in I$.
Then $f_\leftarrow$ is right-continuous.
Further, if $f$ is right-continuous at $x$, then $\map {f_\leftarrow} x = \map f x$. 
== Proof ==
Let $\epsilon > 0$. 
Let $x \in I$. 
By the definition of the right limit at $x$, there exists $\delta > 0$ such that for all $y \in I$ satisfying $x < y < x + \delta$, we have:
:$\ds \size {\map f y - \map {f_\leftarrow} x} < \frac \epsilon 3$
Take $y \in I$ such that $x < y < x + \delta$. 
Let $\sequence {x_n}_{n \in \N}$ be a sequence with $x < x_n < \delta$ and $x_n \to x$. 
For example, we may take:
:$\ds x_n = x + \frac \delta {n + 1}$
for each $n \ge 1$. 
Also let $\sequence {y_n}_{n \in \N}$ be a sequence with $y < y_n < x + \delta$ and $y_n \to y$. 
For example, we may take:
:$\ds y_n = y + \frac {\paren {x + \delta} - y} {n + 1}$
for each $n \ge 1$.
By Limit of Function by Convergent Sequences: Corollary, we have:
:$\map f {x_n} \to \map {f_\leftarrow} x$
and:
:$\map f {y_n} \to \map {f_\leftarrow} y$
as $n \to \infty$. 
We can now write:
Taking $n \to \infty$ and using the Squeeze Theorem, we have:
:$\ds \size {\map {f_\leftarrow} x - \map {f_\leftarrow} y} \le \frac {2 \epsilon} 3 < \epsilon$
for all $y \in \openint x {x + \delta}$. 
Hence we have shown that if $x \in I$, then there exists $\delta > 0$ such that whenever $y \in \openint x {x + \delta}$ we have:
:$\size {\map {f_\leftarrow} x - \map {f_\leftarrow} y} < \epsilon$
Finally, if right-continuous at $x$, then:
:$\ds \lim_{y \mathop \to x^+} \map f y = \map f x$
That is:
:$\map {f_\leftarrow} x = \map f x$
by the definition of $f_\leftarrow$.
Category:Right-Continuous Functions
Right Module Does Not Necessarily Induce Left Module over Ring

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct{G, +_G, \circ}$ be a right module over $\struct {R, +_R, \times_R}$.
Let $\circ' : R \times G \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ' x = x \circ \lambda$
Then $\struct{G, +_G, \circ'}$ is not necessarily a left module over $\struct {R, +_R, \times_R}$
== Proof ==
Proof by Counterexample:
Let $\struct {S, +, \times}$ be a ring with unity
Let $\struct {\map {\MM_S} 2, +, \times}$ denote the ring of square matrices of order $2$ over $S$.
From Ring of Square Matrices over Ring is Ring, $\struct {\map {\MM_S} 2, +, \times}$ is a ring.
Let $G = \set {\begin{bmatrix}
x & y \\
0 & 0
\end{bmatrix} : x, y \in S }$
=== Lemma ===
From Right Ideal is Right Module over Ring, $\struct {G, +, \times}$ is a right module over $\struct {\map {\MM_S} 2, +, \times}$.
Let $R = \map {\MM_S} 2$.
Let $\mathbf A = \begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}, 
\mathbf B = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}, 
\mathbf C = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}$
Then:
And:
Hence for $\mathbf A, \mathbf B \in R, \mathbf C \in G$:
:$ \mathbf A \times \paren{\mathbf B \times \mathbf C} \neq \mathbf A \times \paren {\mathbf C \times \mathbf B}$
Let $\circ : G \times R \to R$ be the binary operation defined by:
:$\forall \mathbf \Lambda \in R: \forall \mathbf X \in G: \mathbf \Lambda \circ \mathbf X = \mathbf X \times \mathbf \Lambda$
From Right Module induces Left Module over same Ring iff Actions are Commutative, $\struct {G, +, \circ}$ is not a left module over $\struct {R, +, \times}$
== Also see ==
* Left Module Does Not Necessarily Induce Right Module over Ring
Category:Module Theory
Category:Ring Theory
Category:Right Module Does Not Necessarily Induce Left Module over Ring
Right Module Does Not Necessarily Induce Left Module over Ring/Lemma

== Theorem ==
Let $\struct {S, +, \times}$ be a ring with unity.
Let $\struct {\map {\MM_S} 2, +, \times}$ denote the ring of square matrices of order $2$ over $S$.
Let $G = \set {\begin {bmatrix}
x & y \\
0 & 0
\end {bmatrix} : x, y \in S}$.
Then:
:$G$ is a right ideal of $\struct {\map {\MM_S} 2, +, \times}$.
== Proof ==
From Test for Right Ideal, the following need to be proved:
:$(1): \quad G \ne \O$
:$(2): \quad \forall \mathop {\mathbf X}, \mathop{\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$
:$(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\MM_S} 2: \mathbf J \times \mathbf R \in G$
=== Condition $(1): \quad G \ne \O$ ===
By definition of $G$:
:$\begin {bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix} \in G$
=== Condition $(2): \quad \forall \mathop {\mathbf X}, \mathop{\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$ ===
Let:
:$\mathbf X = \begin{bmatrix}
x_1 & x_2 \\
0 & 0
\end{bmatrix}, \quad \mathbf Y = \begin{bmatrix}
y_1 & y_2 \\
0 & 0
\end{bmatrix} \in G$
Then:
:$\mathbf X - \mathbf Y = \begin {bmatrix}
x_1 - y_1 & x_2 - y_2 \\
0 & 0
\end{bmatrix} \in G$
=== Condition $(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\MM_S} 2: \mathbf J \times \mathbf R \in G$ ===
Let:
:$\mathbf J = \begin{bmatrix}
j_1 & j_2 \\
0 & 0
\end{bmatrix} \in G, \quad \mathbf R = \begin{bmatrix}
r_{1 1} & r_{2 1} \\
r_{1 2} & r_{2 2}
\end{bmatrix} \in \map {\MM_S} 2$
Then:
:$\mathbf J \times \mathbf R = \begin{bmatrix}
j_1 \times r_{1 1} + j_2 \times r_{1 2} & j_1 \times r_{2 1} + j_2 \times r_{2 2} \\
0 & 0
\end{bmatrix} \in G$
Category:Right Module Does Not Necessarily Induce Left Module over Ring
Right Module induces Left Module over same Ring iff Actions are Commutative

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}$ be a right module over $\struct {R, +_R, \times_R}$.
Let $\circ' : R \times G \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ' x = x \circ \lambda $
Then $\struct {G, +_G, \circ'}$ is a left module over $\struct {R, +_R, \times_R}$ :
:$\forall \lambda, \mu \in R: \forall x \in G: x \circ \paren{ \lambda \times_R \mu} = x \circ \paren {\mu \times_R \lambda}$ 
== Proof ==
=== Necessary Condition ===
Let $\struct {G, +_G, \circ'}$ be a left module over $\struct {R, +_R, \times_R}$.
Then:
=== Sufficient Condition ===
Let the scalar multiplication $\circ$ satisfy:
:$\forall \lambda, \mu \in R: \forall x \in G: x \circ \paren {\lambda \times_R \mu} = x \circ \paren {\mu \times_R \lambda}$ 
It needs to be shown that $\struct {G, +_G, \circ'}$ satisfies the left module axioms.
==== ====
Let $\lambda, \mu \in R, x \in G$.
Then:
==== ====
Let $\lambda \in R, x, y \in G$.
Then:
==== ====
Let $\lambda, \mu \in R, x \in G$.
Then:
== Also see ==
* Left Module induces Right Module over same Ring iff Actions are Commutative
Category:Module Theory
Category:Ring Theory
Right Module over Commutative Ring induces Bimodule

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a commutative ring.
Let $\struct {G, +_G, \circ}$ be a right module over $\struct {R, +_R, \times_R}$.
Let $\circ' : R \times G \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ' x = x \circ \lambda $
Then $\struct {G, +_G, \circ', \circ}$ is a bimodule over $\struct {R, +_R, \times_R}$.
== Proof ==
From Right Module over Commutative Ring induces Left Module, $\struct {G, +_G, \circ'}$ is a left module.
Let $\lambda, \mu \in R$ and $x \in G$.
Then:
Hence $\struct{G, +_G, \circ', \circ}$ is a bimodule over $\struct {R, +_R, \times_R}$ by definition.
== Also see ==
* Left Module over Commutative Ring induces Bimodule
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Module Theory
Category:Ring Theory
Right Module over Commutative Ring induces Left Module

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a commutative ring.
Let $\struct{G, +_G, \circ}$ be a right module over $\struct {R, +_R, \times_R}$.
Let $\circ' : R \times G \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ’ x = x \circ \lambda$
Then $\struct{G, +_G, \circ'}$ is a left module over $\struct {R, +_R, \times_R}$.
== Proof ==
From Ring is Commutative iff Opposite Ring is Itself, $\struct {R, +_R, \times_R}$ is its own opposite ring.
From Right Module over Ring Induces Left Module over Opposite Ring, $\struct{G, +_G, \circ'}$ is a left module over $\struct {R, +_R, \times_R}$.
== Also see ==
* Left Module over Commutative Ring induces Right Module
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Module Theory
Category:Ring Theory
Right Module over Ring Induces Left Module over Opposite Ring

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {R, +_R, *_R}$ be the opposite ring of $\struct {R, +_R, \times_R}$.
Let $\struct{G, +_G, \circ}$ be a right module over $\struct {R, +_R, \times_R}$.
Let $\circ' : R \times G \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ' x = x \circ \lambda$
Then $\struct {G, +_G, \circ'}$ is a left module over $\struct {R, +_R, *_R}$.
== Proof ==
It is shown that $\struct {G, +_G, \circ'}$ satisfies the left module axioms.
By definition of the opposite ring:
:$\forall x, y \in S: x *_R y = y \times_R x$.
=== === 
Let $\lambda \in R$ and $x, y \in G$.
=== === 
Let $\lambda, \mu \in R$ and $x \in G$.
=== ===
Let $\lambda, \mu \in R$ and $x \in G$.
== Also see ==
* Left Module over Ring Induces Right Module over Opposite Ring
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Module Theory
Category:Ring Theory
Right Operation has no Right Identities

== Theorem ==
Let $S$ be a set with more than $1$ element.
Let $\struct {S, \to}$ be an algebraic structure in which the operation $\to$ is the right operation.
Then $\struct {S, \to}$ has no right identities.
== Proof ==
From Element under Right Operation is Left Identity, every element of $\struct {S, \to}$ is a left identity.
Because there are at least $2$ elements in $\struct {S, \to}$, it follows that $\struct {S, \to}$ has more than one left identity.
From More than one Left Identity then no Right Identity, it follows that $\struct {S, \to}$ has no right identity.
== Also see ==
* Left Operation has no Left Identities
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.3 \ \text{(b)}$
Category:Identity Elements
Category:Right Operation
Right Operation is Anticommutative

== Theorem ==
The right operation is anticommutative:
:$\forall x, y: x \rightarrow y = y \rightarrow x \iff x = y$
== Proof ==
Immediate from the definition of the right operation.
== Also see ==
* Left Operation is Anticommutative
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.17 \ \text{(a)}$
Category:Right Operation
Category:Examples of Anticommutativity
Right Operation is Associative

== Theorem ==
The right operation is associative:
:$\forall x, y, z: \paren {x \to y} \to z = x \to \paren {y \to z}$
== Proof ==
== Also see ==
* Left Operation is Associative
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.4$
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.9$
Category:Right Operation
Category:Examples of Associative Operations
Right Operation is Closed for All Subsets

== Theorem ==
Let $S$ be a set.
Let $\rightarrow$ be the right operation on $S$.
That is:
:$\forall x, y \in S: x \rightarrow y = y$
Let $\powerset S$ be the power set of $S$.
Then for all $T \in \powerset S$, $\rightarrow$ is closed on $T$.
Thus, for all $T \in \powerset S$:
:$\struct {T, \rightarrow}$ is a subsemigroup of $\struct {S, \rightarrow}$.
== Proof ==
From Structure under Right Operation is Semigroup we have that $\struct {S, \rightarrow}$ is a semigroup, whatever the nature of $S$.
Let $T \in \powerset S$.
Then:
:From Structure under Right Operation is Semigroup, $\struct {T, \rightarrow}$ is a semigroup, and therefore a subsemigroup of $\struct {S, \rightarrow}$.
This applies whatever $S$ is and whatever the subset $T$ is.
== Also see ==
* Left Operation is Closed for All Subsets
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.1$
Category:Right Operation
Right Operation is Distributive over Idempotent Operation

== Theorem ==
Let $\struct {S, \circ, \rightarrow}$ be an algebraic structure where:
:$\rightarrow$ is the right operation
:$\circ$ is any arbitrary binary operation.
Then:
:$\rightarrow$ is distributive over $\circ$
:$\circ$ is idempotent.
== Proof ==
From Right Operation is Left Distributive over All Operations:
:$\forall a, b, c \in S: a \rightarrow \paren {b \circ c} = \paren {a \rightarrow b} \circ \paren {a \rightarrow c}$
for all binary operations $\circ$.
It remains to show that $\rightarrow$ is right distributive over $\circ$ $\circ$ is idempotent.
=== Necessary Condition ===
Let $\circ$ be idempotent.
Then:
Thus $\rightarrow$ is right distributive over $\circ$.
=== Sufficient Condition ===
Let $\rightarrow$ be right distributive over $\circ$.
Let $c \in S$ be arbitrary.
Then:
Hence $\circ$ is idempotent.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.23 \ \text{(b)}$
Category:Right Operation
Category:Distributive Operations
Category:Idempotence
Right Operation is Entropic

== Theorem ==
The right operation is entropic:
:$\forall a, b, c, d: \paren {a \to b} \to \paren {c \to d} = \paren {a \to c} \to \paren {b \to d}$
== Proof ==
== Also see ==
* Left Operation is Entropic
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.12 \ \text{(c)}$
Category:Right Operation
Category:Examples of Entropic Operations
Right Operation is Idempotent

== Theorem ==
The right operation is idempotent:
:$\forall x: x \rightarrow x = x$
== Proof ==
Immediate from the definition of the right operation.
== Also see ==
* Left Operation is Idempotent
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.17 \ \text{(a)}$
Category:Right Operation
Category:Examples of Idempotence
Right Operation is Left Distributive over All Operations

== Theorem ==
Let $\struct {S, \circ, \rightarrow}$ be an algebraic structure where:
:$\rightarrow$ is the right operation
:$\circ$ is any arbitrary binary operation.
Then $\rightarrow$ is left distributive over $\circ$.
== Proof ==
By definition of the right operation:
The result follows by definition of left distributivity.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.23 \ \text{(a)}$
Category:Right Operation
Right Operation is not Commutative

== Theorem ==
Let $S$ be a finite set.
Let $\rightarrow$ denote the right operation on $S$.
Then $\rightarrow$ is not commutative on $S$ unless $S$ is a singleton.
== Proof ==
Let $S$ be a singleton, $S = \set s$, say.
Then:
:$s \rightarrow s = s$
and so $\rightarrow$ is trivially commutative on $S$
Otherwise, $\exists s, t \in S$ such that $s \ne t$.
Then:
:$s \rightarrow t = t$
but:
:$t \rightarrow s = s$
and the result follows by definition of commutative operation.
== Also see ==
* Right Operation is Anticommutative
* Left Operation is not Commutative
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.4$
Category:Right Operation
Category:Examples of Commutative Operations
Right Order Topology on Real Numbers is Topology

== Theorem ==
Let $\tau$ be the right order topology on $\R$.
Then $\tau$ forms a topology on $\R$.
That is:
:$T = \struct {\R, \tau}$ is a topological space.
== Proof ==
Write $\O = \openint \infty \infty$ and $\R = \openint {-\infty} \infty$.
Then $\tau$ can be written as $\set {\openint j \infty: j \in \overline \R}$.
First we note that:
:$m \le n \implies \openint n \infty \subseteq \openint m \infty$
By definition we have that:
:$\O \in \tau$
Then each of the open set axioms is examined in turn:
=== ===
Let $\family {\openint j \infty}_{j \mathop \in S}$ be an indexed family of open sets of $T$, where $S \subseteq \overline \R$.
Let $\ds V = \bigcup_{j \mathop \in S} \openint j \infty$ be the union of $\family {\openint j \infty}_{j \mathop \in S}$.
Let $m := \inf S \in \overline \R$.
We claim that:
:$\ds \openint m \infty = \bigcup_{j \mathop \in S} \openint j \infty$
Indeed, for each $x \in \openint m \infty$ we have $x > m$.
By Characterizing Property of Infimum of Subset of Real Numbers:
:$\exists y \in S: y < x$
and thus $\ds x \in \openint y \infty \subseteq \bigcup_{j \mathop \in S} \openint j \infty$.
This shows that $\ds \openint m \infty \subseteq \bigcup_{j \mathop \in S} \openint j \infty$.
For the other direction, note that $j \ge m$ for each $j \in S$.
Therefore $\openint j \infty \subseteq \openint m \infty$ for each $j \in S$.
By Union of Subsets is Subset, $\ds \bigcup_{j \mathop \in S} \openint j \infty \subseteq \openint m \infty$.
Thus our claim is true by definition of set equality.
Hence $V = \openint m \infty$ is open by definition.
=== ===
Let $A = \openint m \infty$ and $B = \openint n \infty$, where $m, n \in \overline \R$.
 let $m < n$.
Then $\openint n \infty \subseteq \openint m \infty$.
Hence by Intersection with Subset is Subset:
:$\openint n \infty \cap \openint m \infty = \openint n \infty \in \tau$
Hence $\openint n \infty \cap \openint m \infty$ is open by definition.
Therefore the intersection of any two elements of $\tau$ is an element of $\tau$.
=== ===
$\R \in \tau$ follows directly from definition.
All the open set axioms are fulfilled, and the result follows.
Category:Right Order Topology
Right Order Topology on Strictly Positive Integers as Neighborhood Space

== 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 $n$ 
:$\forall m \in \Z: m \ge n \implies m \in U$
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 $\text N 1$.
=== 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 $\text N 2$.
=== 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 $\text N 3$.
=== 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 $\text N 4$.
=== 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 $\text N 5$.
It follows from Correspondence between Neighborhood Space and Topological Space that $\NN$ is the right order space on $\Z_{>0}$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces
Category:Neighborhood Spaces
Category:Right Order Topology
Right Order Topology on Strictly Positive Integers is Topology

== Theorem ==
Let $\Z_{>0}$ be the set of strictly positive integers.
Let $\tau$ be the right order topology on $\Z_{>0}$.
Then $\tau$ forms a topology on $\Z_{>0}$.
That is:
:$T = \struct {\Z_{>0}, \tau}$ is a topological space.
== Proof ==
Let $S := \Z_{>0}$ to ease notational clutter.
First we note that:
:$m \le n \implies O_n \subseteq O_m$
where $O_n := \set {x \in \Z_{>0}: x \ge n}$.
By definition we have that:
:$\O \in \tau$
Then each of the open set axioms is examined in turn:
=== ===
Let $\family {O_j}_{j \mathop \in S}$ be an indexed family of open sets of $T$.
Let $\ds V = \bigcup_{j \mathop \in S} O_j$ be the union of $\family {O_j}_{j \mathop \in S}$.
Let $m \in S$ be defined as:
:$\ds m = \inf \set {n \in S: O_n \in \bigcup_{j \mathop \in S} O_j}$
Then $O_m = \bigcup_{j \mathop \in S} O_j$
Hence $V = O_m$ is open by definition.
=== ===
Let $O_m$ and $O_n$ be open sets of $T$.
 let $m < n$.
Then $O_n \subseteq O_m$.
Hence by Intersection with Subset is Subset
:$O_n \cap O_m = O_n$
Hence $O_n \cap O_m$ is open by definition.
=== ===
Then we note that:
:$O_1 = S$
Hence $S \in \tau$ as required.
All the open set axioms are fulfilled, and the result follows.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Example $7$
Category:Right Order Topology
Right Order Topology on Strictly Positive Integers is not Metrizable

== Theorem ==
Let $\Z_{>0}$ be the set of strictly positive integers.
Let $T = \struct {\Z_{>0}, \tau}$ denote the right order space on $\Z_{>0}$.
Then $T = \struct {\Z_{>0}, \tau}$ is not a metrizable space.
== Proof ==
Let $m, n \in \Z_{>0}$ such that $m < n$.
Let $O_m$ and $O_n$ be arbitrary non-empty open sets of $T$.
Then:
:$O_m \cap O_n = O_m$
As $O_m$ and $O_n$ are arbitrary, it follows that there exist no $O_m$ and $O_n$ in $\tau$ such that $O_m \cap O_n = \O$.
Hence $T$ is not Hausdorff.
The result follows from Metrizable Space is Hausdorff.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $1$
Category:Right Order Topology