nyuuzyou/wikis huggingface dataset:
Null Function/Examples/Example 2

== Example of Null Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \begin {cases} 1 & : t = 1 \\ 0 & : \text {otherwise} \end {cases}$
Then $f$ is a null function.
== Proof ==
Let $x > 1$.
The cases where $x < 1$ is treated similarly.
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Impulse Functions. The Dirac Delta Function: $44 \ \text{(a)}$
Category:Examples of Null Functions
Null Function/Examples/Example 3

== Example of Null Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \begin {cases} 1 & : t = 1 \le t \le 2 \\ 0 & : \text {otherwise} \end {cases}$
Then $f$ is not a null function.
== Proof ==
Let $x > 2$.
Hence the result by definition of null function.
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Impulse Functions. The Dirac Delta Function: $44 \ \text{(b)}$
Category:Examples of Null Functions
Null Measure is Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Then the null measure $\mu$ on $\struct {X, \Sigma}$ is a measure.
== Proof ==
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$.
=== Proof of $(1)$ ===
Let $S \in \Sigma$.
Then $\map \mu S = 0 \ge 0$.
=== Proof of $(2)$ ===
It is to be shown that (for a sequence $\sequence {S_n}_{n \in \N}$ of pairwise disjoint sets):
:$\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$
Now by definition of $\mu$:
:$\ds \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = 0$
Thus, the desired equation becomes:
:$\ds \sum_{n \mathop = 1}^\infty 0 = 0$
which trivially holds.
=== Proof of $(3')$ ===
Note that $\O \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra.
Hence $\map \mu \O = 0$.
The axioms are fulfilled, and it follows that $\mu$ is a measure.
Category:Null Measure
Null Module Submodule of All

== Theorem ==
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Then the null module:
:$\struct {\set {e_G}, +_G, \circ}_R$
is a submodule of $\struct {G, +_G, \circ}_R$.
== Proof ==
From Trivial Subgroup is Subgroup, the trivial subgroup is a subgroup of the group $\struct {G, +_G}$.
The result follows from the Submodule Test.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.1$
Category:Submodules
Category:Module Theory
Null Module is Module

== Theorem ==
Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $G$ be the trivial group.
Let $\struct {G, +_G, \circ}_R$ be the null module.
Then $\struct {G, +_G, \circ}_R$ is a module.
== Proof ==
Follows from the fact that $\struct {G, +_G, \circ}_R$ has to be, by definition, a trivial module:
$\circ$ can only be defined as:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$
Category:Module Theory
Null Polynomial is Additive Identity

== Theorem ==
The set of polynomial forms has an additive identity.
== Proof ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity with zero $0_R$.
Let $\set {X_j: j \in J}$ be a set of indeterminates.
Let $Z$ be the set of all multiindices indexed by $\set {X_j: j \in J}$.
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
be an arbitrary polynomial form in the indeterminates $\set {X_j: j \in J}$ over $R$.
Let:
:$\ds N = \sum_{k \mathop \in Z} 0_R \mathbf X^k$ 
be the null polynomial.
Then:
 
 
Therefore, $N + f = f$ for all polynomial forms $f$.
Therefore, $N$ is an additive identity for the set of polynomial forms.
Category:Polynomial Theory
Null Relation is Antireflexive, Symmetric and Transitive

== Theorem ==
Let $S$ be a set which is non-empty.
Let $\RR \subseteq S \times S$ be the null relation.
Then $\RR$ is antireflexive, symmetric and transitive.
If $S = \O$ then Relation on Empty Set is Equivalence applies.
== Proof ==
From the definition of null relation:
:$\RR = \O$
=== Antireflexivity ===
This follows directly from the definition:
:$\RR = \O \implies \forall x \in S: \tuple {x, x} \notin \RR$
and so $\RR$ is antireflexive.
=== Symmetry ===
It follows vacuously that:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR$
and so $\RR$ is symmetric.
=== Transitivity ===
It follows vacuously that:
:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$
and so $\RR$ is transitive.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $24$
* : Appendix $\text{A}.2$: Cartesian Products and Relations: Problem Set $\text{A}.2$: $11$
Category:Examples of Antireflexive Relations
Category:Examples of Symmetric Relations
Category:Examples of Transitive Relations
Category:Null Relation
Null Relation is Mapping iff Domain is Empty Set

== Theorem ==
Let $S$ and $T$ be sets.
The null relation $\RR = \O \subseteq S \times T$ is a mapping $S = \O$.
== Proof ==
=== Sufficient Condition ===
Let $S = \O$.
Then the null relation $\RR = \O \subseteq S \times T$ is a mapping from Empty Mapping is Mapping.
=== Necessary Condition ===
Suppose $S \ne \O$.
From the definition of an empty set, $S \ne \O \implies \exists x \in S$.
Thus:
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Functions
Category:Null Relation
Category:Empty Mapping
Category:Empty Set
Null Ring iff Zero and Unity Coincide

== Theorem ==
The null ring is the only ring in which the unity and zero coincide.
== Proof ==
The single element of the null ring serves as an identity for both of the operations.
So, in this particular ring, the unity and the zero are the same element.
A non-null ring contains some non-zero element $a$.
Since $1_R \circ a = a \ne 0_R = 0_R \circ a$, then $1_R \ne 0_R$.
So if a ring is non-null, its unity cannot be zero.
So a ring is null its unity is also its zero.
== Also see ==
* Group with Zero Element is Trivial
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains
* : Chapter $1$: Rings - Definitions and Examples: Exercise $6$
Category:Null Ring
Null Ring is Commutative Ring

== Theorem ==
Let $R$ be the null ring.
That is, let:
:$R := \struct {\set {0_R}, +, \circ}$
where ring addition and ring product are defined as:
Then $R$ is a commutative ring.
== Proof ==
From Null Ring is Trivial Ring, we have that $R$ is a trivial ring.
The result follows from Trivial Ring is Commutative Ring.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.4$
Category:Null Ring
Category:Commutative Rings
Null Ring is Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Then the null ring $\struct {\set {0_R}, +, \circ}$ is an ideal of $\struct {R, +, \circ}$.
== Proof ==
From Null Ring is Subring of Ring, $\struct {\set {0_R}, +, \circ}$ is a subring of $\struct {R, +, \circ}$.
Also, from Ring Product with Zero:
:$\forall x \in R: x \circ 0_R = 0_R = 0_R \circ x \in \set {0_R}$
thus fulfilling the condition for $\struct {\set {0_R}, +, \circ}$ to be an ideal of $\struct {R, +, \circ}$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
* : Chapter $5$: Rings: $\S 22$. Quotient Rings: Example $40$
* : $\S 2.2$: Homomorphisms: Definition $2.5$
* : $\S 58.1$ Ideals
Category:Ideal Theory
Null Ring is Ring

== Theorem ==
Let $R$ be the null ring.
That is, let:
:$R := \struct {\set {0_R}, +, \circ}$
where ring addition and ring product are defined as:
Then $R$ is a ring.
== Proof ==
A null ring is a trivial ring.
So, by Trivial Ring is Commutative Ring, the result follows.
Category:Null Ring
Null Ring is Ring with Unity

== Theorem ==
Let $R$ be the null ring.
Then $R$ is a ring with unity.
== Proof ==
We have that $R$ is the null ring.
That is, by definition it has a single element, which can be denoted $0_R$, such that:
:$R := \struct {\set {0_R}, +, \circ}$
where ring addition and the ring product are defined as:
Hence we have that the algebraic structure $\struct {\set {0_R}, \circ}$ is a trivial group.
Thus we see that:
:$\forall a, b \in R: a \circ b = 0_R$
Thus by definition, $0_R$ is a unity.
Category:Rings with Unity
Category:Null Ring
Null Ring is Subring of Ring

== Theorem ==
Let $R$ be a ring.
Then the null ring is a subring of $R$.
== Proof ==
From Null Ring is Ring, the null ring $\struct {\set {0_R}, +, \circ}$ is a ring.
We also have that $\set {0_R}$ is a subset of $R$
Hence the result by definition of subring.
== Sources ==
* : Chapter $5$: Rings: $\S 19$. Subrings: Example $30$
* : $\S 56.1$ Subrings and subfields
Category:Subrings
Null Ring is Trivial Ring

== Theorem ==
Let $R$ be the null ring.
Then $R$ is a trivial ring.
== Proof ==
We have that $R$ is the null ring.
That is, by definition it has a single element, which can be denoted $0_R$, such that:
:$R := \struct {\set {0_R}, +, \circ}$
where ring addition and the ring product are defined as:
Consider the operation $+$.
By definition, the algebraic structure $\struct {\set {0_R}, +}$ is a trivial group.
Then:
:$\forall a, b \in R: a \circ b = 0_R$
Thus by definition, $R$ is a trivial ring.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.5$
Category:Trivial Rings
Category:Null Ring
Null Sequence Test for Convergence

== Theorem ==
Let $\struct {R, \norm{\,\cdot\,}}$ be a normed division ring.
Let $\sequence{x_n}$ be a convergent sequence in $\struct {R, \norm{\,\cdot\,}}$ with limit $l$.
Let $\sequence{y_n}$ be a sequence.
Then:
:$\sequence{y_n}$ converges to the limit $l$ the sequence $\sequence{y_n - x_n}$ is a null sequence
== Proof ==
=== Necessary Condition ===
Let $\sequence{y_n}$ converge to the limit $l$.
From Difference Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} y_n - x_n = l - l = 0$
Hence $\sequence{y_n - x_n}$ is a null sequence by definition.
=== Sufficient Condition ===
Let $\sequence{y_n - x_n}$ be a null sequence.
By definition of a null sequence:
:$\ds \lim_{n \mathop \to \infty} y_n - x_n = 0$
From Sum Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} x_n + \paren{y_n - x_n} = l + 0 = l$
For all $n \in \N$:
:$x_n + \paren{y_n - x_n} = y_n$
Hence:
:$\ds \lim_{n \mathop \to \infty} y_n = l$
It follows that $\sequence{y_n}$ converges to the limit $l$ by definition.
Category:Convergence
Category:Normed Division Rings
Null Sequence in Exponential Sequence

== Theorem ==
Let $\sequence {a_n}_{n \mathop \in \N} \in \C$ be a sequence of complex numbers such that:
:$\ds \lim_{n \mathop \to +\infty}a_n = 0$
Then:
:$\ds \lim_{n \mathop \to +\infty} \paren {1 + \dfrac {a_n} n}^n = 1$
== Proof 1 ==
== Proof 2 ==
Category:Exponential Function
Category:Sequences
Category:Null Sequence in Exponential Sequence
Null Sequence in Exponential Sequence/Proof 2

== Theorem ==
== Proof ==
Let $\sequence {E_n}$ be the sequence of complex functions $E_n: \C \to \C$ defined by:
:$\map {E_n} z = \paren {1 + \dfrac z n}^n$
We have that:
:$\ds \lim_{n \mathop \to \infty} \map {E_n} z = \map \exp z$
where $\map \exp z$ is the complex exponential.
We also have that:
:$E_n \paren {a_n} = \paren {1 + \dfrac {a_n} n}^n$
By Convergent Sequence in Metric Space is Bounded, we have that $\sequence {a_n}$ is Bounded Complex Sequence.
Let this bound be $M$.
Let $K \subseteq \C$ be the closed disk of radius $M$.
By Closed Disk is Compact, $K$ is compact.
By Exponential Sequence is Uniformly Convergent on Compact Sets, $\sequence {E_n}$ is uniformly convergent on $K$.
Now the hypotheses of Uniformly Convergent Sequence Evaluated on Convergent Sequence are satisfied, so:
:$\ds \lim_{n \mathop \to \infty} \map {E_n} {a_n} = \map \exp 0 = 1$
Hence the result.
Category:Null Sequence in Exponential Sequence
Null Sequence in Metrizable Topological Vector Space Dominates some Sequence of Scalars Tending to Infinity

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a metrizable topological vector space over $\GF$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence with $x_n \to {\mathbf 0}_X$. 
Then there exists a sequence of positive real numbers $\sequence {\gamma_n}_{n \mathop \in \N}$ such that:
:$\gamma_n \to \infty$
and:
:$\gamma_n x_n \to {\mathbf 0}_X$
== Proof ==
From Birkhoff-Kakutani Theorem: Topological Vector Space, there exists an invariant metric $d$ on $X$ that induces $\tau$.
Then:
:$\map d {x_n, { {\mathbf 0}_X} } \to 0$
Pick $n_1 \in \N$ such that:
:$\map d {x_n, { {\mathbf 0}_X} } < 1$ for $n > n_1$.
Inductively, for $k \ge 2$, pick $n_k > n_{k - 1}$ such that:
:$\map d {x_n, { {\mathbf 0}_X} } < k^{-2}$ for $n > n_k$.
For $n < n_2$, set $\gamma_n = 1$. 
For $k \ge 2$ and $n_k \le n < n_{k + 1}$, set $\gamma_n = k$.
For each $n$, let $\map k n$ be the unique $k \in \N$ such that $n_k \le n < n_{k + 1}$. 
Since $n_k < n_{k + 1}$, we have $\map k n \to \infty$ as $n \to \infty$.
Now, for $n \ge n_2$:
Taking $n \to \infty$, we obtain:
:$\map d {\gamma_n x_n, { {\mathbf 0}_X} } \to 0$
and hence:
:$\gamma_n x_n \to {\mathbf 0}_X$
== Sources ==
* : $1.28$: Theorem
Category:Topological Vector Spaces
Null Sequence induces Local Basis in Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\sequence {x_n}$ be a real null sequence such that:
:$\forall n \in N: x_n > 0$
Let $\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$ in $M$.
Then:
:$\BB_{\sequence {x_n} } = \set{\map {B_{x_n} } a : n \in \N}$ is a local basis at $a$.
=== Corollary ===
== Proof ==
By Open Ball of Metric Space is Open Set, every element of $\BB_{\sequence {x_n} }$ is an open neighborhood of $a$.
Let $U$ be an open neighborhood of $a$.
By definition of an open set, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} a \subseteq U$.
By definition of a real null sequence:
:$\exists N \in \N : \forall n > N : \size {x_n} < \epsilon$
Let $m = N + 1$.
Then $\size {x_m} < \epsilon$.
Since $x_m > 0$ then $x_m = \size {x_m} < \epsilon$, and so $\map {B_{x_m} } a \subseteq \map {B_\epsilon} a \subseteq U$.
The result follows from Subset Relation is Transitive.
Category:Metric Spaces
Null Sequence induces Local Basis in Metric Space/Sequence of Reciprocals

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$ in $M$.
Then:
:$\BB = \set {\map {B_{1/n}} a : n \in \N}$ is a local basis at $a$.
== Proof ==
Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
:$x_n = \dfrac 1 n$
From Sequence of Reciprocals is Null Sequence, $\sequence {x_n}$ is a real null sequence.
From Null Sequence induces Local Basis in Metric Space:
:$\BB = \set {\map {B_{1/n} } a : n \in \N}$ is a local basis at $a$.
Category:Metric Spaces
Null Sequence induces Neighborhood Basis of Closed Sets in Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\sequence {x_n}$ be a real null sequence such that:
:$\forall n \in N: x_n > 0$
Let $\map {B^-_\epsilon} a$ denote the closed $\epsilon$-ball of $a$ in $M$.
Then:
:$\NN_{\sequence {x_n} } = \set{\map {B^-_{x_n} } a : n \in \N}$ is a neighborhood basis of $a$ consisting of closed sets.
== Proof ==
From Closed Ball in Metric Space is Closed Neighborhood:
:every element of $\NN_{\sequence {x_n} }$ is an closed neighborhood of $a$.
By definition of closed neighborhood of $a$:
:every element of $\NN_{\sequence {x_n} }$ is an neighborhood of $a$ and a closed set.
Let $U$ be an open neighborhood of $a$.
By definition of an open set, there exists a strictly positive real number $\epsilon$:
:$\map {B_\epsilon} a \subseteq U$
From Open Ball contains Strictly Smaller Closed Ball:
:$\map {B_{\epsilon/2}^-} a \subseteq \map {B_\epsilon} a$
By definition of a real null sequence:
:$\exists N \in \N : \forall n > N : \size {x_n} < \dfrac \epsilon 2$
Let $m = N + 1$.
Then $\size {x_m} < \dfrac \epsilon 2$.
Since $x_m > 0$ then $x_m = \size {x_m} < \dfrac \epsilon 2 $, and so:
:$\map {B_{x_m}^- } a \subseteq \map {B_{\epsilon/2}^-} a \subseteq \map {B_\epsilon} a \subseteq U$
The result follows from Subset Relation is Transitive.
Category:Metric Spaces
Null Sequences form Maximal Left and Right Ideal

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$.
Let $\NN$ be the set of null sequences.
That is:
:$\ds \NN = \set {\sequence {x_n}: \lim_{n \mathop \to \infty} x_n = 0 }$
Then $\NN$ is a ring ideal of $\CC$ that is a maximal left ideal and a maximal right ideal.
 
=== Corollary ===
Let $\struct {R, \norm {\, \cdot \,} }$ be a valued field.
== Proof ==
By every convergent sequence is a Cauchy sequence then $\NN \subseteq \CC$.
The proof is completed in these steps:
:$(1): \quad \NN$ is an ideal of $\CC$.
:$(2): \quad \NN$ is a maximal left ideal.
:$(3): \quad \NN$ is a maximal right ideal.
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
== Sources ==
* : $\S 3.2$: Completions: Lemma $3.2.8$, Problem $80$
* : $\S 1.3$ Construction of the completion of a normed field
Category:Cauchy Sequences
Category:Normed Division Rings
Category:Null Sequences form Maximal Left and Right Ideal
Null Sequences form Maximal Left and Right Ideal/Corollary 1

== Theorem ==
Let $\struct {F, \norm {\, \cdot \,} }$ be a valued field.
Let $\CC$ be the ring of Cauchy sequences over $F$.
Let $\NN$ be the set of null sequences in $F$.
That is:
:$\NN = \set {\sequence {x_n}: \ds \lim_{n \mathop \to \infty} x_n = 0}$
Then $\NN$ is a maximal ring ideal of $\CC$.
== Proof ==
By Null Sequences form Maximal Left and Right Ideal then $\NN$ is a maximal left ideal of $\CC$. 
A field is by definition a commutative ring.
In a commutative ring, a maximal left ideal is by definition a maximal ideal.
== Sources ==
* : $\S 3.2$: Completions
* : $\S 1.3$ Construction of the completion of a normed field
Category:Null Sequences form Maximal Left and Right Ideal
Null Sequences form Maximal Left and Right Ideal/Lemma 1

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Then:
:$\NN$ is an ideal of $\CC$.
 
== Proof ==
The Test for Ideal is applied to prove the result.
=== Lemma 4 ===
=== Lemma 5 ===
=== Lemma 6 ===
By Test for Ideal then the result follows.
== Sources ==
* : $\S 3.2$: Completions
* : $\S 1.2$: Normed Fields: Exercise $11 \ (3)$
Category:Null Sequences form Maximal Left and Right Ideal
Null Sequences form Maximal Left and Right Ideal/Lemma 2

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Then:
:$\NN$ is a maximal left ideal.
== Proof ==
By Lemma 1 of Null Sequences form Maximal Left and Right Ideal:
:$\NN$ is an ideal of $\CC$.
Hence $\NN$ is a left ideal of $\CC$.
It remains to show that $\NN$ is maximal.
=== Lemma 7 ===
=== Lemma 8 ===
The result follows by definition of maximal left ideal.
== Sources ==
* : $\S 3.2$: Completions
Category:Null Sequences form Maximal Left and Right Ideal
Null Sequences form Maximal Left and Right Ideal/Lemma 3

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Then:
:$\NN$ is a maximal right ideal.
 
== Proof ==
By Lemma 1 of Null Sequences form Maximal Left and Right Ideal then $\NN$ is an ideal of $\CC$.
Hence $\NN$ is a right ideal of $\CC$.
It remains to show that $\NN$ is maximal.
By Lemma 7 of Null Sequences form Maximal Left and Right Ideal then $\NN \subsetneq \CC$.
By maximal right ideal it needs to be shown that:
:There is no right ideal $\JJ$ of $\CC$ such that $\NN \subsetneq \JJ \subsetneq \CC$
Let $\JJ$ be a Right ideal of $\CC$ such that $\NN \subsetneq \JJ \subseteq \CC$.
It will be shown that $\JJ$ = $\CC$, from which the result will follow.
Let $\sequence {x_n} \in \JJ \setminus \NN$
By Inverse Rule for Cauchy sequences then 
:$\exists K \in \N: \sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ is a Cauchy sequence.
Let $\sequence {y_n}$ be the sequence defined by:
:$y_n = \begin{cases} 0 & : n \le K \\ \paren {x_n}^{-1} & : n > K \end{cases}$
By Cauchy Sequence with Finite Elements Prepended is Cauchy Sequence then $\sequence {y_n} \in \CC$
By the definition of a right ideal the product $\sequence {x_n} \sequence {y_n} = \sequence {x_n y_n} \in \JJ$
By the definition of $\sequence {y_n}$ then:
:$x_n y_n = \begin{cases} 0 & : n \le K \\ 1 & : n > K \end{cases}$
Let $\mathcal 1 = \tuple {1, 1, 1, \dots}$ be the unity of $\CC$
Then $\mathcal 1 - \sequence {x_n} \sequence {y_n}$ is the sequence $\sequence {w_n}$ defined by:
:$w_n = \begin {cases} 1 & : n \le K \\ 0 & : n > K \end {cases}$
By Convergent Sequence with Finite Elements Prepended is Convergent Sequence then $\sequence {w_n}$ is convergent to 0.
So $\sequence {w_n} \in \NN \subsetneq \JJ$
Since \sequence {x_n} $\sequence {y_n}, \sequence {w_n} \in \JJ$ by the definition of a ring ideal then:
:$\sequence {w_n} + \sequence {x_n} \sequence {y_n} = \mathcal 1 \in \JJ$
By the definition of a right ideal then:
:$\forall \sequence {a_n} \in \CC, \mathcal 1 \circ \sequence {a_n} = \sequence {a_n} \in \JJ$
Hence $\JJ = \CC$
== Sources ==
* : $\S 3.2$: Completions
Category:Null Sequences form Maximal Left and Right Ideal
Null Sequences form Maximal Left and Right Ideal/Lemma 4

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Then:
:$\NN \ne \O$
 
== Proof ==
From Constant Sequence Converges to Constant in Normed Division Ring, the zero $\tuple {0, 0, 0, \dots}$ of $\CC$ to converges $0 \in R$.
Therefore $\tuple {0, 0, 0, \dots} \in \NN$.
== Sources ==
* : $\S 3.2$: Completions
Category:Null Sequences form Maximal Left and Right Ideal
Null Sequences form Maximal Left and Right Ideal/Lemma 5

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Then:
:$\forall \sequence {x_n}, \sequence {y_n} \in \NN: \sequence {x_n} + \paren {-\sequence {y_n} } \in \NN$
 
== Proof ==
Let $\ds \lim_{n \mathop \to \infty} x_n = 0$ and $\ds \lim_{n \mathop \to \infty} y_n = 0$.
The sequence $\sequence {x_n} + \paren {-\sequence {y_n} } = \sequence {x_n - y_n}$.
From Difference Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} x_n - y_n = 0 - 0 = 0$
The result follows.
== Sources ==
* : $\S 3.2$: Completions
Category:Null Sequences form Maximal Left and Right Ideal
Null Sequences form Maximal Left and Right Ideal/Lemma 6

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Then:
:$\quad \forall \sequence {x_n} \in \NN, \sequence {y_n} \in \CC: \sequence {x_n} \sequence {y_n} \in \NN, \sequence {y_n} \sequence {x_n} \in \NN$
 
== Proof ==
Let $\ds \lim_{n \mathop \to \infty} x_n = 0$
By the definition of the product on the ring of Cauchy sequences then:
:$\sequence {x_n} \sequence {y_n} = \sequence {x_n y_n}$
:$\sequence {y_n} \sequence {x_n} = \sequence {y_n x_n}$
By Product of Sequence Converges to Zero with Cauchy Sequence Converges to Zero:
:$\ds \lim_{n \mathop \to \infty} x_n y_n = 0$
:$\ds \lim_{n \mathop \to \infty} y_n x_n = 0$
The result follows.
== Sources ==
* : $\S 3.2$: Completions
Category:Null Sequences form Maximal Left and Right Ideal
Null Sequences form Maximal Left and Right Ideal/Lemma 7

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Then:
:$\NN \subsetneq \CC$.
 
== Proof ==
By every convergent sequence is a Cauchy sequence then $\NN \subseteq \CC$.
From Constant Sequence Converges to Constant in Normed Division Ring, the unity $\tuple {1, 1, 1, \dotsc}$ of $\CC$ converges to $1 \in R$, and therefore $\tuple {1, 1, 1, \dotsc} \in \CC \setminus \NN$
So $\NN \subsetneq \CC$.
== Sources ==
* : $\S 3.2$: Completions
Category:Null Sequences form Maximal Left and Right Ideal