Null Sequences form Maximal Left and Right Ideal/Lemma 3

Theorem
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.

Let $\mathcal {C}$ be the ring of Cauchy sequences over $R$

Let $\mathcal {N}$ be the set of null sequences.

Then:
 * $\mathcal {N}$ is a maximal right ideal.

Proof
By Lemma 1 of Null Sequences form Maximal Left and Right Ideal then $\mathcal {N}$ is an ideal of $\mathcal {C}$.

Hence $\mathcal {N}$ is a right ideal of $\mathcal {C}$.

It remains to show that $\mathcal {N}$ is maximal.

By Lemma 2.1 of Null Sequences form Maximal Left and Right Ideal then $\mathcal {N} \subsetneq \mathcal {C}$.

By maximal right ideal it needs to be shown that:
 * There is no right ideal $\mathcal J$ of $\mathcal {C}$ such that $\mathcal {N} \subsetneq \mathcal J \subsetneq \mathcal {C}$

Let $\mathcal J$ be a Right ideal of $\mathcal {C}$ such that $\mathcal {N} \subsetneq \mathcal J \subseteq \mathcal {C}$.

It will be shown that $\mathcal J$ = $\mathcal {C}$, from which the result will follow.

Let $\sequence {x_n} \in \mathcal {J} \setminus \mathcal {N}$

By Inverse Rule for Cauchy sequences then
 * $\exists K \in \N: \sequence { \paren {x_{K+n}}^{-1} }_{n \in \N}$ 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 $\sequence {y_n} \in \mathcal {C}$

By the definition of a right ideal the product $\sequence {x_n} \sequence {y_n} = \sequence {x_n y_n} \in \mathcal {J}$

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 $\mathcal {C}$

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 $\sequence {w_n} \in \mathcal {N} \subsetneq \mathcal {J}$

Since \sequence {x_n} $\sequence {y_n}, \sequence {w_n} \in \mathcal {J}$ by the definition of a ring ideal then:
 * $\sequence {w_n} + \sequence {x_n} \sequence {y_n} = \mathcal {1} \in \mathcal{J}$

By the definition of a right ideal then:
 * $\forall \sequence {a_n} \in \mathcal {C}, \mathcal {1} \circ \sequence {a_n} = \sequence {a_n} \in \mathcal{J}$

Hence $\mathcal {J} = \mathcal {C}$