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 $\JJ = \CC$