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
 * $\exists K \in \N: \sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ 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 $\sequence {y_n} \in \mathcal C$

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 $\mathcal J = \mathcal C$