\section{Null Sequences form Maximal Left and Right Ideal/Lemma 3}
Tags: Null Sequences form Maximal Left and Right Ideal

\begin{theorem}
:$\NN$ is a [[Definition:Maximal Right Ideal of Ring|maximal right ideal]].
\end{theorem}

\begin{proof}
By [[Null Sequences form Maximal Left and Right Ideal/Lemma 1|Lemma 1 of Null Sequences form Maximal Left and Right Ideal]] then $\NN$ is an [[Definition:Ideal of Ring|ideal]] of $\CC$.
Hence $\NN$ is a [[Definition:Right Ideal of Ring|right ideal]] of $\CC$.
It remains to show that $\NN$ is [[Definition:Maximal Right Ideal of Ring|maximal]].
By [[Null Sequences form Maximal Left and Right Ideal/Lemma 2/Lemma 2.1|Lemma 2.1 of Null Sequences form Maximal Left and Right Ideal]] then $\NN \subsetneq \CC$.
By [[Definition:Maximal Right Ideal of Ring|maximal right ideal]] it needs to be shown that:
:There is no [[Definition:Right Ideal of Ring|right ideal]] $\JJ$ of $\CC$ such that $\NN \subsetneq \JJ \subsetneq \CC$
Let $\JJ$ be a [[Definition:Right Ideal of Ring|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 [[Combination Theorem for Cauchy Sequences/Inverse Rule|Inverse Rule for Cauchy sequences]] then 
:$\exists K \in \N: \sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|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 [[Definition:Right Ideal of Ring|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 [[Definition:Unity of Ring|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 [[Definition:Ideal of Ring|ring ideal]] then:
:$\sequence {w_n} + \sequence {x_n} \sequence {y_n} = \mathcal 1 \in \JJ$
By the definition of a [[Definition:Right Ideal of Ring|right ideal]] then:
:$\forall \sequence {a_n} \in \CC, \mathcal 1 \circ \sequence {a_n} = \sequence {a_n} \in \JJ$
Hence $\JJ = \CC$
{{qed}}
\end{proof}


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