Null Sequences form Maximal Left and Right Ideal

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.

That is:
 * $\mathcal {N} = \set {\sequence {x_n}: \displaystyle \lim_{n \mathop \to \infty} x_n = 0 }$

Then $\mathcal {N}$ is a ring ideal of $\mathcal {C}$ that is a maximal left ideal and a maximal right ideal.

Proof
The proof is completed in these steps:
 * (1): $\mathcal {N} \subset \mathcal {C}$


 * (2): $\mathcal {N}$ is an ideal of $\mathcal {C}$.


 * (3): $\mathcal {N}$ is a maximal left ideal.


 * (4): $\mathcal {N}$ is a maximal right ideal.

(1) $\mathcal {N} \subset \mathcal {C}$
Follows directly from every convergent sequence is a Cauchy sequence.

(2) $\mathcal {N}$ is an ideal of $\mathcal {C}$
By Test for Ideal it is sufficient to prove:
 * $(a): \quad \mathcal {N} \ne \varnothing$


 * $(b): \quad \forall \sequence {x_n}, \sequence {y_n} \in \mathcal {N}: \sequence {x_n} + \paren {-\sequence {y_n} } \in \mathcal {N}$


 * $(c): \quad \forall \sequence {x_n} \in \mathcal {N}, \sequence {y_n} \in \mathcal {C}: \sequence {x_n} \sequence {y_n} \in \mathcal {N}, \sequence {y_n} \sequence {x_n} \in \mathcal {N}$

$(a): \quad \mathcal {N} \ne \varnothing$
The zero $\tuple {0,0,0,\dots}$ of $\mathcal {C}$ converges to $0 \in R$, and therefore $\tuple {0,0,0,\dots} \in \mathcal {N}$

$(b): \quad \forall \sequence {x_n}, \sequence {y_n} \in \mathcal {N}: \sequence {x_n} + \paren {-\sequence {y_n} } \in \mathcal {N}$
Let $\displaystyle \lim_{n \mathop \to \infty} x_n = 0$ and $\displaystyle \lim_{n \mathop \to \infty} y_n = 0$.

The sequence $\sequence {x_n} + \paren {-\sequence {y_n} } = \sequence {x_n - y_n}$.

By Difference Rule for Sequences, $\displaystyle \lim_{n \mathop \to \infty} x_n - y_n = 0 - 0 = 0.$

The result follows.

$(c): \quad \forall \sequence {x_n} \in \mathcal {N}, \sequence {y_n} \in \mathcal {C}: \sequence {x_n} \sequence {y_n} \in \mathcal {N}, \sequence {y_n} \sequence {x_n} \in \mathcal {N}$
Let $\displaystyle \lim_{n \mathop \to \infty} x_n = 0$

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 then:
 * $\displaystyle \lim_{n \mathop \to \infty} x_n y_n = 0$
 * $\displaystyle \lim_{n \mathop \to \infty} y_n x_n = 0$

The result follows.

(3) $\mathcal {N}$ is a maximal left ideal
By maximal left ideal it needs to be shown that:
 * $(a): \quad \mathcal {N} \subsetneq \mathcal {C}$
 * $(b): \quad$ There is no left ideal $\mathcal J$ of $\mathcal {C}$ such that $\mathcal {N} \subsetneq \mathcal J \subsetneq \mathcal {C}$

$(a): \quad \mathcal {N} \subsetneq \mathcal {C}$
The unity $\tuple {1,1,1,\dots}$ of $\mathcal {C}$ converges to $1 \in R$, and therefore $\tuple {1,1,1,\dots} \in \mathcal {C} \setminus \mathcal {N}$

$(b): \quad$ There is no left ideal $\mathcal J$ of $\mathcal {C}$ such that $\mathcal {N} \subsetneq \mathcal J \subsetneq \mathcal {C}$
Let $\mathcal J$ be a left 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 left ideal the product $\sequence {y_n} \sequence {x_n} = \sequence {y_n x_n} \in \mathcal {J}$

By the definition of $\sequence {y_n}$ then:
 * $y_n x_n = \begin{cases} 0 & : n \le K \\ 1 & : n > K \end{cases}$

Let $\mathcal {1} = \tuple {1,1,1,\dots}$ be the unity of $\mathcal {C}$

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

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

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

(4) $\mathcal {N}$ is a maximal right ideal
This proof proceeds along the same lines as in (3) above, the difference being that all products are reversed.