Null Sequences form Maximal Left and Right Ideal/Lemma 1

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 an ideal of $\mathcal {C}$.

Proof
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.