Quotient of Cauchy Sequences is Metric Completion/Lemma 2

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Theorem

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.

Let $d$ be the metric induced by $\struct {R, \norm {\, \cdot \,} }$.


Let $\CC$ be the ring of Cauchy sequences over $R$

Let $\NN$ be the set of null sequences.

Let $\CC \,\big / \NN$ be the quotient ring of Cauchy sequences of $\CC$ by the maximal ideal $\NN$.

Let $\norm {\, \cdot \,}: \CC \,\big / \NN \to \R_{\ge 0}$ be the norm on the quotient ring $\CC \,\big / \NN$ defined by:

$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN} = \lim_{n \mathop \to \infty} \norm {x_n}$

Let $d'$ be the metric induced by $\struct {\CC \,\big / \NN, \norm {\, \cdot \,} }$


Let $\sim$ be the equivalence relation on $\CC$ defined by:

$\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$

Let $\tilde {\CC} = \CC / \sim$ denote the set of equivalence classes under $\sim$.

For $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the equivalence class containing $\sequence {x_n}$.

Let $\tilde d: \tilde {\CC} \times \tilde {\CC} \to \R_{\ge 0}$ be the metric defined by:

$\map {\tilde d} {\eqclass {x_n} {}, \eqclass {y_n} {} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$


Then:

$\quad d' = \tilde d$


Proof

By Lemma 1 of Quotient of Cauchy Sequences is Metric Completion we have that:

$\CC \,\big / \NN = \tilde {\CC}$

Let $\eqclass {x_n} {}$ and $\eqclass {x_n} {}$ be equivalence classes in $\CC \,\big / \NN = \tilde {\CC}$.

Then:

\(\ds \map {d'} {\eqclass {x_n}{}, \eqclass {x_n}{} }\) \(=\) \(\ds \norm {\eqclass {x_n}{} - \eqclass {x_n}{} }\) Definition of $d'$
\(\ds \) \(=\) \(\ds \norm {\eqclass {x_n - y_n}{} }\) Definition of Quotient Ring Addition
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \norm {x_n - y_n}\) Definition of Norm on Division Ring: $\norm {\,\cdot\,}$ on $\CC \,\big / \NN$
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \map d {x_n, y_n}\) Definition of Metric Induced by Norm on Division Ring: by the norm $\norm {\,\cdot\,}$ on $R$
\(\ds \) \(=\) \(\ds \map {\tilde d} {\eqclass {x_n}{}, \eqclass {x_n}{} }\) Definition of $\tilde d$

The result follows.

$\blacksquare$


Sources

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