Quotient of Cauchy Sequences is Metric Completion

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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 $\mathcal C$ be the ring of Cauchy sequences over $R$.

Let $\mathcal N$ be the set of null sequences in $R$.

Let $\mathcal C \,\big / \mathcal N$ be the quotient ring of Cauchy sequences of $\mathcal C$ by the maximal ideal $\mathcal N$.

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

$\displaystyle \forall \sequence {x_n} + \mathcal N: \norm {\sequence {x_n} + \mathcal N} = \lim_{n \mathop \to \infty} \norm{x_n}$

Let $d'$ be the metric induced by $\struct {\mathcal C \,\big / \mathcal N, \norm {\, \cdot \,} }$


Let $\phi: R \to \mathcal C \,\big / \mathcal N$ be the mapping from $R$ to the quotient ring $\mathcal C \,\big / \mathcal N$ defined by:

$\forall a \in R: \map \phi a = \sequence {a, a, a, \ldots} + \mathcal N$

where $\sequence {a, a, a, \ldots} + \mathcal N$ is the left coset in $\mathcal C \, \big / \mathcal N$ that contains the constant sequence $\sequence {a, a, a, \ldots}$.


Then:

$\struct {\mathcal C \,\big / \mathcal N, d'}$ is the metric completion of $\struct {R,d}$

and:

$\map \phi R$ is a dense subset of $\mathcal C \,\big / \mathcal N$


Proof

By the definition of the metric induced by a norm then:

a sequence $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\, \cdot \,} }$ if and only if $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, d}$.

So $\mathcal C$ is the set of Cauchy sequences in $\struct {R, d}$.


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

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

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

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


Lemma 1

$\quad \mathcal C \,\big / \mathcal N = \tilde {\mathcal C}$

$\Box$


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

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

By Completion of a Metric Space then:

$\struct {\tilde {\mathcal C}, \tilde d}$ is the metric completion of $\struct {R, d}$.


Lemma 2

$\quad d' = \tilde d$

$\Box$


Let $\tilde \phi: R \to \tilde {\mathcal C}$ be the mapping defined by:

$\map {\tilde \phi} a = \eqclass {a, a, a, \dotsc} {}$

where $\eqclass {a, a, a, \dotsc} {}$ denotes the equivalence class containing the constant sequence $\sequence {a, a, a, \dotsc}$.

By Completion of a Metric Space then:

$\map {\tilde \phi} R$ is a dense subset of $\tilde {\mathcal C}$.


By Lemma 1 then:

$\forall \sequence {x_n} \in \mathcal C: \sequence {x_n} + \mathcal N = \eqclass {x_n} {}$

In particular:

$\forall a \in R: \sequence {a, a, a, \dotsc} + \mathcal N = \eqclass {a, a, a, \dotsc} {}$

That is:

$\forall a \in R: \map \phi a = \map {\tilde \phi} a$

Hence $\phi = \tilde \phi$.

The result follows.

$\blacksquare$


Sources