# Equivalence of Definitions of Convergence in Normed Division Rings

## Theorem

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

Let $\sequence {x_n}$ be a sequence in $R$.

The following definitions of the concept of Convergent Sequence in Normed Division Ring are equivalent:

### Definition 1

The sequence $\sequence {x_n}$ converges to $x \in R$ in the norm $\norm {\, \cdot \,}$ if and only if:

$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N: n > N \implies \norm {x_n - x} < \epsilon$

### Definition 2

The sequence $\sequence {x_n}$ converges to $x \in R$ in the norm $\norm {\, \cdot \,}$ if and only if:

$\sequence {x_n}$ converges to $x$ in the metric induced by the norm $\norm {\, \cdot \,}$

### Definition 3

The sequence $\sequence {x_n}$ converges to $x \in R$ in the norm $\norm {\, \cdot \,}$ if and only if:

the real sequence $\sequence {\norm {x_n - x} }$ converges to $0$ in the reals $\R$

## Proof

### Definition 1 iff Definition 2

By definition, the metric induced by the norm $\norm {\, \cdot \,}$ is the mapping $d: R \times R \to \R_{\ge 0}$ defined as:

$\map d {x, y} = \norm {x - y}$

By definition of a convergent sequence in a metric space, $\sequence{x_n}$ converges to $x \in R$ if and only if:

$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N: n > N \implies \map d {x_n, x} < \epsilon$
$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N: n > N \implies \norm {x_n - x} < \epsilon$

The result follows.

$\Box$

### Definition 1 iff Definition 3

Let $x \in R$.

By norm on a division ring, $\norm {\, \cdot \,}$ is a mapping $\norm {\, \cdot \,}:R \to \R_{\ge 0}$.

Then:

$\forall n \in \N: \size {\norm{x_n - x} - 0} = \size {\norm{x_n - x}} = \norm{x_n - x}$

By definition of convergence of a real sequence, the real sequence $\sequence {\norm {x_n - x} }$ converges to $0$ in the reals $\R$ if and only if

$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: n > N \implies \size {\norm{x_n - x} - 0} < \epsilon$
$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: n > N \implies \norm{x_n - x} < \epsilon$

The result follows.

$\blacksquare$