Sequence is Bounded in Norm iff Bounded in Metric
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Theorem
Let $\struct {R, \norm {\,\cdot\,} } $ be a normed division ring.
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a sequence in $R$.
Then:
- $\sequence {x_n} $ is a bounded sequence in the normed division ring $\struct {R, \norm {\,\cdot\,} }$ if and only if $\sequence {x_n} $ is a bounded sequence in the metric space $\struct {R, d}$.
Proof
Necessary Condition
Let $\sequence {x_n} $ be a bounded sequence in $\struct {R, \norm {\,\cdot\,} }$.
Then:
- $\exists K \in \R_{\gt 0} : \forall n : \norm {x_n} \le K$
Then $\forall n, m \in \N$:
\(\ds \map d { x_n , x_m }\) | \(=\) | \(\ds \norm {x_n - x_m}\) | Definition of Metric Induced by Norm on Division Ring | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_n} + \norm {x_m}\) | Norm of Difference | |||||||||||
\(\ds \) | \(\le\) | \(\ds K + K\) | Definition of Bounded Sequence in Normed Division Ring | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 K\) |
Hence the sequence $\sequence {x_n} $ is bounded by $2 K$ in the metric space $\struct {R, d}$.
$\Box$
Sufficent Condition
Let $\sequence {x_n} $ be a bounded sequence in the metric space $\struct {R, d}$.
Then:
- $\exists K \in \R_{> 0} : \forall n, m : \map d {x_n , x_m} \le K$
By the definition of the metric induced by a norm this is equivalent to:
- $\exists K \in \R_{> 0} : \forall n, m : \norm {x_n - x_m} \le K$
Then $\forall n \in \N$:
\(\ds \norm {x_n}\) | \(=\) | \(\ds \norm {x_n - x_1 + x_1}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_n - x_1} + \norm {x_1}\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds K + \norm {x_1}\) | by hypothesis: Definition of Bounded Sequence in Metric Space |
Hence the sequence $\sequence {x_n}$ is bounded by $K + \norm {x_1}$ in the normed division ring $\struct {R, \norm {\,\cdot\,} }$.
$\blacksquare$