Convergent Sequence in Normed Division Ring is Bounded

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Theorem

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

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

Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limit:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l$

Then $\sequence {x_n}$ is bounded.


Proof 1

Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the limit $l$, then:

$\forall \epsilon \in \R_{\gt 0}: \exists N \in \N : \forall n \ge N: \norm {x_n - l} \lt \epsilon$.


Let $n_1$ satisfy:

$\forall n \ge n_1: \norm {x_n - l} < 1$


Then $\forall n \ge n_1$:

\(\displaystyle \norm {x_n}\) \(=\) \(\displaystyle \norm {x_n - l + l}\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n - l } + \norm l\) Norm axiom (N3) (Triangle Inequality)
\(\displaystyle \) \(\le\) \(\displaystyle 1 + \norm l\) as $n \ge n_1$


Let $K = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_{n_1 - 1}}, 1 + \norm {l} }$.

Then:

$\forall n \lt n_1: \norm {x_n} \le K$
$\forall n \ge n_1: \norm {x_n} \le 1 + \norm l \le K$

By the definition of a bounded sequence in a normed division ring then $\sequence {x_n}$ is bounded.

$\blacksquare$


Proof 2

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

Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.

By the definition of a convergent sequence in a normed division ring, $\sequence {x_n} $ is convergent to the limit $l$ in $\struct {R, d}$.

By Convergent Sequence in Metric Space is Bounded, $\sequence {x_n} $ is a bounded sequence in $\struct {R, d}$.

By Sequence is Bounded in Norm iff Bounded in Metric, $\sequence {x_n} $ is a bounded sequence in $\struct {R, \norm {\,\cdot\,} }$.

$\blacksquare$


Proof 3

Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,} }$.

By modulus of limit in normed division ring, $\sequence {\norm {x_n} }$ is a convergent sequence in $\R$.

By Convergent Real Sequence is Bounded, $\sequence {\norm {x_n} }$ is bounded.

That is:

$\exists M \in \R_{> 0}: \forall n, \norm {x_n} = \size {\norm {x_n} } \le M$

By the definition of a bounded sequence in a normed division ring, $\sequence {x_n}$ is bounded.

$\blacksquare$


Proof 4

Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.

By Convergent Sequence is Cauchy Sequence in Normed Division Ring, $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$.

By Cauchy Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is a bounded sequence in $\struct {R, \norm {\,\cdot\,}}$.

$\blacksquare$


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