Cauchy Sequence is Bounded/Normed Vector Space
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Theorem
Let $V = \struct {X, \norm {\,\cdot\,} }$ be a normed vector space.
Every Cauchy sequence in $X$ is bounded.
Proof
Let $\sequence {x_n} $ be a Cauchy sequence in $V$.
Then by definition:
- $\forall \epsilon \in \R_{>0}: \exists N \in \N : \forall n, m \ge N: \norm {x_n - x_m} < \epsilon$
Let $N$ satisfy:
- $\forall n, m \ge N: \norm {x_n - x_m} < 1$
Let $m = N + 1 > N$.
Then $\forall n \ge N$:
\(\ds \norm {x_n}\) | \(=\) | \(\ds \norm {x_n - x_{N + 1} + x_{N + 1} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_n - x_{N + 1} } + \norm {x_{N + 1} }\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds 1 + \norm {x_{N + 1} }\) |
Let $M = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_N}, 1 + \norm {x_{N + 1} } }$.
Then:
- $\forall n < N: \norm {x_n} \le M$
- $\forall n \ge N: \norm {x_n} \le 1 + \norm {x_{N + 1} } \le M$
It follows by definition that $\sequence {x_n}$ is bounded.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces