Convergent Sequence is Cauchy Sequence/Normed Vector Space
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Theorem
Let $\struct{X, \norm{\,\cdot\,} }$ be a normed vector space.
Every convergent sequence in $X$ is a Cauchy sequence.
Proof
Let $\sequence {x_n}$ be a sequence in $X$ that converges to the limit $L \in X$.
Let $\epsilon > 0$.
Then also $\dfrac \epsilon 2 > 0$.
Because $\sequence {x_n}$ converges to $L$, we have:
- $\exists N: \forall n > N: \norm {x_n - L} < \dfrac \epsilon 2$
So if $m > N$ and $n > N$, then:
\(\ds \norm {x_n - x_m}\) | \(=\) | \(\ds \norm {x_n - L + L - x_m}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_n - L} + \norm {x_m - L}\) | Triangle Inequality Axiom of Norm | |||||||||||
\(\ds \) | \(<\) | \(\ds \frac \epsilon 2 + \frac \epsilon 2\) | (by choice of $N$) | |||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
Thus $\sequence {x_n}$ is a Cauchy sequence.
$\blacksquare$
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