Convergent Sequence in Normed Vector Space is Bounded
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Theorem
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence.
Then there exists $M > 0$ such that:
- $\norm {x_n} \le M$
for each $n \in \N$.
Proof
Suppose that $x_n \to x$.
From Convergent Sequence is Cauchy Sequence, $\sequence {x_n}_{n \in \N}$ is a Cauchy sequence.
So there exists $N \in \N$ such that:
- $\norm {x_n - x_m} < 1$
for each $n, m \ge N$.
That is:
- $\norm {x_n - x_N} < 1$
for $n \ge N$.
From Reverse Triangle Inequality: Normed Vector Space, we have:
- $\norm {x_n} - \norm {x_N} < 1$
so that:
- $\norm {x_n} < 1 + \norm {x_N}$
for $n \ge N$.
Now set:
- $M = \max \set {\norm {x_1}, \norm {x_2}, \ldots, \norm {x_{N - 1} }, 1 + \norm {x_N} }$
Then we have:
- $\norm {x_n} \le M$
for each $n \in \N$.
$\blacksquare$