Convergent Sequence in Normed Vector Space is Bounded

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 $n \in \N$.