Convergent Real Sequence is Bounded/Proof 2

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

Let $l \in A$ such that $\ds \lim_{n \mathop \to \infty} x_n = l$.

Then $\sequence {x_n}$ is bounded.


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

Let $x_n \to l$ as $n \to \infty$.

To show that $\sequence {x_n}$ is bounded sequence, we need to find $K$ such that:

$\forall n \in \N: \size {x_n} \le K$

Because $\sequence {x_n}$ converges:

$\forall \epsilon > 0: \exists N: n > N \implies \size {x_n - l} < \epsilon$

In particular, this is true when $\epsilon = 1$.

That is:

$\exists N_1: \forall n > N_1: \size {x_n - l} < 1$

By Backwards Form of Triangle Inequality:

$\forall n > N_1: \size {x_n} - \size l \le \size {x_n - l} < 1$

That is:

$\size {x_n} < \size l + 1$

So we set:

$K = \max \set {\size {x_1}, \size {x_2}, \ldots, \size {x_{N_1} }, \size l + 1}$

and the result follows.