Convergent Real Sequence is Bounded

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Theorem

Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

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


Then $\left \langle {x_n} \right \rangle$ is bounded.


That is, all convergent real sequences are bounded.


Proof 1

From Real Number Line is Metric Space, the set $\R$ under the usual metric is a metric space.

By Convergent Sequence in Metric Space is Bounded it follows that:

$\exists M > 0: \forall n, m \in \N: \size {x_n - x_m} \le M$

Then for $n \in \N$, by the Triangle Inequality for Real Numbers:

\(\displaystyle \size {x_n}\) \(=\) \(\displaystyle \size {x_n - x_1 + x_1}\)
\(\displaystyle \) \(\le\) \(\displaystyle \size {x_n - x_1} + \size {x_1}\)
\(\displaystyle \) \(\le\) \(\displaystyle M + \size {x_1}\)

Hence $\sequence {x_n}$ is bounded.

$\blacksquare$


Proof 2

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.

$\blacksquare$


Sources