Limit of Bounded Convergent Sequence is Bounded

Theorem

Let $\sequence {x_n}$, $\sequence {a_n}$, and $\sequence {b_n}$ be convergent sequences in $\R$.

Let $\sequence {x_n}$, $\sequence {a_n}$, and $\sequence {b_n}$ converge to $x, a, b \in \R$, respectively.

Suppose that:

$\exists N \in \N: n \ge N \implies a_n \le x_n \le b_n$

Then:

$a \le x \le b$

Proof

Aiming for a contradiction, suppose that $x < a$.

Let $\epsilon = \dfrac {a - x} 2 > 0$

From the convergence of $\sequence {x_n}$:

$\exists M_1 \in \N : n \ge M \implies x - \epsilon < x_n < x + \epsilon$

Or, equivalently:

$\exists M_1 \in \N : n \ge M \implies \dfrac {3 x - a} 2 < x_n < \dfrac {x + a} 2$

From the convergence of $\sequence {a_n}$:

$\exists M_2 \in \N : n \ge M \implies a - \epsilon < a_n < a + \epsilon$

Or, equivalently:

$\exists M_2 \in \N : n \ge M \implies \dfrac {x + a} 2 < a_n < \dfrac {3 a - x} 2$

Let $M = \max \set {N, M_1, M_2}$

Then, for any $n \ge M$:

\(\ds x_n\)

\(<\)

\(\ds \dfrac {x + a} 2\)

\(\ds \)

\(<\)

\(\ds a_n\)

This contradicts the hypothesis that:

$\forall n \ge N : a_n \le x_n$

The same argument, mutatis mutandis, brings us to a contradiction if we suppose $x > b$.

Hence the result, by Proof by Contradiction.

$\blacksquare$