Limit of Bounded Convergent Sequence is Bounded
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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.
$\blacksquare$