Suppose that $\{a_n\}, \{b_n\}$ are both Cauchy sequences of real numbers, and that $a_n \le b_n \ \forall n$. Prove that $\lim_{n \to \infty} a_n \le \lim_{n \to \infty} b_n$.
The definition of a Cauchy sequence I am using is
A sequence $\{a_n\}$ is Cauchy if, given $\varepsilon > 0, \ \exists N$ so that $$ |a_n – a_m| \le \varepsilon \text{ if } n,m \ge N$$
My Work
Since $\{a_n\}, \{b_n\}$ are both Cauchy, they have finite limits $A, B$, so considering the sequence $\{c_n\}$ where $c_n = b_n – a_n$, we have
$$ a_n \le b_n \ \forall n \Rightarrow 0 \le b_n – a_n$$
$$\lim_{n\to \infty} c_n = \lim_{n \to \infty} b_n – \lim_{n \to \infty} a_n = B – A \ge 0 \Rightarrow A \le B$$
Therefore, $\lim_{n \to \infty} a_n \le \lim_{n \to \infty} b_n$.
Steps Left Out: By a limit theorem, I know the limit of the difference of two convergent sequences is the difference of their limits, and I know that $\lim_{n\to \infty} c_n \ge 0$ because by assumption each term of $\{c_n\} \ge 0$.
Have I done this proof correctly?
Best Answer
Your proof is correct, and here's another proof.
Let $a = \lim_{n \to \infty} a_n$ and $b = \lim_{n \to \infty} b_n$. Suppose by contradiction that $a > b$. Set $\epsilon = \frac{a-b}{2}$, and choose $N \in \mathbb{N}$ such that $n \ge N$ implies $|a_n - a| < \epsilon$ and $|b_n - b| < \epsilon$. Then for this $n$,
$$ a_n > a - \epsilon = a - \frac{a -b}{2} = b + \frac{a - b}{2} = b + \epsilon > b_n$$
which is a contradiction.