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?

  • $\begingroup$ It is not true that every Cauchy sequence is convergent. This is the case for complete spaces. $\endgroup$
    – Sigur
    Aug 13, 2012 at 22:27
  • $\begingroup$ My mistake, I added the note that we are in $\mathbb R$ $\endgroup$
    – Moderat
    Aug 13, 2012 at 22:29
  • $\begingroup$ It is a good idea to write that $A$ and $B$ are the limits, even this being so obvious. I guess that it is enough. $\endgroup$
    – Sigur
    Aug 13, 2012 at 22:32
  • $\begingroup$ Can you see that the sum (or difference) of two Cauchys is again a Cauchy, with limit that’s sum (difference) of the original sequences? Once you know that, you're dealing with a Cauchy sequence of nonnegative terms... $\endgroup$
    – Lubin
    Aug 13, 2012 at 22:34
  • $\begingroup$ Seems good, note the typo in your definition of Cauchy sequence, you have $\varepsilon \lt 0$. $\endgroup$ Aug 13, 2012 at 23:07

1 Answer 1


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.


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