One interesting thing about the Cauchy sequence definition is that it can be stated without reference to real numbers at all: We say a sequence of rational numbers $\{r_n\}$ is Cauchy (in the absolute value metric on the rational numbers) if for every $\epsilon \in \mathbb{Q}$ satisfying $\epsilon > 0$, there exists some $M \in \mathbb{N}$ such that for all $n$, $k ≥ M$, we have $|r_n − r_k| < \epsilon$. Prove that $\mathbb{Q}$ is not Cauchy complete, that is, show that there exists a Cauchy sequence {r_n} which does not converge to some limit $r \in \mathbb{Q}$.

Previously, I have proved that there exists a bounded sequence of rational numbers such that no subsequence converges to a rational number by an example $r_n = (1+\frac{1}{n})^n$ where the sequence converges to $e$. I think I am supposed to do similar in this proof, but I am not sure what an example in $\mathbb{Q}$ would be.

  • 3
    $\begingroup$ Pick your favorite irrational $x \in \mathbb{R}$ and then pick any sequence of rationals that converges to $x$. $\endgroup$
    – Mason
    Oct 18, 2022 at 1:57
  • $\begingroup$ I think you should work this out with your classical Greek hat on - you should proceed pretending you do not know a thing about $\Bbb R$. $\endgroup$ Oct 18, 2022 at 2:39
  • $\begingroup$ Can I use the same sequence $r_n=(1+\frac{1}{n}) $ since it converges to $e$? $\endgroup$ Oct 18, 2022 at 3:08
  • $\begingroup$ $(1+\frac{1}{n})$ definitely does not converge to $e$... You're thinking of $(1+\frac{1}{n})^n$. $\endgroup$
    – C-RAM
    Oct 18, 2022 at 3:30
  • 2
    $\begingroup$ Any representation in decimal of any $x\in\Bbb Q\cap [0,1]$ is eventually periodic. Let $r_n=\sum_{j=1}^nd_j 10^{-j}$ where the sequence $(d_j)_{j\in\Bbb N}$ of digits never becomes periodic, e.g. a $1$ followed by two $0$'s, then three $1$'s, then four $0$'s, etc. $\endgroup$ Oct 18, 2022 at 3:52

2 Answers 2


Let $a_1=2$ and $a_{n+1}=a_n/2+1/a_n$ for $n \ge 1$. Then $a_n^2>2$ for all $n$, the sequence $a_n$ is decreasing, and $a_n^2 \to 2$ as $n \to \infty$, so $a_n$ cannot tend to a rational limit.


Any s̶t̶r̶i̶c̶t̶l̶y̶ increasing sequence that is bounded above is a Cauchy sequence in $\Bbb Q$.

Define a sequence $(a_n)_{\,n\ge1}$ as follows,

$\quad a_1 = 1$
$\quad \text{For } n \ge 1\text{, }\; a_{n+1} = \frac{2+2a_n}{2+a_n}$

We leave it to the OP to use induction to verify that

$\tag 1 \text{For all } n \ge 1, a_n^2 \lt 2$

Note: This implies that our sequence is bounded above.


$\quad \frac{2+2a_n}{2+a_n} \gt a_n \text{ iff}$
$\quad 2+2a_n \gt 2a_n + a_n^2 \text{ iff}$
$\quad 2 \gt a_n^2$

and so $(a_n)_{\,n\ge1}$ is also increasing.

So our sequence is indeed a Cauchy sequence.

Since $a_{n+1} - a_n = \frac{2-a_n^2}{2+a_n}$ must converge to $0$, the numerator $2 - a_n^2$ must converge to $0$.
It follows that in $\Bbb Q$,

$$ \lim_{n\to 0} a_n^2 = 2$$

If we assume (to get a contradiction) that

$$ \lim_{n\to 0} a_n = \alpha$$

then, since the limit of a product is the product of the limits, we could write

$$ \alpha^2 = 2$$

but that is impossible over $\Bbb Q$.

We have to conclude that the space $\Bbb Q$ is not Cauchy complete.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .