# Prove that $\mathbb{Q}$ is not Cauchy complete

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.

• Pick your favorite irrational $x \in \mathbb{R}$ and then pick any sequence of rationals that converges to $x$. Oct 18, 2022 at 1:57
• I think you should work this out with your classical Greek hat on - you should proceed pretending you do not know a thing about $\Bbb R$. Oct 18, 2022 at 2:39
• Can I use the same sequence $r_n=(1+\frac{1}{n})$ since it converges to $e$? Oct 18, 2022 at 3:08
• $(1+\frac{1}{n})$ definitely does not converge to $e$... You're thinking of $(1+\frac{1}{n})^n$. Oct 18, 2022 at 3:30
• 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. Oct 18, 2022 at 3:52

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.

Continuing,

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

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.