Show that the rationals are an incomplete metric space without reference to reals

I know that you can create rational sequences that converge to irrationals, but is there a simple way to do this without explicit assumption of the existence of the reals?

I'm thinking of something along the lines of

1) Show that there exists a particular Cauchy sequence of rationals

2) Assume that the cauchy sequence converges to a rational.

Hence rationals are incomplete.

I had an idea about using the property discussed in this question Choice of $q$ in Baby Rudin's Example 1.1 but couldn t make anything of it.

I would like to know this because I feel that this is an inherent property of the rationals and that a proof of it should not need to reference anything else.

• How about the sequence of rationals that converges to $\sqrt{2}$? Aug 16, 2015 at 4:14
• @Chou How would you write that? Aug 16, 2015 at 4:15
• I'm not sure if this fits your "avoiding irrationals" criterion, but you can show that the sequence 0.1, 0.1011, 0.10110111, ... does not converge to a rational number, since the decimals do not repeat, and moreover the sequence is Cauchy. Aug 16, 2015 at 4:16
• You could prove that $\dfrac{F_{n+1}}{F_n}$ is a Cauchy sequence, and that a limit — if it existed — would satisfy $x=1+\dfrac1x$, which no rational does. ($F_n$ is the $n$th Fibonacci number.) Aug 16, 2015 at 4:17
• @columbus8myhw Thanks, this was the idea I was looking for. And is your dog's name 'columbus'? Aug 16, 2015 at 4:20

Your approach is spot on, and you don't need to assume the reals exist to do it. Define a Cauchy sequence of rationals such that each one squared is closer to $2$ than the one before. Note that every number in this argument is rational. Then if the rationals are complete, the sequence converges to a rational. Now prove that no rational squared equals $2$. You have a Cauchy sequence that does not converge to an element of the space.

• Sir, can you write your sequence, in mathematical form Feb 10, 2021 at 3:09
• @AkashPatalwanshi: you can use the result of the Babylonian approach. $x_{i+1}=\frac 12(x_i+\frac 2{x_i})$ Feb 10, 2021 at 3:21

You don’t even have to deal directly with rationals (except as radii of metric balls): you can prove the superficially more general result that a countable metric space without isolated points is not complete. (The generality is only superficial, because such a space is homeomorphic to $$\Bbb Q$$.)

Let $$\langle X,d\rangle$$ be a countable metric space without isolated points, and index $$X=\{x_k:k\in\Bbb N\}$$. Let $$m_0=0$$, $$r_0=1$$, $$B_0=\{x\in X:d(x_{m_0},x), $$C_0=\operatorname{cl}B_0$$, and $$M_0=\{k\in\Bbb N:x_k\in B_0\}$$; $$M_0$$ is infinite, since $$X$$ has no isolated points, and $$m_0=\min M_0$$.

Suppose that $$n\in\Bbb N$$, and we have an $$m_n\in\Bbb N$$, an open set $$B_n$$ containing $$x_{m_n}$$, $$C_n=\operatorname{cl}B_n$$, and an infinite $$M_n=\{k\in\Bbb N:x_k\in B_n\}$$ such that $$m_n=\min M_n$$. Let

$$m_{n+1}=\min(M_n\setminus\{m_n\})\;;$$

then $$x_{m_{n+1}}\in B_n\setminus\{x_{m_n}\}$$. We can therefore choose a positive rational $$r_{n+1}\le\min\left\{d(x_{m_n},x_{m_{n+1}}),r_n-d(x_{m_n},x_{m_{n+1}}),2^{-(n+1)}\right\}$$ and set

\begin{align*} &B_{n+1}=\{x\in X:d(x_{m_{n+1}},x)

It’s not hard to verify that $$C_{n+1}\subseteq B_n$$, $$M_{n+1}$$ is infinite, and $$m_{n+1}=\min M_{n+1}$$, so the recursive construction can continue. Note that we always have $$m_{n+1}>m_n$$.

In the end we have a sequence $$\langle C_n:n\in\Bbb N\rangle$$ of closed sets such that $$C_n\supset C_{n+1}$$ for each $$n\in\Bbb N$$. Moreover, $$\operatorname{diam}C_n\le 2r_n\le2\cdot2^{-(n+1)}=2^{-n}$$, so if $$X$$ were complete, the Baire category theorem would ensure that $$\bigcap_{n\in\Bbb N}C_n=\varnothing$$. Suppose that some $$x_k\in\bigcap_{n\in\Bbb N}C_n$$. Then $$k\in\bigcap_{n\in\Bbb N}M_n$$. But $$\langle m_n:n\in\Bbb N\rangle$$ is strictly increasing, so there is an $$n\in\Bbb N$$ such that $$k, and hence $$x_k\notin B_n$$. Finally, $$C_{n+1}\subseteq B_n$$, so $$x_k\notin C_{n+1}$$, and $$X$$ therefore cannot be complete.

Construct a sequence of rationals that would converge to, say, $\sqrt{2}$. For example, $x_1 = 1$, $x_{n+1} =\frac{x_n + 2/x_n}{2} =\frac{x_n^2+2}{2 x_n}$.

Then, if this converges to a rational, let the limit be $r = a/b$, and let $x_n = a_n/b_n$. Then $a_{n+1}/b_{n+1} =\frac{a_n/b_n + 2b_n/a_n}{2} =\frac{a_n^2 + 2b_n^2}{2a_nb_n}$ so $a_{n+1} = a_n^2 + 2b_n^2$ and $b_{n+1} = 2a_nb_n$.

$\begin{array}\\ a_{n+1}^2-2b_{n+1}^2 &=(a_n^2 + 2b_n^2)^2-2(2a_nb_n)^2\\ &=a_n^4 +4a_n^2b_n^2+ 4b_n^4-8a_n^2b_n^2\\ &=a_n^4 -4a_n^2b_n^2+ 4b_n^4\\ &=(a_n^2 - 2b_n^2)^2\\ \end{array}$

If we start with $x_1 = 1$, then $a_1 = b_1 = 1$, so $(a_1^2 - 2b_1^2)^2 = 1$. Therefore $a_n^2 -2b_n^2 = 1$ for all $n > 1$.

Since $r$ is rational, $r = a/b$ and $2 = r^2 = a^2/b^2$. Therefore

$\begin{array}\\ 1 &= a_n^2 -2b_n^2\\ &= a_n^2 -(a^2/b^2)b_n^2\\ &= \frac{a_n^2b^2 -a^2b_n^2}{b^2}\\ &= \frac{(a_nb -ab_n)(a_nb +ab_n)}{b^2}\\ &\ge \frac{(a_nb +ab_n)}{b^2} \quad\text{since } (a_nb -ab_n) \ge 1\\ &> \frac{a_nb}{b^2}\\ &= \frac{a_n}{b}\\ \end{array}$

Therefore, $a_n \le b$. But, since $a_{n+1} = a_n^2 + 2b_n^2$, $a_n$ gets arbitrarily large. This is a contradiction, so the assumption that the sequence converges to a rational is false.

Therefore, the rationals are not complete.

Note: This is a rewrite of my answer to this question of mine: What is the most unusual proof you know that $\sqrt{2}$ is irrational?

The line of reason you suggested is exactly "create a rational sequence that converge to an irrational" .

If you want an explicitly sequence, you can create a Cauchy sequence of rationals that converges to $\sqrt2$, the decimal approaches to it works.