# 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 $\sqrt{2}$? – Megadeth Aug 16 '15 at 4:14
• @Chou How would you write that? – ignoramus Aug 16 '15 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. – Ilham Aug 16 '15 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.) – Akiva Weinberger Aug 16 '15 at 4:17
• @columbus8myhw Thanks, this was the idea I was looking for. And is your dog's name 'columbus'? – ignoramus Aug 16 '15 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.

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 $m_0=0$, $r_0=1$, $B_0=\{x\in X:d(x_{m_0},x)<r_0\}$, $C_0=\operatorname{cl}B_0$, 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)<r_{n+1}\}\;,\\ &C_{n+1}=\operatorname{cl}B_{n+1}\;,\text{ and}\\ &M_{n+1}=\{k\in\Bbb N:x_k\in B_{n+1}\}\;. \end{align*}

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 $k<m_n=\min M_n$, 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.

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.

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?