How does one show that the set of rationals is topologically disconnected? Let $\mathbb{Q}$ be the set of rationals with its usual topology based on distance:
$$d(x,y) = |x-y|$$
Suppose we can only use axioms about $\mathbb{Q}$ (and no axiom about $\mathbb{R}$, the set of reals). Then how can we show that $\mathbb{Q}$ is topologically disconnected, i.e.: there exist two open sets $X$ and $Y$ whose union is $\mathbb{Q}$?
If we were allowed to use axioms about $\mathbb{R}$, then we could show that for any irrational number $a$:


*

*if $M$ is the intersection of $]-\infty, a[$ with the rationals, then $M$ is an open set of $\mathbb{Q}$

*if N is the intersection of $]a, +\infty[$ with the rationals, then $N$ is an open set of $\mathbb{Q}$

*$\mathbb{Q}$ is the union of $M$ and $N$. CQFD. 


But if we are not allowed to use axioms about $\mathbb{R}$, just axioms about $\mathbb{Q}$?
 A: The rationals is the union of two disjoint open sets $\{x\in\mathbb{Q}:x^2>2\}$ and $\{x\in\mathbb{Q}:x^2<2\}$.
A: $(\mathbb{Q},d)$ is metric space. Since connceted metric space having more that more than one point is uncountable and $\mathbb{Q}$ is countable, $(\mathbb{Q},d)$ must be disconnected.
A: A la Cantor's first proof of uncountability of $\mathbb R$:
Fix an enumeration of the countable  set $\mathbb Q$. In the following, expressions like "first rational" refer to this enumeration.
Let $a_0$ be the first rational.
Let $b_0$ be the first rational bigger than $a_0$.
For $n=0,1,2\ldots$ do the following:
Given rationals $a_n<b_n$ let $c,d$ be the first two rationals between $a_n$ and $b_n$ (such numbers exist, e.g. $a_n<\frac{2a_n+b_n}3<\frac{a_n+2b_n}3<b_n$).
Let $a_{n+1}=\min\{c,d\}$, $b_{n+1}=\max\{c,d\}$. Note that this implies $$\tag1a_n<a_{n+1}<b_{n+1}<b_n.$$
Let $$\tag2U=\{x\in\mathbb Q\mid \exists n\colon x<a_n\}, \quad V=\{x\in\mathbb Q\mid \exists n\colon x>b_n\}.$$
Then $U,V$ are  nonempty open subsets of $\mathbb Q$ with $U\cap V=\emptyset$, $U\cup V=\mathbb Q$:


*

*nonempty: $a_0-1\in U$, $b_0+1\in V$

*open: $x\in U$ implies $x<a_n$ for some $n$, implies $y\in U$ for all $y$ with $|y-x|<a_n-x$. Similarly for $V$

*disjoint: Assume $x\in U\cap V$. Then $b_n<x<a_m$ for some $n,m$. Repeated application of (1) produces $b_{\max\{n,m\}}\le\ldots \le b_n<a_m\le\ldots\le a_{\max\{n,m\}}$, contradiction

*covering: If $x\in \mathbb Q$, then there is an index $m$ at which it occurs in the enumeration of $\mathbb Q$. If $x$ is not among the first $2n$ numbers $a_0,\ldots a_{n-1}, b_0,\ldots,b_{n-1}$ selected, then either $m>2n$ or $x$ is not between $a_{n-1}$ and $b_{n-1}$. The first option is ruled out for $n$ big enough, the second implies $x<a_n$ or $x>b_n$, hence $x\in U\cup V$.

