# prove that the set of rational numbers is not connected on the real line [closed]

Could someone help me through this problem? Prove that the set of rational numbers is not connected on the real line

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## closed as off-topic by Andres Caicedo, amWhy, Ivo Terek, Rick Decker, Adam HughesJul 26 at 18:57

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Consider the set of rational numbers $q < \sqrt{2}$. Why is it open? Why is it closed? –  Pedro Apr 23 '12 at 3:00
Something even stronger is true: The rationals (with the usual interval topology) are totally disconnected. That is, the only connected subsets of the rationals are the singletons. –  Austin Mohr Apr 23 '12 at 3:04
Consider two separate one with q> root of 2 and the other with q <2 for q root of the real and their union is the set of rational –  Breton Apr 23 '12 at 3:05

$(-\infty,\pi),(\pi,\infty)$ are two open disjoint sets which cover $\mathbb{Q}$.
The same statement could be made for any two intervals of the form $(-\infty, \alpha), (\alpha, \infty)$, where $\alpha\in\mathbb{R}$ is irrational. That being said, I like the elegance of your having chosen a transcendental number. –  Nicholas Stull Apr 23 '12 at 5:17
It is not enough to show that there are two disjoint open sets $U$ and $V$ which cover the rationals. The definition of connectedness requires that the union of the disjoint open sets $U$ and $V$ actually equals the set of rationals $\mathbb{Q}$. For this we can take $U = (-\infty, \pi) \cap \mathbb{Q}$ and $V = (\pi, \infty) \cap \mathbb{Q}$. Note that $U$ and $V$ are open sets of the subspace topology on $\mathbb{Q}$ as a subset of $\mathbb{R}$, and satisfy the given conditions.
And if two open set are disjoint and they cover $\Bbb Q$, then their intersection with $\Bbb Q$ are two disjoint and relatively open sets whose union is the set of rational numbers. Again, cover means exactly that. –  Asaf Karagila Jul 26 at 23:06