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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 Hughes Jul 26 at 18:57

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4  
Consider the set of rational numbers $q < \sqrt{2}$. Why is it open? Why is it closed? –  Pedro Apr 23 '12 at 3:00
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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

2 Answers 2

up vote 11 down vote accepted

$(-\infty,\pi),(\pi,\infty)$ are two open disjoint sets which cover $\mathbb{Q}$.

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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.

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Cover means exactly that. –  Asaf Karagila Jul 26 at 17:46
    
@AsafKaragila Formally a collection of sets C serving cover of a given set X is only required to contain the given set X as a subset of the union of the elements of C. My point is that the definition of connectedness requires that X actually equal the the union of the disjoint open sets U and V. A fine distinction but a distinction nonetheless. –  JonathanWard Jul 26 at 23:05
    
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

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