Suppose we take all the rationals and take any neighbourhood around each of them. Will they cover whole $\Bbb R$. I think so as rationals are dense. So, for each irrational we can find a rational which is arbitrary close to it. Is it correct ?
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$\begingroup$ Yes, this is correct. $\endgroup$– fgpApr 15, 2014 at 10:22
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$\begingroup$ the density implies that if you consider all the rationals, no matter what the scale at which you are looking $\mathbb{R}$ at, there will be no "holes" $\endgroup$– T_OApr 15, 2014 at 10:23
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2$\begingroup$ They won't necessarily cover. Consider $(-\infty,\pi)$, and $(\pi,\infty)$. $\endgroup$– David MitraApr 15, 2014 at 10:23
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$\begingroup$ @TomCollinge The OP wanted nhoods about the rationals. $\endgroup$– David MitraApr 15, 2014 at 10:38
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$\begingroup$ @David Mitra. Thanks, I missread that part. $\endgroup$– Tom CollingeApr 15, 2014 at 10:41
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2 Answers
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No to the first statement, yes to the second.
As you probably know, there are only countably many rationals, so we can enumerate them in some order $q_1$, $q_2$, etc. Now let us look at the set:
$$U := \bigcup_{i \in \mathbb{N}} B(q_i, 2^{-i-1})$$
where $B(q,r)$ is the open ball with center $q$ and radius $r$. This set has measure $\leq 0.5$ (which you can see by summing up all the radii), is open and contains all rational numbers.
So despite the fact that the rationals are dense in $\mathbb{R}$, there are very small open neighborhoods of all the rationals.
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If you talk about arbitrarily close constant neighbourhood then yes.
However, it might be of interest, that if you allow certain restrictions this is not always true. Look at the Roth Theorem (Fields Medal) which states that there are finitely many rationals $\frac pq$ such that : $$|\alpha - \frac pq| < \frac{c}{q^{2+\epsilon}}$$ where $\alpha$ is an algebraic irrational, $\epsilon > 0$ and $c$ is any constant. Taking $c$ to be small enough, the "finitely many" becomes "no rational".
