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Cover every rational number on [0,1] with an interval of radius $\epsilon/2^n$, the total length of these intervals is $\epsilon$, so the measure of rational numbers on [0,1] is 0. My question is that, can you prove there exist a irrational number on [0,1] that is not covered by these intervals without the use of prove by contradiction?

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Why do anyone vote down this without leaving a comment? – Ziqian Xie Nov 17 '12 at 16:47
Although I did not vote on this question (yet), let me note that you say nothing about the approaches you tried, or the similar questions you can solve, or the context in which this question arose. All of which are recommended pieces of information. – Did Nov 17 '12 at 17:07
@did The approach is a standard trick in measure theory, I cannot solve it, I ask it out of curiosity. – Ziqian Xie Nov 17 '12 at 17:14
Do as you wish but do not be surprised next time. – Did Nov 17 '12 at 17:17
up vote 5 down vote accepted

To do what you ask more concretely, one needs to be more specific. Let's consider the enumeration of the rationals according to the following sequence: $$\tag0 0, 1, \frac12, \frac13, \frac23,\frac14,\frac34,\frac15,\frac25,\frac35,\frac45,\frac16,\frac56,\ldots, \frac{p_n}{q_n},\ldots $$ i.e. all reduced frections ordered first by denominator, then by numerator. I claim that $\phi=\frac{\sqrt 5-1}2$, which is a root of $f(X)=X^2+X-1$, is not covered. Assume on the contrary that the $n$th element $\frac {p_n}{q_n}$ of the above sequence covers $\phi$, i.e. we have $$\tag1\left|\phi-\frac {p_n}{q_n}\right|<2^{-n}.$$ Then $q_n^2f\left(\frac{p_n}{q_n}\right)=p_n^2+p_nq_n-q_n^2$ is a nonzero integer, hence $$\tag2\left|f\left(\frac{p_n}{q_n}\right)\right|\ge \frac1{q_n^2}.$$ On the other hand $f\left(\frac{p_n}{q_n}\right)=f\left(\frac{p_n}{q_n}\right)-f(\phi)=\left(\frac{p_n}{q_n}-\phi\right)f'(\xi)=\left(\frac{p_n}{q_n}-\phi\right)(2\xi+1)$ for some $\xi$ between $\phi$ and $\frac{p_n}{q_n}$. From $(1)$, $(2)$ and $\xi\in[0,1]$ we conclude $$\tag3 \frac1{q_n^2}<\frac{3}{2^n}.$$ For each denominator $q>2$, there appear at least two fractions in $(0)$, namely $\frac1q$ and $\frac{q-1}q$. Taking the initial exceptions for $q\le2$ into account, we see that $q_n\le \frac n2+1$. Together with $(3)$ this implies $$ 2^n<3\left(\frac n2+1\right)^2,$$ which holds only for $n\le 5$. Checking the first five fractions explicitly, we find that $(1)$ does not hold for them either.

Remark: Formally, this is a proof by contradiction (though it can be converted to a direct one), but presumably not the kind you really wanted to exclude (i.e. simply using that measure $<1$ implies proper subset). Also note that I actually used $\epsilon=1$. Of course, smaller values of $\epsilon$ will also fail to cover $\phi$.

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Wonderful, how do you come up with this idea! I tried your method with lots of irrational numbers, none of them can be covered. – Ziqian Xie Nov 17 '12 at 17:57

My friend told me another method: get an enumeration of all rational numbers, like the one given by Hagen, denote it by {$q_1,q_2,q_3...$}, there is an interval of radius $\epsilon/2^i$ centered at each $q_i$. Choose an irrational number $s$, it can be either algebraic or transcendental, now look at $q_1$, if the distance $d_1$ between $q_1$ and $s$ is less than $\epsilon$/2, then make $q_1$ the $n$th term of this sequence such that $n=inf\{n\in\mathbb{N^+}:\epsilon/2^n<d_1\}$, if $d_1>\epsilon/2$, leave $q_1$ alone and look at $q_2$. Proceed like this, it is easy to show that this sequence will finally converge, because if you move the $n$th term of this sequence, you will not change the order of sequence before n (which is apparent that you can only move a term forward). The intervals which cover the final sequence of all rational numbers will not cover the irrational number $s$.

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this method can only show that there exist ONE irrational number which cannot be covered by all those intervals, but in fact there are infinitely many. – Ziqian Xie Nov 18 '12 at 5:22

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