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Consider $Q$ a interval in $R$. Number all the rationals in this interval. Then for rational $q_i$ consider interval $[q_j-\epsilon /2^{j+2} ,q_j+\epsilon/2^{j+2}]$. Now the whole interval lies inside the union of these intervals, sum of whose volumes is $ < \epsilon $.
Why I am getting this contradiction? I know (with proof) that an interval does not have measure 0. What is wrong with the above argument?
I only know the concept of zero measure from analysis.

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    $\begingroup$ You've given a proof by contradiction that the union of the balls contains the interval. Also maybe consider the set [1,2] and then consider the union of open balls around rationals $q\in[1,2]$ with radius $|q-\sqrt 2|$. This set won't have $\sqrt 2$ $\endgroup$ – Calvin Khor Feb 2 at 9:40
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The gap in the proof is that you do not know that the interval $[a,b]$ lies in the union of these balls. In fact, if $\epsilon < b-a,$ you have shown that there are points in $[a,b]$ which are not in any of the balls.

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You've given a proof by contradiction that the union of the balls contains the interval. As a simpler example of this sort of phenomenon, consider the set $Q=[1,2]$ and the union $U$ of open balls around rationals $q∈[1,2]$with radius $|q-\sqrt 2|$. $U$ doesn't have the point $\sqrt 2 \in Q$.

If the above does not convince you, one can directly prove that this set has other points: First consider the variant with open balls around each $q_i$, $$B_i := B_i(\epsilon) := (q_i - \epsilon2^{-i-2}, q_i +\epsilon2^{-j-2})$$ As each $B_i$ is open, the complement $C_i=Q\setminus B_i$ is closed. For $\epsilon \ll 1$, they are not empty. Moreover, $$ D_i := Q \setminus \left (\bigcup_{j=1}^i B_j\right) = \bigcap_{j=1}^i C_j$$ is a sequence of nested, non-empty, closed, and bounded sets. By Cantor's Intersection Theorem, the set $\bigcap_{j=1}^\infty C_i$ is not empty, and thus $\bigcup_{j=1}^\infty B_i$ does not contain all of $Q$.

If you insist on using closed balls, you can use some $\epsilon'<\epsilon$; then notice that $\overline{B_i(\epsilon')} \subset B_i(\epsilon)$, so by the above, their union does not contain all of $Q$.

Further, the above works for every subset of $Q$ of measure more than $\epsilon$.

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