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.
Interval in $R$ does not have zero measure.
analysismeasure-theoryreal-analysis
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Best Answer
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.