Does there exists intervals \( I_1, I_2, \ldots \), each of which are subintervals of \( [ 0,1] \), such that

1. the total length of these subintervals is strictly less than 1.

2. For each rational number \( r \in [0,1] \), there exists a \(j\) such that \( r \in I_j \).

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TopNewestWe can write the set of rational numbers in \([0,1]\) as \(\{m/n : 0\leq m \leq n, n \geq 1\}\). Now exclude the point \(1\) from all these sets and enclose each rational point with denominator \(n\) by a sub interval of length \(3^{-n}\), lying completely in \([0,1]\) (e.g. enclose the point \(m/n\) by the sub-interval \([m/n, m/n + 3^{-n}]\)). The total length of all these sub-intervals is \(\sum_{n=1}^{\infty} n3^{-n}=\frac{3}{4}\). Finally enclose the point \(1\) by the sub-interval \([9/10,1]\). Hence all rational points are enclosed and total length of the sub-intervals = \(3/4+1/10 <1\). – Abhishek Sinha · 2 years, 2 months ago

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– Jonathan Wong · 2 years, 2 months ago

Quick note that the total length is less than \(17/20\) because of overlapping intervals.Log in to reply

– Ethan Robinett · 2 years, 2 months ago

This was really good.Log in to reply

Indeed, the rationals are measure-zero and countable (despite being dense in \(\mathbb{R}\)); they are a null set. The total length of the required subintervals can be any \(\epsilon > 0\). – Jonathan Wong · 2 years, 2 months ago

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– Abhishek Sinha · 2 years, 2 months ago

That does not prove existence of sub intervals of the required properties.Log in to reply

If anyone is able to prove my original statement (especially constructively), by all means, post it here. The "overlaps" of the intervals are bugging me quite a bit. – Jonathan Wong · 2 years, 2 months ago

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0.999999999999999.................................=1 – Amritanshu Kumar · 2 years, 2 months ago

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– Sharky Kesa · 2 years, 2 months ago

Hey, it isn't good to self-advertise your problems on completely unrelated notes.Log in to reply