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# Can $$< 1 = 1$$?

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 $$r \in I_j$$.

Note by Calvin Lin
3 years, 1 month ago

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We 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 = $$3/4+1/10 <1$$. · 3 years, 1 month ago

Quick note that the total length is less than $$17/20$$ because of overlapping intervals. · 3 years, 1 month ago

This was really good. · 3 years, 1 month ago

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 $$\epsilon > 0$$. · 3 years, 1 month ago

That does not prove existence of sub intervals of the required properties. · 3 years, 1 month ago

The proof of my initial statement eludes me. A proof that the total length is less than $$\epsilon$$ for any $$\epsilon>0$$ is much easier (order the rationals and construct intervals around them with lengths of the terms of a converging infinite geometric series of your choice).

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. · 3 years, 1 month ago

0.999999999999999.................................=1 · 3 years, 1 month ago

Comment deleted Aug 15, 2014