Can you cover all of the rational numbers in [0, 1] with tiny intervals?
This problem was inspired by an awesome video by 3blue1brown.
For a given \(\epsilon > 0\), is there a way for you to cover all the rational numbers in the interval \([0, 1]\) with small intervals \(I_k\), such that the sum of the lengths of the intervals \(I_k\) is less than or equal to \(\epsilon\)?
In other words (with almost no words), for what values of \(\epsilon > 0\) is there a collection \(\{I_k\}\) of intervals such that
\($\left(\mathbb{Q}\cap [0,1]\right) \subseteq \left(\cup_k I_k \right) \wedge \sum_k |I_k| < \epsilon\)$
Such a family of intervals always exists, for any value of \(\epsilon > 0\). We start by noticing that the rational numbers in the interval \([0, 1]\) are countably many, which means I can order them as \(q_1, q_2, q_3, \cdots\). If you haven't solved the problem yet, take the hint I just gave you and try to solve it.
After enumerating the rationals inside \([0, 1]\), we define \(I_k\) to be \([q_k - \epsilon2^{-k-1}, q_k + \epsilon2^{-k-1}]\). By defining \(I_k\) this way, we get that
\($|I_k| = (q_k + \epsilon2^{-k-1}) - (q_k - \epsilon2^{-k-1}) = \frac{\epsilon}{2^k}\)$
We now are left with showing that with the intervals defined this way, we have the two desired properties. In fact, we have that \(\left(\mathbb{Q}\cap [0,1]\right) \subseteq \left(\cup_k I_k \right)\).
If \(q\) is some rational number in \([0, 1]\), then because they are countably many and I listed all of them, \(q\) is equal to some \(q_k\), but then \(q = q_k \in I_k \subset \cup_k I_k\).
As to whether the sum of the lengths is smaller than or equal to \(\epsilon\), we just have to compute the series
\($\sum_{k=1}^\infty |I_k| = \sum_{k=1}^\infty \frac{\epsilon}{2^k} = \epsilon \sum_{k=1}^\infty \frac1{2^k} = \epsilon\)$
And that was it! Pretty simple proof, but then I feel like the result is absolutely amazing if we consider \(\epsilon < 1\).
Don't forget that the rationals are dense in \([0, 1]\), which in a not-so-rigorous way, means that there are rationals everywhere in the interval \([0, 1]\)... But we can cover them up with smaller intervals that clearly don't cover the whole interval, because the whole interval has size \(1\) and the sizes of our intervals only add up to \(\epsilon\)... And we didn't even take into account that some of the smaller \(I_k\) intervals overlap with each other!
In case you are wondering, the video from which I got this problem is this one.
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