I’m not sure if this is a genuine paradox but it does seem very unintuitive and I’d appreciate if someone can explain it so that it seems reasonable and/or explain the flaw in the reasoning.

Consider the following countable collection of open intervals: I1 is ( 0, 1/2 ), I2 is ( 1/2, 1/4 ), I3 is ( 1/4, 1/8 ), etc. The sum of the lengths of the intervals is 1.

Now cover each rational number with one of the intervals.

There can be at most one uncovered irrational between any two successive open intervals.

There are a countable number of open intervals so there can be at most a countable number of irrationals left uncovered.

So we’ve covered the entire real line, except for a countable number of points, with a countable number of open intervals the sum of whose lengths adds up to 1.

Or have we?

P.S. I asked a similar question a year ago but didn't write it up clearly. I hope this is better.

  • $\begingroup$ except for the first your intervals are backwards.. $\endgroup$ – Thoth Nov 12 '15 at 2:49
  • $\begingroup$ You're not covering the real line, only (all but countably many rationals in) the unit interval $[0,1]$ $\endgroup$ – BrianO Nov 12 '15 at 2:56
  • 1
    $\begingroup$ The first interval contains all the rest, and the sum of the lengths is 1/2. Do you mean the second to be (1/2, 3/4), the third to be (3/4, 7/8), and so on? Assuming you do, then yes, the union of all the intervals omits the countable collection of rationals $(2^n-1)/2^n$, and yet its length, or measure, is 1. Measure theory gives a precise definition to this generalized notion of "length" (namely, measure), and it turns out that two sets that have a measure and differ by a countable set in fact have the same measure. So the measure of the union of your collection is 1. $\endgroup$ – BrianO Nov 12 '15 at 2:57

What I think you're asking is this:

Let $\{ r_n : n \in \mathbb N \}$ be an enumeration of the rationals. Then the intervals $I_n = (r_n - 2^{-n}, r_n + 2^{-n})$ are a covering of the rationals and the measure of their union is bounded above

$$\mu\left( {\Large\cup}_{n=1}^\infty I_n \right) \leq \sum_{n=1}^\infty \mu(I_n) = \sum_{n=1}^\infty \frac{1}{2^{n-1}} = 2$$

If $\cup_n I_n = \mathbb R$ then isn't this is a contradiction?

Instead, I would turn the logic around and conclude that $\cup_n I_n$ does not cover the reals.

And yet, isn't every irrational within an arbitrary $\epsilon > 0$ of a rational? Yes. However we cannot use that to deduce that $\cup_n I_n = \mathbb R$. (Try to write that argument down and see what goes wrong.)

  • $\begingroup$ This is pretty much for sure what he was hoping to say. If you don't already know this fact, however, you would never guess it from the way it's expressed above. The apparent paradox then, to write it up correctly is this: on the one hand you have an open set that is huge, it contains all rationals. it is everywhere dense. But not so huge: it has total length only $2$, yet covers the whole real line, at least as close as one looks it does. The usual version of this paradox produces a dense open set with arbitrarily small total length, but if $2$ is enough to dazzle you, go for it. $\endgroup$ – B. S. Thomson Nov 12 '15 at 3:33
  • $\begingroup$ Right. Instead make the intervals $J_n = (r_n - 2^{-(n+1)}\delta, r_n + 2^{-(n+1)}\delta )$ for any $\delta > 0$, then the measure of their union is bounded above by $\delta$. $\endgroup$ – Simon S Nov 12 '15 at 3:39
  • $\begingroup$ Comments: "except for the first your intervals are backwards" Yes, they are. And yes, my intervals add up to 1 but the same could be done with intervals summing to an any length. But is it true they cover all of the real except for a countable number of irrationals? And can anyone tell me what "the usual version of the paradox" is called or provide a reference for further investigation. (Simon, I hit return too soon. This is the corrected comment.) $\endgroup$ – ArtD Nov 12 '15 at 11:55
  • $\begingroup$ @ArtD I don't know what you mean $\endgroup$ – Simon S Nov 12 '15 at 11:58
  • $\begingroup$ "is it true they cover all of the real except for a countable number of irrationals?" No, because the measure of any countable set is zero. Thus the $I_n$ or the $J_n$ don't cover uncountably many irrationals. The usual version is the version with the $J_n$ and arbitrarily measure bound $\delta$. You can find this in many first books on real analysis and measure theory. If I find a good reference I'll post it. But I encourage you to search yourself. $\endgroup$ – Simon S Nov 12 '15 at 12:06

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