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I have followed the argument that rationals, being countable and ordered, can be covered by a convergent sequence of decreasing intervals.

I am trying to understand why the same argument can’t be applied to the reals, on the basis that axiom of choice leads to the well ordering theorem which says that the reals can be well ordered. In that case why can’t I cover the first real with an interval length ½ etc in the same way as the rationals ?

I suspect that there aren’t enough natural numbers to complete the process since the reals are uncountable, but can the process not extend into the ordinal ω1. If it does then surely the series ½ + ¼ ... should still sum to 1 ? I’d appreciate any feedback or web reference you could offer to clarify this for me.

Tom Collinge
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    A sum over uncountably many positive numbers, no matter how small, will always be infinite. – Clayton Oct 02 '13 at 12:19
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    Yes, "there aren't enough natural numbers to complete the process". There's your answer. – Prahlad Vaidyanathan Oct 02 '13 at 12:20
  • This takes an argument, and it is somewhat surprising at first that the proof is a bit delicate. The point is that if a union of open intervals covers an interval of length $\epsilon$, then the sum of the lengths of the individual intervals used in the covering is at least $\epsilon$ as well, that is, we cannot reduce length even if we allow overlaps. This follows from the answers to this question. – Andrés E. Caicedo Oct 02 '13 at 14:04
  • That there is something subtle going on is even mentioned explicitly by Tim Gowers here. – Andrés E. Caicedo Oct 02 '13 at 14:07

2 Answers2

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Every well order of the real numbers must have at least $\omega_1$ terms. This means that the same argument would have to amount to a series with an uncountable number of terms.

However a series with uncountably many terms can only be convergent if all but countably many are zero.

Asaf Karagila
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  • Thanks, I need to read up a bit on uncountable series: I've never looked at the idea previously. – Tom Collinge Oct 02 '13 at 12:47
  • Well. It's a really simple idea. If the series is convergent, then there can't be infinitely many terms over $\frac1n$, for every $n\in\Bbb N$. So there are only a countable-union-of-finite-sets of non-zero terms. – Asaf Karagila Oct 02 '13 at 12:52
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The same argument can't be applied to the reals since a sum over uncountably many positive numbers will always be infinite. Thus, regardless of the size of these "open intervals" under the well ordering, there are necessarily uncountably many with a positive radius. Therefore, the sum of their measures will be infinite.

Edit: At @user43208's suggestion, the details to prove the above statement goes as follows: if $S$ is a set of positive reals and the supremum over all sums of finite subsets is finite, then each set $S_n=\{x\in S:x>1/n\}$ is finite, hence $S$ (the union of the $S_n$) is countable.

Clayton
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    It might be worthwhile to the OP to give the details of this easy argument: if $S$ is a set of positive reals and the supremum over all sums of finite subsets is finite, then each set $S_n = {x \in S: x > 1/n}$ is finite, hence $S$ (the union of the $S_n$) is countable. – user43208 Oct 02 '13 at 12:25
  • Thanks, I need to read up a bit on uncountable series: I've never looked at the idea previously – Tom Collinge Oct 02 '13 at 12:48
  • @user43208: I've provided the details you proposed. Thanks. – Clayton Oct 02 '13 at 13:21