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I can think of one way to do this: Singletons are Borel, and since a countable subset of $\mathbb{R}$ is a countable union of singletons, the countable subset is Borel.

Is there a way to prove this without using the fact that singletons are Borel?

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    Why would you want to avoid using that singletons are Borel? – davidlowryduda Sep 23 '14 at 20:13
  • @mixedmath Frankly, to expand my understanding of Borel sets –  Sep 23 '14 at 20:17
  • @longcat: to understand Borel sets, go for more and more examples. For example (open/close/halfopen) intervals are Borel, and this already includes points ${x}$ as closed intervals $[x,x]$, and complement and countable union/intersection of these are also Borel. – Berci Sep 23 '14 at 21:15

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The fact that singletons are Borel gives an immediate proof of this fact. This suggests that singletons being Borel is very tightly related to the fact, and so any proof avoiding using that singletons are Borel might feel like a "guise."

But let us assume that we have forgotten that singletons are Borel. Then using De Morgan's Law, since a countable subset of $\mathbb{R}$ is a countable union of singletons, the complement of the countable subset is a countable intersection of open intervals. As open intervals are the prototypical Borel set, and Borel sets are preserved under countable union and relative complements, we see that a countable union of singletons is Borel. $\diamondsuit$

But you might notice that what we really did was replicate a proof that singletons are Borel. So it goes.

davidlowryduda
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