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I was wondering why Borel algebra $B(X)$ is defined to be the smallest sigma algebra containing all open subsets of X. If it contains all the subsets, then how can any other sigma algebra have more subsets? I'm asking why the word 'smallest' is used in definition.

user4205580
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1 Answers1

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It doesn't have to include all the subsets of $X$. We can prove, in the case of $\Bbb R$ for example, that the Borel sets is a collection whose cardinality is the same as $\Bbb R$. Since Cantor's theorem tells us that there are many more subsets to $\Bbb R$ it follows that most subsets are not Borel sets.

To see a much more degenerate example, consider any space with the trivial topology. It's Borel subsets are the open sets, $\varnothing$ and $X$ itself.

Asaf Karagila
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  • Quoting your post. "Borel sets are defined to be the smallest σ-algebra which contains all the subintervals". You didn't use the word 'open' here. Could you tell me why? – user4205580 Jan 04 '15 at 09:23
  • You can construct closed and half closed from the open intervals using complements, unions and intersections. And vice versa too. – Asaf Karagila Jan 04 '15 at 09:25
  • So basically I could add a non-Borel set to borel sigma algebra and then it won't be the smallest σ-algebra containing all open subsets? – user4205580 Jan 04 '15 at 09:51
  • It won't be a $\sigma$-algebra, but if take the $\sigma$-algebra this collection generates, then it will be strictly larger than the Borel sets, yes. – Asaf Karagila Jan 04 '15 at 09:53
  • If Cantor's theorem was wrong, then I wouldn't be able to do it and the word smallest would be useless again? – user4205580 Jan 04 '15 at 09:56
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    (1) Cantor's theorem is not wrong. (2) No, there are many $\sigma$-algebras which are the same size as the Borel sets, but are in fact larger (with respect to inclusion). (3) Cantor's theorem is just a quick argument why this thing holds, if you really want you can use the axiom of choice to engineer a non-Borel set. (4) It is consistent, however, that the axiom of choice fails, and $\Bbb R$ is a countable union of countable sets, in which case every set is Borel. (Note that it does not mean that $\Bbb R$ is countable!) – Asaf Karagila Jan 04 '15 at 10:22
  • What part of Cantor's theorem says anything about how many subsets of $\mathbb{R}$ there are? – user4205580 Jan 04 '15 at 12:18
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    Cantor's theorem says that for any set $X$, $|X|<|\mathcal P(X)|$. In particular $\Bbb R$. – Asaf Karagila Jan 04 '15 at 12:24
  • "We can prove, in the case of R for example, that the Borel sets is a collection whose cardinality is the same as R". And how to prove this is true? – user4205580 Jan 04 '15 at 12:28
  • Or is the proof easy at least? – user4205580 Jan 04 '15 at 14:25
  • It's not trivial if you don't know about transfinite induction and cardinal arithmetic. It's quite easy if you do know about it. – Asaf Karagila Jan 04 '15 at 15:19
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    @user4205580 You can find some proofs (which use transfinite induction, as Asaf said) on this site. For example: http://math.stackexchange.com/questions/694908/cardinality-of-borel-sets and http://math.stackexchange.com/questions/70880/cardinality-of-borel-sigma-algebra – Martin Sleziak Jan 04 '15 at 18:56