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The borel algebra on the topological space R is defined as the σ-algebra generated by the open sets (or, equivalently, by the closed sets). Logically, I thought that since this includes all the open sets (a,b) where a and b are real numbers, then, this would be equivalent to the power set. For example, the set (0.001, 0.0231) would be included as well as (-12, 19029) correct? I can't think of any set that would not be included. However, I have read that the Borel σ-algebra is not, in general, the whole power set.

Can anyone give a gentle explanation as to why this is the case?

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There are lots of subsets of $\mathbb{R}$. What makes you think that every subset of $\mathbb{R}$ can be written as a countable union of countable intersections of open and closed subsets? –  Zhen Lin Oct 20 '12 at 9:46
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For a set which isn't even lebesgue measureable, check the vitali set. Its construction requires the axiom of choice. –  k.stm Oct 20 '12 at 9:50
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And to see why it's difficult to give a completely explicit example of a set which isn't in the Borel $\sigma$-algebra, see this MO answer. –  Zhen Lin Oct 20 '12 at 9:52
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Why should having all of the open sets give you all subsets of $\Bbb R$? Most subsets of $\Bbb R$ are neither open nor closed. Indeed, $\Bbb R$ has only $2^\omega$ open and closed subsets, but it has $2^{2^\omega}>2^\omega$ subsets. –  Brian M. Scott Oct 20 '12 at 9:54
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The set of Borel sets has the same cardinality as $\mathbb R$, but the cardinality of the powerset of $\mathbb R$ is strictly larger. –  Andrew Oct 20 '12 at 10:14
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1 Answer

up vote 8 down vote accepted

You can show that there are $\mathfrak{c} = 2^{\aleph_0}$ Borel subsets of the real line, and so by Cantor's Theorem ($|X| < | \mathcal{P} (X)|$) it follows that there are non-Borel subsets of $\mathbb{R}$.

To see that there are $\mathfrak{c}$-many Borel subsets of $\mathbb{R}$, we can proceed as follows:

  1. define $\Sigma_1^0$ to be the family of all open subsets of $\mathbb{R}$;
  2. for $0 < \alpha < \omega_1$ define $\Pi_\alpha^0$ to be the family of all complements of sets in $\Sigma_\alpha^0$ (so that $\Pi_1^0$ consists of all closed subsets of $\mathbb{R}$);
  3. for $1 < \alpha < \omega_1$ define $\Sigma_\alpha^0$ to be the family of all countable unions of sets in $\bigcup_{\xi < \alpha} \Pi_\xi^0$.

Then you can show that $B = \bigcup_{\alpha < \omega_1} \Sigma_\alpha^0 = \bigcup_{\alpha < \omega_1} \Pi_\alpha^0$ is the family of all Borel subsets of $\mathbb{R}$. Furthermore, transfinite induction will show that $| \Sigma_\alpha^0 | = \mathfrak{c}$ for all $\alpha < \omega_1$, which implies that $\mathfrak{c} \leq | B | \leq \aleph_1 \cdot \mathfrak{c} = \mathfrak{c}$.

Specific examples of non-Borel sets are in general difficult to describe. Perhaps the easiest to describe is a Vitali set, obtained by taking a representative from each equivalence class of the relation $x \sim y \Leftrightarrow x -y \in \mathbb{Q}$. Such a set is not Lebesgue measurable, and hence not Borel. Another example, due to Lusin, is given in Wikipedia.

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Have you been to the Wein & Co. recently? –  Asaf Karagila Oct 22 '12 at 19:38
    
@Asaf: No Wein & Co. yet. But I guess I'll have to make a pickle run soon. –  Arthur Fischer Oct 23 '12 at 20:01
    
@Asaf: I arrive in J-town on 20 Cheshvan 5773. –  Arthur Fischer Oct 26 '12 at 11:24
    
Excellent. If you're up to some bus rides, the day after that I am giving a lecture in the BGU seminar. I'll talk about ordering of cardinals and how disorderly they could be. The plan is to continue a week after that with a result from my thesis (about antichains of cardinals). Either way I will see you in the Holy University of the oracle. –  Asaf Karagila Oct 26 '12 at 11:42
    
I hear you have a boarding pass to give me, along with the pickles that I trust you have bought. Also congrats on the Fanatic badge. Took you long enough... –  Asaf Karagila Oct 31 '12 at 22:46
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