Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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
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
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
The set of Borel sets has the same cardinality as $\mathbb R$, but the cardinality of the powerset of $\mathbb R$ is strictly larger. – Yang Zhou Oct 20 '12 at 10:14
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.

share|cite|improve this answer
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. – inactive... for now Oct 23 '12 at 20:01
@Asaf: I arrive in J-town on 20 Cheshvan 5773. – inactive... for now 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.