# How many sigma algebras exist on $\mathbb{R}$?

On the one hand, there are at least $$2^\mathfrak{c}$$ sigma algebras on $$\mathbb{R}$$: one can take any subset $$A$$ of $$\mathbb{R}$$ and consider a sigma algebra $$\{\emptyset, A, \bar A, \mathbb{R}\}$$

On the other hand, number of sigma algebras can be bounded as $$2^{2^\mathfrak{c}}$$ since a sigma algebra is a family of subsets of $$\mathbb{R}$$.

Let's assume generalized continuum hypothesis to simplify. In this case there are only two possible answers. Unfortunately I can't either find $$2^{2^\mathfrak{c}}$$ different sigma algebras, or prove their number is exactly $$2^\mathfrak{c}$$.

• Is it obvious that assuming the continuum hypothesis there are only possible answers? – Aryeh Kontorovich Nov 24 '18 at 18:20
• @Aryeh Yes. It is immediate from the fact that there are at least $2^{\mathfrak c}$ $\sigma$-algebras, as indicated in the post, and at most $2^{2^{\mathfrak c}}$ such algebras, as the post also explains. What GCH gives you is that $2^{2^{\mathfrak c}}=(2^{\mathfrak c})^+$. – Andrés E. Caicedo Nov 24 '18 at 20:03

Since $$|\mathbb{R}|^{\aleph_0}= |\mathbb{R}|$$, there is a sigma-independent (all countable boolean cominations are non-empty) family $$F \subseteq \mathcal{P}(\mathbb{R})$$ of size $$2^{|\mathbb{R}|}$$. Now for each subfamily $$G$$ of $$F$$ consider the sigma-ideal that contains each member of $$G$$ and the complement of each member of $$F\setminus G$$. So there are $$2^{2^{2^{\aleph_0}}}$$ different sigma-ideals on $$\mathbb{R}$$.

• But does this answer the question about sigma-algebras? – Nik Weaver Nov 25 '18 at 5:34

The other answer gives many $$\sigma$$-ideals, but I don't think it settles the question about $$\sigma$$-algebras.

Assume GCH. Find a family $$\mathcal{F}$$ of $$\aleph_2$$ many subsets of $$\mathbb{R}$$, any two of which have countable intersection. There are $$\aleph_3 = 2^{2^c}$$ distinct subsets of $$\mathcal{F}$$ which contain more than one element.

Say that $$A \subseteq \mathbb{R}$$ essentially contains $$B$$ if $$B \setminus A$$ is countable. Then for each $$\mathcal{F}_0 \subseteq \mathcal{F}$$ with more than one element the family of subsets of $$\mathbb{R}$$ which either essentially contain every set in $$\mathcal{F}_0$$ or no set in $$\mathcal{F}_0$$, is a $$\sigma$$-algebra which contains no set in $$\mathcal{F}_0$$ and every set in $$\mathcal{F}\setminus\mathcal{F}_0$$. So there are as many distinct $$\sigma$$-algebras as there are subsets of $$\mathcal{F}$$ with more than one element.

(To construct $$\mathcal{F}$$, identify $$\mathbb{R}$$ with $$\{f: \alpha \to \{0,1\}\, |\, \alpha$$ is a countable ordinal $$\}$$. For each function from $$\aleph_1$$ to $$\{0,1\}$$, its restrictions to all countable ordinals is a subset of this set, and any two of these have countable intersection.)

• I had posted some comments earlier, but they actually just rehashed Thanos's original proof. – Douglas Ulrich Nov 25 '18 at 16:04
• Take a sigma ideal with its coideal, and you've got yourself a merry little sigma algebra! – Asaf Karagila Nov 26 '18 at 6:42

This is a continuation of my previous answer.

Suppose $$F$$ is a family of $$2^{2^{|\mathbb{R}|}}$$ sigma-ideals on $$\mathbb{R}$$. By throwing away the principal ideals, we can assume that they are all non-principal. Let $$I, J \in F$$ be distinct and towards a contradiction suppose they generate the same sigma-algebra. Note that the sigma-algebra generated by a sigma-ideal is just the ideal together with the dual filter. Say $$X \in I \setminus J$$. Then $$\mathbb{R} \setminus X \in J$$ and $$J$$ restricted to $$X$$ is a non-principal prime sigma-ideal on $$X$$. But this is impossible as there is no measurable cardinal below $$|Y|$$. Am I missing something obvious?

• It is not true that the $\sigma$-algebra generated by a $\sigma$-ideal is just the ideal together with the dual filter – Douglas Ulrich Nov 25 '18 at 16:10
• @DouglasUlrich- The filter plus the ideal is closed under complements and countable unions, no? – Monroe Eskew Nov 25 '18 at 16:20
• I see, I was just confused – Douglas Ulrich Nov 25 '18 at 16:28
• Why does $X \in I\setminus J$ imply $\mathbb{R}\setminus X \in J$? – Nik Weaver Nov 25 '18 at 17:24
• @Nik Weaver: Since I and J generate the same $\sigma$-algebra, i.e. $I \cup I^* = J \cup J^*$ – Douglas Ulrich Nov 25 '18 at 17:43