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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 $2^\mathfrak{c}$.

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  • $\begingroup$ Is it obvious that assuming the continuum hypothesis there are only possible answers? $\endgroup$ – Aryeh Kontorovich Nov 24 '18 at 18:20
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    $\begingroup$ @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 $2^{2^{\mathfrak c}}=(2^{\mathfrak c})^+$. $\endgroup$ – Andrés E. Caicedo Nov 24 '18 at 20:03
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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 $\mathbb{R}$.

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    $\begingroup$ But does this answer the question about sigma-algebras? $\endgroup$ – Nik Weaver Nov 25 '18 at 5:34
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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.)

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

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

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